Calculus II Workbook

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How to use this workbook

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Quizzes are not announced in advance, so keep the current week's topic quiz-ready. This box shows the week and the predicted quiz for today.

In plain words

The idea

Every quiz in this course has the same shape: two short problems, 1.5 points each, 6–20 minutes, no calculator. Each problem is one of a small number of problem types (find a limit, test a series, find an interval of convergence, compute a cross product …). Each chapter below teaches those types one at a time, with a recipe, the order of steps that earns full marks.

Read the chapter: the plain-words summary first, then definitions, theorems and the worked examples. Watch-out boxes list the mistakes that cost points on the real quizzes.
Learn the recipe for each problem type. The steps follow the way the solution sheets award points (0.5 + 0.5 + 0.5).
Solve on paper, then type only the final answer in the box. The box says right or wrong without giving the answer away, and often says what kind of mistake you made (sign error, wrong endpoint, forgot to divide by \(k!\) …).
Stuck? Open Hint, then Answer, then Full solution, in that order. Every solution is written step by step.
Tick “done”. Progress is saved in this browser, and the contents rail counts it for every practice set.
Before a quiz: do the past quizzes on the topic with the timer on, then the predicted quizzes, then endless practice (new random problems every click).

The semester, week by week

From the syllabus (Thomas section numbers). Week numbers assume classes began on Monday 14 September 2026. The highlighted row is the current week.

Wk 1Sequences10.1
Wk 2Infinite series10.2
Wk 3Integral, comparison, ratio, root, alternating tests10.3–10.6
Wk 4Power series, Taylor & Maclaurin series10.7–10.8
Wk 5Parametric curves and their calculus11.1–11.2
Wk 63D coordinates, vectors, dot & cross product, lines & planes12.1–12.5
Wk 7Curves in space, integrals of vector functions13.1–13.2
Wk 8Functions of several variables, limits & continuity14.1–14.2
Wk 9Partial derivatives, chain rule14.3–14.4
Wk 10Gradient, directional derivatives, tangent planes14.5–14.6
Wk 11Extreme values and saddle points14.7
Wk 12Double integrals over rectangles and general regions15.1–15.2
Wk 13Area by double integration15.3
Wk 14Line integrals, vector fields, work, circulation, flux16.1–16.2
Wk 15Path independence, conservative fields, potentials16.3
FinalFinal examination (practice set)all

The syllabus labels weeks 10–11 as “14.4–14.5” and “14.6–14.7”. In the 13th edition these topics are §14.5 (gradient), §14.6 (tangent planes) and §14.7 (extreme values), which is the numbering used on this page.

Overview

The big picture

The course has two halves. The first asks when adding infinitely many numbers gives a finite answer, and uses that to write functions as infinite polynomials. The second takes Calculus I one dimension up: curves and vectors in space, functions of several variables, and integrals over regions and along curves.

I · Series (Ch 10)

Sequences \(a_n\) and limits
Series = limit of partial sums \(S_N\)
Tests: does \(\sum a_n\) converge?
Power series \(\sum c_n(x-a)^n\): for which \(x\)?
Taylor series: functions as power series

II · Curves (Ch 11)

Curves as \(x=f(t),\ y=g(t)\)
Slope \(dy/dx = \dfrac{dy/dt}{dx/dt}\)
Arc length, area under a curve

III · Space (Ch 12–13)

Points, distance, spheres in \(\mathbb R^3\)
Vectors, dot product (angles, projections)
Cross product (areas, normals, volumes)
Lines and planes
Vector functions \(\mathbf r(t)\): velocity, acceleration

IV · Several variables (Ch 14)

Graphs and level curves of \(f(x,y)\)
Limits along paths
Partial derivatives, chain rule
Gradient, directional derivative, tangent plane
Max, min, saddle (2nd-derivative test)

V · Integrals (Ch 15–16)

Double integrals as iterated integrals
General regions, reversing the order, area
Line integrals \(\int_C f\,ds\)
Work, circulation and flux of a field
Conservative fields and potentials

In plain words

How the parts connect

Part I reuses Calculus I limits (L'Hôpital, \(\lim (1+\frac1n)^n = e\)) and integrals (the integral test). Taylor series come back whenever a function has to be approximated. In Parts III–V each Calculus I idea gets a several-variable version: derivative → partial derivatives and the gradient; tangent line → tangent plane; \(f'=0\) → critical points with the test \(D = f_{xx}f_{yy}-f_{xy}^2\); \(\int_a^b f\,dx\) → double and line integrals; the Fundamental Theorem → \(\int_C \nabla f\cdot d\mathbf r = f(B)-f(A)\).

Convergence tests

Test	When to use it	Condition	Conclusion
\(n\)th-term test	Always first: a 5-second check	\(\lim a_n \ne 0\) or does not exist	\(\sum a_n\) diverges. If \(a_n\to0\): no conclusion
Geometric	\(\sum a r^{n}\): constant ratio between terms	\(\|r\|<1\)	Converges to \(\dfrac{\text{first term}}{1-r}\); diverges if \(\|r\|\ge1\)
\(p\)-series	\(\sum 1/n^p\), also the comparison series	\(p>1\)	Converges; diverges if \(p\le1\)
Telescoping	\(\sum (b_n - b_{n+1})\), partial fractions	\(b_n \to L\)	\(S = b_1 - L\)
Integral test	\(a_n=f(n)\) with \(f\) easy to integrate: \(\ln n\), \(\frac{1}{n\ln n}\), \(ne^{-n^2}\)	\(f\) positive, continuous, decreasing; \(\int_N^\infty f\,dx\)	Series and integral both converge or both diverge
Direct comparison	Positive terms, a simple bound is visible (\(\sin^2 n \le 1\))	\(0\le a_n\le b_n\)	\(\sum b_n\) conv \(\Rightarrow \sum a_n\) conv; \(a_n\ge b_n\ge0\), \(\sum b_n\) div \(\Rightarrow\) div
Limit comparison	Rational / root expressions: keep the dominant terms	\(\lim \frac{a_n}{b_n} = c\), \(0<c<\infty\)	Both converge or both diverge
Ratio test	Factorials and exponentials: \(n!,\ 3^n,\ n^n\)	\(\rho = \lim \left\|\frac{a_{n+1}}{a_n}\right\|\)	\(\rho<1\) abs. conv.; \(\rho>1\) div.; \(\rho=1\) no info
Root test	Whole term is an \(n\)th power: \(\left(\frac{3n+2}{4n+1}\right)^n\)	\(\rho = \lim \sqrt[n]{\|a_n\|}\)	Same as ratio test
Alternating series test	\(\sum (-1)^{n} b_n\), \(b_n>0\)	\(b_{n+1}\le b_n\) and \(b_n\to0\)	Converges (then check \(\sum b_n\) for absolute)
Absolute convergence	Any series with mixed signs	\(\sum \|a_n\|\) converges	\(\sum a_n\) converges absolutely; conv. but not abs. = conditionally

Common Maclaurin series

Function	Series	First terms	Converges on
\(\dfrac{1}{1-x}\)	\(\sum_{n=0}^\infty x^n\)	\(1+x+x^2+x^3+\cdots\)	\((-1,1)\)
\(\dfrac{1}{1+x}\)	\(\sum_{n=0}^\infty (-1)^n x^n\)	\(1-x+x^2-x^3+\cdots\)	\((-1,1)\)
\(e^x\)	\(\sum_{n=0}^\infty \dfrac{x^n}{n!}\)	\(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\)	\((-\infty,\infty)\)
\(\sin x\)	\(\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\)	\(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\)	\((-\infty,\infty)\)
\(\cos x\)	\(\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}\)	\(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\)	\((-\infty,\infty)\)
\(\ln(1+x)\)	\(\sum_{n=1}^\infty \dfrac{(-1)^{n-1}x^n}{n}\)	\(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\)	\((-1,1]\)
\(\arctan x\)	\(\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{2n+1}\)	\(x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots\)	\([-1,1]\)
\((1+x)^m\)	\(\sum_{k=0}^\infty \binom{m}{k}x^k\)	\(1+mx+\frac{m(m-1)}{2!}x^2+\cdots\)	\((-1,1)\) at least

Derivatives next to their integrals

Read each row left to right: differentiate \(F\) to get \(f\), so \(\int f\,dx = F + C\). These are the integrals the integral test and the line-integral problems need.

\(F(x)\)	\(F'(x)=f(x)\)	Typical use
\(\dfrac{x^{p+1}}{p+1}\ (p\ne-1)\)	\(x^p\)	\(p\)-series: \(\int_1^\infty x^{-p}dx\) finite iff \(p>1\)
\(\ln\|x\|\)	\(\dfrac1x\)	harmonic series diverges
\(\dfrac{e^{kx}}{k}\)	\(e^{kx}\)	exponential terms
\(-\dfrac{1}{2}e^{-x^2}\)	\(x e^{-x^2}\)	\(\sum n e^{-n^2}\) by the integral test
\(\ln\|\ln x\|\)	\(\dfrac{1}{x\ln x}\)	\(\sum \frac{1}{n\ln n}\) diverges
\(-\dfrac{1}{\ln x}\)	\(\dfrac{1}{x(\ln x)^2}\)	\(\sum \frac{1}{n(\ln n)^2}\) converges
\(\dfrac{(\ln x)^{1-p}}{1-p}\)	\(\dfrac{1}{x(\ln x)^p}\)	converges iff \(p>1\)
\(\dfrac12 (\ln x)^2\)	\(\dfrac{\ln x}{x}\)	\(\sum \frac{\ln n}{n}\) diverges
\(\dfrac{x^{1-p}\ln x}{1-p} - \dfrac{x^{1-p}}{(1-p)^2}\)	\(\dfrac{\ln x}{x^p}\)	by parts; converges iff \(p>1\)
\(\arctan x\)	\(\dfrac{1}{1+x^2}\)	\(\sum \frac1{1+n^2}\); Maclaurin series of \(\arctan\)
\(\dfrac{1}{2}\ln(x^2+a^2)\)	\(\dfrac{x}{x^2+a^2}\)	\(\sum \frac{n}{n^2+4}\) diverges
\(\arcsin x\)	\(\dfrac{1}{\sqrt{1-x^2}}\)
\(-\cos x,\ \sin x\)	\(\sin x,\ \cos x\)	parametric curves, line integrals
\(\tan x\)	\(\sec^2 x\)
\(\sec x\)	\(\sec x\tan x\)

Integration by parts \(\int u\,dv = uv - \int v\,du\) (choose \(u=\ln x\)). Trig identities used in integrals: \(\sin^2x+\cos^2x=1\), \(\sin^2x = \frac{1-\cos 2x}{2}\), \(\cos^2x=\frac{1+\cos2x}{2}\), \(\sin 2x = 2\sin x\cos x\), \(1+\tan^2x=\sec^2x\).

Vectors, lines and planes (Ch 12)

Quantity	Formula
Length, unit vector	\(\|\mathbf v\| = \sqrt{v_1^2+v_2^2+v_3^2}\), \(\ \mathbf u = \mathbf v/\|\mathbf v\|\)
Dot product, angle	\(\mathbf u\cdot\mathbf v = u_1v_1+u_2v_2+u_3v_3 = \|\mathbf u\|\|\mathbf v\|\cos\theta\); \(\ \mathbf u\perp\mathbf v \iff \mathbf u\cdot\mathbf v=0\)
Projection of \(\mathbf u\) onto \(\mathbf v\)	\(\operatorname{proj}_{\mathbf v}\mathbf u = \dfrac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\,\mathbf v\), scalar component \(\dfrac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|}\)
Cross product	\(\mathbf u\times\mathbf v = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\), \(\ \|\mathbf u\times\mathbf v\| = \|\mathbf u\|\|\mathbf v\|\sin\theta\)
Areas	parallelogram \(\|\mathbf u\times\mathbf v\|\); triangle \(PQR\): \(\tfrac12\|\overrightarrow{PQ}\times\overrightarrow{PR}\|\)
Volume of parallelepiped	\(\|\mathbf u\cdot(\mathbf v\times\mathbf w)\|\) = \|determinant of the three rows\|
Sphere	\((x-x_0)^2+(y-y_0)^2+(z-z_0)^2=a^2\): complete the square to find center and radius
Line through \(P_0\) along \(\mathbf v\)	\(\mathbf r(t)=\mathbf r_0+t\mathbf v\): \(x=x_0+tv_1,\ y=y_0+tv_2,\ z=z_0+tv_3\)
Plane through \(P_0\) with normal \(\mathbf n=\langle A,B,C\rangle\)	\(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\)
Distances	point \(S\) to plane: \(\dfrac{\|\overrightarrow{PS}\cdot\mathbf n\|}{\|\mathbf n\|}\); point \(S\) to line: \(\dfrac{\|\overrightarrow{PS}\times\mathbf v\|}{\|\mathbf v\|}\)

Several variables (Ch 13–14)

Idea	Formula
Motion along \(\mathbf r(t)\)	\(\mathbf v=\mathbf r'(t)\), speed \(\|\mathbf v\|\), \(\mathbf a=\mathbf r''(t)\), \(\mathbf T=\mathbf v/\|\mathbf v\|\); length \(\int_a^b\|\mathbf v\|\,dt\)
Partial derivatives	\(f_x\): differentiate in \(x\) with \(y\) held constant; Clairaut \(f_{xy}=f_{yx}\)
Chain rule	\(\dfrac{dw}{dt} = w_x\dfrac{dx}{dt}+w_y\dfrac{dy}{dt}\); implicit: \(\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}\)
Gradient, directional derivative	\(\nabla f=\langle f_x,f_y,f_z\rangle\), \(\ D_{\mathbf u}f = \nabla f\cdot\mathbf u\) (\(\mathbf u\) a unit vector); fastest increase \(\|\nabla f\|\) along \(\nabla f\)
Tangent plane	to \(F(x,y,z)=c\): \(\nabla F(P_0)\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0\); to \(z=f(x,y)\): \(z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\)
Linearization, differential	\(L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\), \(\ df=f_x\,dx+f_y\,dy\)
Second-derivative test	at a critical point, \(D=f_{xx}f_{yy}-f_{xy}^2\): \(D>0,f_{xx}>0\) min; \(D>0,f_{xx}<0\) max; \(D<0\) saddle; \(D=0\) no info

Double and line integrals (Ch 15–16)

Idea	Formula
Fubini (rectangle)	\(\iint_R f\,dA=\int_a^b\!\int_c^d f\,dy\,dx=\int_c^d\!\int_a^b f\,dx\,dy\)
General regions	vertical slices: \(\int_a^b\int_{g_1(x)}^{g_2(x)} f\,dy\,dx\); horizontal slices: \(\int_c^d\int_{h_1(y)}^{h_2(y)} f\,dx\,dy\)
Area	\(A=\iint_R 1\,dA\)
Line integral of a function	\(\int_C f\,ds=\int_a^b f(\mathbf r(t))\,\|\mathbf r'(t)\|\,dt\)
Work / circulation	\(\int_C \mathbf F\cdot d\mathbf r=\int_a^b \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt=\int_C M\,dx+N\,dy+P\,dz\)
Flux across a plane curve	\(\oint_C M\,dy-N\,dx\) (counterclockwise)
Conservative field	\(\mathbf F=\nabla f\iff\) \(M_y=N_x\) (2D), and also \(P_y=N_z,\ M_z=P_x\) (3D); then \(\int_A^B\mathbf F\cdot d\mathbf r=f(B)-f(A)\), and \(\oint=0\)

Before you start

Typing math on a plain keyboard

Type answers as you would on a calculator. A live preview under each box shows how your input was read, so check it before pressing Enter. Multiplication can be implicit: 2x, x sin x, (x+1)(x-1).

Math	Type	Notes
\(x^2\)	`x^2`	also `x²`
\(\sqrt{x}\)	`sqrt(x)`	`cbrt(x)` for \(\sqrt[3]x\)
\(\pi\)	`pi`	also `π`
\(e^x\)	`e^x` or `exp(x)`	`e^2x` means \(e^2\cdot x\): type `e^(2x)`
\(\ln\|x\|\)	`ln\|x\|` or `ln(abs(x))`	`log` means base 10 (as in Thomas); `log_2(x)` for base 2
\(\sin^2x\)	`sin^2(x)` or `(sin x)^2`	`sin^-1 x` means \(\arcsin x\)
\(\arctan x\)	`arctan(x)` or `atan(x)`	also `tan^-1(x)`
\(\infty\)	`inf`	`-inf` for \(-\infty\)
a fraction	`(top)/(bottom)` with parentheses	`1/2x` means \(\tfrac12x\); type `1/(2x)` for \(\tfrac{1}{2x}\)
\(+\,C\)	optional	antiderivatives and potentials are checked up to a constant
diverges	`diverges` or `DNE`	`inf` also counts when the limit is \(+\infty\)
\(n!\)	`n!`, `(2n)!`
\(\theta\)	`theta`
interval	`[-1, 3)`, `(-inf, inf)`, `{2}`	or `-1 <= x < 3`
vector \(\langle1,-2,3\rangle\)	`<1, -2, 3>` or `i - 2j + 3k`	also `(1, -2, 3)`
line	`<1,0,2> + t<3,-1,1>`	or `x = 1+3t, y = -t, z = 2+t`
plane, sphere	`2x - y + 3z = 5`	any equivalent equation is accepted
region	`x^2 + y^2 + z^2 <= 4`	chains like `1 <= x^2+y^2 <= 4`, joined with `and`
points	`(0, 0), (1, -1)`	`none` if there are none
angle	`pi/3` or `60 deg`	degrees need the word `deg`

Try it: type anything below and watch the preview. Nothing is graded here.

Chapter 1 · Thomas §10.1 · Week 1

Sequences

In plain words

The idea

A sequence is an endless list \(a_1, a_2, a_3, \dots\). The only question we ask is where the list is heading. If the terms settle down to one number \(L\), the sequence converges to \(L\); otherwise it diverges. Every series question in the next chapters is really a question about a sequence: either the terms \(a_n\) or the partial sums \(S_N\).

Definition§10.1

Convergence

The sequence \(\{a_n\}\) converges to \(L\) if for every \(\varepsilon>0\) there is an integer \(N\) such that \(|a_n - L| < \varepsilon\) for all \(n>N\). We write \(\lim_{n\to\infty} a_n = L\). If no such \(L\) exists the sequence diverges. It diverges to \(\infty\) if for every \(M\) there is an \(N\) with \(a_n>M\) for all \(n>N\).

The sequence \(a_n = n(\sqrt{n^2+4}-n)\) from Quiz 1 (CRNs 10895–10897) with its limit \(L=2\). Move \(\varepsilon\): the smaller the band, the later the terms enter it for good, but they always do. That is exactly the definition.

Theorem§10.1

Limit laws and the Sandwich Theorem

If \(a_n\to A\) and \(b_n\to B\) then \(a_n\pm b_n\to A\pm B\), \(a_nb_n\to AB\), \(ka_n\to kA\), and \(a_n/b_n\to A/B\) when \(B\ne0\).

Sandwich: if \(a_n\le b_n\le c_n\) for all \(n\) beyond some index and \(a_n\to L\), \(c_n\to L\), then \(b_n\to L\). In particular, if \(|b_n|\le c_n\) and \(c_n\to0\) then \(b_n\to0\).

Theorem§10.1

Continuous functions and L'Hôpital

If \(a_n\to L\) and \(f\) is continuous at \(L\), then \(f(a_n)\to f(L)\). That is why \(\lim \ln a_n = \ln(\lim a_n)\) and \(\lim e^{a_n}=e^{\lim a_n}\).

If \(a_n = f(n)\) and \(\lim_{x\to\infty} f(x) = L\), then \(a_n\to L\). So you may use L'Hôpital's rule on \(f(x)\), the function of a real variable.

Key facts§10.1

Commonly occurring limits

\[\frac{\ln n}{n}\to0,\qquad \sqrt[n]{n}\to1,\qquad x^{1/n}\to1\ (x>0),\qquad x^n\to0\ (|x|<1),\] \[\left(1+\frac{x}{n}\right)^n\to e^x,\qquad \frac{x^n}{n!}\to0.\]

Growth ladder (each beats the one before): \(\ln n \ll n^p \ll a^n\ (a>1) \ll n! \ll n^n\).

Theorem§10.1

Monotone sequences

A sequence that is bounded and monotonic (always increasing or always decreasing) converges.

The six tools for a limit of a sequence

Recipe

Limit of a sequence, step by step

Plug in “\(n=\infty\)” mentally and name the form: \(\frac{\infty}{\infty}\), \(\infty-\infty\), \(0\cdot\infty\), \(1^\infty\), or (bounded)\(\times\)(something \(\to0\)).
Pick the tool for that form:
- \(\frac{\infty}{\infty}\) with powers → divide top and bottom by the highest power in the denominator.
- \(\infty-\infty\) with roots → multiply and divide by the conjugate.
- \(\sin, \cos, (-1)^n\) → sandwich using \(-1\le\sin\le1\).
- \(1^\infty\) → rewrite with \(\left(1+\frac{x}{n}\right)^n\to e^x\).
- difference of logs → combine into one log, then use continuity of \(\ln\).
- factorials and exponentials → the growth ladder (or the ratio \(a_{n+1}/a_n\)).
Name the theorem you used (Sandwich, continuity, \(e\)-limit). The solution sheets give a point for the justification.
Conclude in words: “the sequence converges to \(L\)” or “diverges”.

Watch out

Common mistakes

\(1^\infty\) is not 1: \(\left(1+\frac1n\right)^n\to e\).
Bounded does not mean convergent: \((-1)^n\) is bounded and diverges.
\((-1)^n b_n\) with \(b_n\to c\ne0\) diverges: the even terms go to \(c\), the odd terms to \(-c\).
In \(n(\sqrt{n^2+4}-n)\) you cannot take the limit of each piece separately: that gives \(\infty\cdot 0\). Use the conjugate first.
\(\sqrt{n^2+4n}\) is not \(n+2\). After rationalizing, divide by \(n\) using \(\sqrt{n^2}=n\) (for \(n>0\)).

Worked examples (from the real quizzes)

Example∞ − ∞

Conjugate (Quiz 1, CRN 10892, part b)

\(x_n = n^{3/2}\left(\sqrt{n^3+1}-\sqrt{n^3-2}\right)\).

Multiply and divide by the conjugate: \(x_n = n^{3/2}\dfrac{(n^3+1)-(n^3-2)}{\sqrt{n^3+1}+\sqrt{n^3-2}} = \dfrac{3n^{3/2}}{\sqrt{n^3+1}+\sqrt{n^3-2}}\).
Divide top and bottom by \(n^{3/2}=\sqrt{n^3}\): \(x_n = \dfrac{3}{\sqrt{1+1/n^3}+\sqrt{1-2/n^3}}\).
So \(x_n\to \dfrac{3}{1+1}=\dfrac32\): the sequence converges to \(\tfrac32\).

Example(−1)ⁿ

Subsequences (Quiz 1, CRN 10892, part a)

\(x_n=(-1)^n\dfrac{2n}{n+1}\). Since \(\dfrac{2n}{n+1}\to2\), the even terms tend to \(2\) and the odd terms to \(-2\). Two different limits, so the sequence diverges.

Example1^∞

The e-limit (Quiz 1, CRNs 10892 and 10893)

\(\left(\dfrac{n}{n+1}\right)^n = \left(\dfrac{n+1}{n}\right)^{-n} = \left[\left(1+\dfrac1n\right)^{n}\right]^{-1}\to e^{-1}\).

\(\left(1+\dfrac{1}{n(n+2)}\right)^n = \left[\left(1+\dfrac{1}{n(n+2)}\right)^{n(n+2)}\right]^{\frac{1}{n+2}}\): the bracket tends to \(e\) and the exponent to \(0\), so the limit is \(e^0=1\).

Examplebounded × small

Sandwich (Quiz 1, CRN 10890 and the unlabelled Quiz 1)

\(-\dfrac1n\le \dfrac{\sin 4n}{n}\le\dfrac1n\) and both sides tend to \(0\), so \(\dfrac{\sin 4n}{n}\to0\). In the same way \(0\le\dfrac{\sin^2 n}{2^n}\le\dfrac{1}{2^n}\to0\).

Quick checks · true or false?

Every bounded sequence converges.

If \(|a_n|\to0\), then \(a_n\to0\).

\(\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^{n} = 1\), because \(1+\frac2n\to1\).

If \(a_n\to3\), then \(\ln(a_n)\to\ln 3\).

\(\dfrac{100^n}{n!}\to0\).

Practice set 1 · Limits of sequences

10 problems in the style of Quiz 1

Rational expression

new · quiz style

Find \(\displaystyle\lim_{n\to\infty}\frac{3n^2-n+4}{5n^2+2}\).

Hint

Divide every term by \(n^2\), the highest power in the denominator.

Answer

\(\tfrac35\)

Full solution

Divide top and bottom by \(n^2\): \(\dfrac{3-\frac1n+\frac4{n^2}}{5+\frac{2}{n^2}}\).
Each fraction with \(n\) in the denominator tends to \(0\), so the limit is \(\tfrac35\): the ratio of the leading coefficients.

∞ − ∞ with a root

new · like Quiz 1, CRN 10891

Find \(\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+6n}-n\right)\).

Hint

Multiply and divide by \(\sqrt{n^2+6n}+n\).

Answer

\(3\)

Full solution

\(\sqrt{n^2+6n}-n = \dfrac{(n^2+6n)-n^2}{\sqrt{n^2+6n}+n} = \dfrac{6n}{\sqrt{n^2+6n}+n}\).
Divide by \(n\) (and use \(\sqrt{n^2}=n\)): \(\dfrac{6}{\sqrt{1+6/n}+1}\to\dfrac{6}{2}=3\).

Difference of two roots

new · like Quiz 1, CRN 10892

Find \(\displaystyle\lim_{n\to\infty} n\left(\sqrt{n^2+1}-\sqrt{n^2-1}\right)\).

Hint

The conjugate turns the difference of roots into \(\dfrac{2}{\sqrt{n^2+1}+\sqrt{n^2-1}}\).

Answer

\(1\)

Full solution

\(a_n = n\cdot\dfrac{(n^2+1)-(n^2-1)}{\sqrt{n^2+1}+\sqrt{n^2-1}} = \dfrac{2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}\).
Divide by \(n\): \(\dfrac{2}{\sqrt{1+1/n^2}+\sqrt{1-1/n^2}}\to\dfrac22=1\).

A 1^∞ form

new · like Quiz 1, CRN 10893

Find \(\displaystyle\lim_{n\to\infty}\left(1+\frac3n\right)^{2n}\).

Hint

\(\left(1+\frac3n\right)^{2n} = \left[\left(1+\frac3n\right)^{n}\right]^2\).

Answer

\(e^6\)

Full solution

Write \(a_n = \left[\left(1+\frac3n\right)^n\right]^2\).
The bracket tends to \(e^3\) (because \((1+\frac xn)^n\to e^x\)), and squaring is continuous.
So \(a_n\to(e^3)^2 = e^6\).

Another e-limit

new · like Quiz 1, CRN 10892

Find \(\displaystyle\lim_{n\to\infty}\left(\frac{n}{n+3}\right)^{n}\).

Hint

\(\dfrac{n}{n+3} = \left(1+\dfrac3n\right)^{-1}\).

Answer

\(e^{-3}=\dfrac1{e^3}\)

Full solution

\(\left(\dfrac{n}{n+3}\right)^n = \left(\dfrac{n+3}{n}\right)^{-n} = \left[\left(1+\dfrac3n\right)^n\right]^{-1}\).
The bracket tends to \(e^3\), so the limit is \(e^{-3}\).

Sandwich

new · like the unlabelled Quiz 1

Find \(\displaystyle\lim_{n\to\infty}\frac{\cos^2 n}{3^n}\).

Hint

\(0\le\cos^2n\le1\).

Answer

\(0\)

Full solution

Since \(0\le\cos^2 n\le1\), we have \(0\le\dfrac{\cos^2n}{3^n}\le\dfrac1{3^n}\).
\(\dfrac{1}{3^n}=\left(\tfrac13\right)^n\to0\) because \(|\tfrac13|<1\).
By the Sandwich Theorem the limit is \(0\).

Factorial beats exponential

new · like Quiz 1, CRN 10890

Find \(\displaystyle\lim_{n\to\infty}\frac{5^n}{n!}\).

Hint

Use \(\dfrac{x^n}{n!}\to0\), or look at \(a_{n+1}/a_n = \dfrac{5}{n+1}\).

Answer

\(0\)

Full solution

This is \(\dfrac{x^n}{n!}\) with \(x=5\), one of the commonly occurring limits: it tends to \(0\).
Why: \(a_{n+1} = a_n\cdot\dfrac{5}{n+1}\), and once \(n\ge 9\) every step multiplies by at most \(\tfrac12\).

Logs: combine first

new · like Quiz 1, CRN 10891

Find \(\displaystyle\lim_{n\to\infty}\left[\ln(2n^2+1)-2\ln n\right]\).

Hint

\(2\ln n = \ln n^2\). Combine into one logarithm.

Answer

\(\ln 2\)

Full solution

\(a_n = \ln(2n^2+1)-\ln n^2 = \ln\dfrac{2n^2+1}{n^2} = \ln\left(2+\dfrac1{n^2}\right)\).
Inside: \(2+\frac1{n^2}\to2\). \(\ln\) is continuous, so \(a_n\to\ln2\).

Alternating sign

new · like Quiz 1, CRN 10892

Does \(a_n = (-1)^n\dfrac{n+1}{n}\) converge? If it does, type the limit; otherwise type diverges.

Hint

Look at the even terms and the odd terms separately.

Answer

The sequence diverges (no limit).

Full solution

\(\dfrac{n+1}{n}\to1\).
So \(a_{2k}\to1\) and \(a_{2k+1}\to-1\): two subsequences with different limits.
Therefore the sequence diverges. (It does not diverge to \(\pm\infty\); it oscillates.)

An nth root

new · lecture style

Find \(\displaystyle\lim_{n\to\infty}\sqrt[n]{3n}\).

Hint

\(\sqrt[n]{3n} = 3^{1/n}\cdot n^{1/n}\).

Answer

\(1\)

Full solution

Split: \(\sqrt[n]{3n}=3^{1/n}\,n^{1/n}\).
\(3^{1/n}\to1\) and \(n^{1/n}\to1\) (both commonly occurring limits).
So the product tends to \(1\cdot1=1\).

Chapter 2 · Thomas §10.2 · Week 2

Infinite series

In plain words

The idea

A series \(\sum a_n\) means “keep adding terms forever”. We cannot literally do that, so we add the first \(N\) terms, call the result \(S_N\), and ask where the sequence of partial sums \(S_1, S_2, S_3,\dots\) is heading. If \(S_N\to S\), the series converges and its sum is \(S\). Only two kinds of series can be summed exactly in this course: geometric and telescoping.

Definition§10.2

Series, partial sums, sum

Given \(\{a_n\}\), the \(N\)th partial sum is \(S_N = a_1+a_2+\cdots+a_N\). The series \(\sum_{n=1}^\infty a_n\) converges to \(S\) if \(S_N\to S\); then \(S\) is its sum. Otherwise the series diverges.

Theorem§10.2

Geometric series

\[\sum_{n=0}^\infty a r^n = a + ar + ar^2 + \cdots = \frac{a}{1-r}\quad\text{if } |r|<1,\] and the series diverges if \(|r|\ge1\) (\(a\ne0\)). Starting at any index: sum \(=\dfrac{\text{first term}}{1-r}\).

Theorem§10.2

The nth-term test for divergence

If \(\sum a_n\) converges then \(a_n\to0\). Equivalently: if \(\lim a_n\ne0\) or the limit does not exist, then \(\sum a_n\) diverges. The converse is false: \(\frac1n\to0\) but \(\sum\frac1n\) diverges.

Theorem§10.2

Combining series

If \(\sum a_n=A\) and \(\sum b_n=B\) both converge, then \(\sum(a_n\pm b_n)=A\pm B\) and \(\sum ka_n = kA\). If \(\sum a_n\) converges and \(\sum b_n\) diverges, then \(\sum(a_n\pm b_n)\) diverges. Adding or removing finitely many terms never changes convergence (it changes the sum).

Partial sums of \(\sum_{n\ge1} 3\left(\frac34\right)^n\) from Quiz 2 (CRN 10892): the terms (grey) shrink by the factor \(\frac34\), and the partial sums (teal) climb to the sum \(9\).

Recipe

Sum of a series, step by step

\(n\)th-term check: if \(a_n\not\to0\), stop: the series diverges.
Geometric? Rewrite every term as (constant)\(\cdot r^n\). Example: \(\dfrac{3^{n+1}}{4^n} = 3\left(\dfrac34\right)^n\). Read off \(r\) and the first term (plug in the starting index!).
Telescoping? Write \(a_n = b_n - b_{n+1}\) (partial fractions, or differences of logs or \(\arctan\)). Write out \(S_N\), cancel, and let \(N\to\infty\).
Sum of two series? Split only if each piece converges. One convergent plus one divergent piece means the whole series diverges.
State the conclusion: “converges to \(S=\dots\)” or “diverges”.

Watch out

Common mistakes

The formula \(\frac{a}{1-r}\) needs the first term, not always \(a\): \(\sum_{n=1}^\infty r^n = \frac{r}{1-r}\), not \(\frac1{1-r}\).
\(r\) is the ratio of consecutive terms. In \(\sum \frac{5^n}{2^{n+1}} = \frac12\sum\left(\frac52\right)^n\), \(r=\frac52\), so it diverges.
\(a_n\to0\) does not prove convergence (harmonic series).
In a telescoping sum, check which terms survive at both ends of \(S_N\).

Worked examples (from the real quizzes)

Exampletelescoping

Telescoping with arctan (Quiz 2, CRN 10892)

\(\sum_{n=1}^\infty\left(\tan^{-1}n-\tan^{-1}(n+1)\right)\).

\(S_N = (\tan^{-1}1-\tan^{-1}2)+(\tan^{-1}2-\tan^{-1}3)+\cdots+(\tan^{-1}N-\tan^{-1}(N+1)) = \tan^{-1}1-\tan^{-1}(N+1)\).
\(\tan^{-1}(N+1)\to\frac\pi2\), so \(S = \frac\pi4-\frac\pi2 = -\frac\pi4\).

Examplegeometric

Geometric (Quiz 1, CRN 10893)

\(\sum_{n=1}^\infty\left(\frac1{\sqrt2}\right)^n\): \(r = \frac1{\sqrt2}\), \(|r|<1\), first term \(\frac1{\sqrt2}\). So \(S=\dfrac{1/\sqrt2}{1-1/\sqrt2}=\dfrac{1}{\sqrt2-1}=\sqrt2+1\).

Examplecareful

Splitting a sum (Quiz 2, CRN 10892)

\(\sum\left[\frac{3^{n+1}}{4^n}+\frac{5^n}{2^{n+1}}\right]\): the first part is \(3\sum\left(\frac34\right)^n = 3\cdot\frac{3/4}{1/4} = 9\), but the second is geometric with \(r=\frac52>1\) and diverges. Convergent + divergent = divergent.

Exampledivergence

nth-term test (Quiz 1, CRN 10893 and unlabelled Quiz 1)

\(\sum\left(1+\frac1n\right)^n\) diverges because the terms tend to \(e\ne0\). \(\sum\frac{e^n}{e^n+n}\) diverges because \(\frac{e^n}{e^n+n} = \frac{1}{1+n/e^n}\to1\). \(\sum\cos n\pi = \sum(-1)^n\) diverges because the terms have no limit.

Quick checks · true or false?

If \(\lim_{n\to\infty} a_n = 0\), then \(\sum a_n\) converges.

\(\sum_{n=1}^\infty r^n = \dfrac{1}{1-r}\) for \(|r|<1\).

If \(\sum a_n\) converges and \(\sum b_n\) diverges, then \(\sum (a_n+b_n)\) diverges.

\(\sum(a_n+b_n)\) can converge even though \(\sum a_n\) and \(\sum b_n\) both diverge.

Deleting the first 1000 terms of a divergent series can make it converge.

Practice set 2 · Sums of series

10 problems in the style of Quiz 2 (CRN 10892) and Quiz 1 (CRN 10893)

Geometric, starting at n = 1

new · quiz style

Find \(\displaystyle\sum_{n=1}^\infty 2\left(\frac25\right)^n\).

Hint

\(r=\frac25\). The first term (at \(n=1\)) is \(2\cdot\frac25\).

Answer

\(\tfrac43\)

Full solution

Geometric with \(r=\frac25\), \(|r|<1\): converges.
First term: \(2\cdot\frac25=\frac45\).
\(S = \dfrac{4/5}{1-2/5} = \dfrac{4/5}{3/5} = \dfrac43\).

Alternating geometric

new · quiz style

Find \(\displaystyle\sum_{n=0}^\infty \frac{(-1)^n\,3}{4^n}\).

Hint

\(\frac{(-1)^n 3}{4^n} = 3\left(-\frac14\right)^n\).

Answer

\(\tfrac{12}{5}\)

Full solution

Geometric with \(r=-\frac14\) and first term \(3\) (at \(n=0\)).
\(S=\dfrac{3}{1-(-1/4)} = \dfrac{3}{5/4} = \dfrac{12}{5}\).

Split into two geometric series

new · like Quiz 2, CRN 10892

Find \(\displaystyle\sum_{n=1}^\infty \frac{2^n+3^n}{6^n}\).

Hint

\(\frac{2^n}{6^n} = \left(\frac13\right)^n\) and \(\frac{3^n}{6^n}=\left(\frac12\right)^n\). Both converge, so you may split.

Answer

\(\tfrac32\)

Full solution

\(\sum\left(\frac13\right)^n = \dfrac{1/3}{1-1/3} = \dfrac12\).
\(\sum\left(\frac12\right)^n = \dfrac{1/2}{1-1/2} = 1\).
Both converge, so the sum is \(\frac12+1 = \frac32\).

Telescoping with partial fractions

new · lecture style

Find \(\displaystyle\sum_{n=1}^\infty \frac{1}{n(n+2)}\).

Hint

\(\dfrac{1}{n(n+2)} = \dfrac12\left(\dfrac1n-\dfrac1{n+2}\right)\). Two terms survive at the start.

Answer

\(\tfrac34\)

Full solution

Partial fractions: \(\dfrac{1}{n(n+2)} = \dfrac12\left(\dfrac1n-\dfrac{1}{n+2}\right)\).
\(S_N = \frac12\left[\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac1N-\frac1{N+2}\right)\right] = \frac12\left[1+\frac12-\frac1{N+1}-\frac1{N+2}\right]\).
As \(N\to\infty\): \(S=\frac12\cdot\frac32 = \frac34\).

Telescoping logs

new · lecture style

Find \(\displaystyle\sum_{n=1}^\infty \ln\frac{n}{n+1}\), or type -inf / diverges.

Hint

\(\ln\frac{n}{n+1} = \ln n - \ln(n+1)\). Write out \(S_N\).

Answer

The partial sums are \(S_N=-\ln(N+1)\to-\infty\): the series diverges (to \(-\infty\)).

Full solution

\(S_N = (\ln1-\ln2)+(\ln2-\ln3)+\cdots+(\ln N-\ln(N+1)) = -\ln(N+1)\).
\(-\ln(N+1)\to-\infty\), so the series diverges. Note that the terms \(\ln\frac{n}{n+1}\to\ln1=0\): the \(n\)th-term test alone could not decide this.

Telescoping roots

new · lecture style

Find \(\displaystyle\sum_{n=1}^\infty \left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)\).

Hint

Everything cancels except the very first term and the last.

Answer

\(1\)

Full solution

\(S_N = 1 - \frac1{\sqrt{N+1}}\) (all middle terms cancel).
\(\frac1{\sqrt{N+1}}\to0\), so \(S=1\).

nth-term test

new · like Quiz 1, CRN 10893

Find \(\displaystyle\sum_{n=1}^\infty \frac{n}{2n+5}\), or type diverges.

Hint

What do the terms tend to?

Answer

The series diverges.

Full solution

\(\dfrac{n}{2n+5} = \dfrac{1}{2+5/n}\to\dfrac12\ne0\).
By the \(n\)th-term test the series diverges.

Shifted index

new · quiz style

Find \(\displaystyle\sum_{n=2}^\infty \frac{5}{3^{\,n-1}}\).

Hint

The first term (at \(n=2\)) is \(\frac53\); the ratio is \(\frac13\).

Answer

\(\tfrac52\)

Full solution

First term: \(n=2\) gives \(\frac{5}{3}\). Ratio \(r=\frac13\).
\(S=\dfrac{5/3}{1-1/3} = \dfrac{5/3}{2/3} = \dfrac52\).

A repeating decimal

new · Thomas style

Write \(0.\overline{27} = 0.272727\ldots\) as a fraction in lowest terms.

Hint

\(0.\overline{27} = \frac{27}{100}+\frac{27}{100^2}+\cdots\)

Answer

\(\tfrac{3}{11}\)

Full solution

Geometric with first term \(\frac{27}{100}\) and \(r=\frac1{100}\).
\(S = \dfrac{27/100}{1-1/100} = \dfrac{27}{99} = \dfrac{3}{11}\).

A ratio that is not a fraction

new · quiz style

Find \(\displaystyle\sum_{n=1}^\infty \left(\frac{e}{\pi}\right)^n\).

Hint

\(e\approx2.718<\pi\), so \(|r|<1\).

Answer

\(\dfrac{e}{\pi-e}\)

Full solution

Geometric with \(r = \frac e\pi\), and \(0<r<1\) because \(e<\pi\).
\(S = \dfrac{e/\pi}{1-e/\pi} = \dfrac{e}{\pi-e}\).

Chapter 3 · Thomas §10.3–10.6 · Week 3

Convergence tests

In plain words

The idea

Most series cannot be summed exactly, so we only ask “does it converge?”. For series with positive terms we compare them with something we already understand: an integral, a \(p\)-series, or a geometric series. For series whose terms change sign, we first test the series of absolute values. If that fails, the alternating series test may still give convergence, which is then called conditional.

Positive-term tests

Theorem§10.3

The Integral Test

Let \(a_n=f(n)\) where \(f\) is positive, continuous and decreasing for \(x\ge N\). Then \(\sum_{n=N}^\infty a_n\) and \(\int_N^\infty f(x)\,dx\) either both converge or both diverge.

Theorem§10.3

p-series

\(\displaystyle\sum_{n=1}^\infty\frac1{n^p}\) converges if \(p>1\) and diverges if \(p\le1\). (\(p=1\): the harmonic series diverges.) The same holds for \(\sum\frac{1}{n(\ln n)^p}\), starting at \(n=2\).

Theorem§10.4

The Comparison Test

Suppose \(0\le a_n\le b_n\) for all \(n\ge N\). If \(\sum b_n\) converges then \(\sum a_n\) converges. If \(\sum a_n\) diverges then \(\sum b_n\) diverges. (Smaller than convergent → converges; bigger than divergent → diverges.)

Theorem§10.4

The Limit Comparison Test

Suppose \(a_n>0\), \(b_n>0\) and \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=c\). If \(0<c<\infty\), both series converge or both diverge. (If \(c=0\) and \(\sum b_n\) converges, so does \(\sum a_n\); if \(c=\infty\) and \(\sum b_n\) diverges, so does \(\sum a_n\).)

Theorem§10.5

The Ratio Test and the Root Test

Let \(\rho=\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\) or \(\rho=\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}\). If \(\rho<1\) the series converges absolutely; if \(\rho>1\) (or \(\rho=\infty\)) it diverges; if \(\rho=1\) the test says nothing.

Series with mixed signs

Definition§10.5–10.6

Absolute and conditional convergence

\(\sum a_n\) converges absolutely if \(\sum|a_n|\) converges. It converges conditionally if \(\sum a_n\) converges but \(\sum|a_n|\) diverges.

Theorem§10.5

Absolute Convergence Test

If \(\sum|a_n|\) converges, then \(\sum a_n\) converges.

Theorem§10.6

The Alternating Series Test (Leibniz)

\(\sum_{n=1}^\infty(-1)^{n+1}b_n\) converges if (1) \(b_n>0\), (2) \(b_{n+1}\le b_n\) for all \(n\ge N\), and (3) \(b_n\to0\).

Error estimate: if the test applies, \(|S-S_N|\le b_{N+1}\): the error is at most the first omitted term.

The alternating harmonic series \(1-\frac12+\frac13-\cdots\): the partial sums zig-zag ever closer to \(\ln 2\approx0.693\), but \(\sum\frac1n\) diverges, so the convergence is conditional.

Choosing the right test

Recipe

Which test? Go down the list

\(n\)th-term test. If \(a_n\not\to0\): diverges. Done.
Known type? Geometric (\(|r|<1\)), \(p\)-series (\(p>1\)), telescoping.
Mixed signs? Test \(\sum|a_n|\) with the steps below. If \(\sum|a_n|\) diverges, try the alternating series test → conditional.
Factorials or \(c^n\)? → ratio test. The whole term is an \(n\)th power \((\cdots)^n\)? → root test.
A rational or root expression in \(n\)? → keep only the dominant terms to guess \(b_n=\frac1{n^p}\), then the limit comparison test.
\(\ln n\) or a function you can integrate (\(\frac{1}{n\ln n}\), \(ne^{-n^2}\))? → integral test (check positive, continuous, decreasing).
A bounded factor like \(\sin^2n\), \(\arctan n\)? → direct comparison using its bound.

Recipe

How to write it up (how the sheets give 0.5 + 0.5 + 0.5)

Integral test: (1) name \(f(x)\) and check it is positive, continuous, decreasing; (2) compute \(\lim_{b\to\infty}\int_N^b f\,dx\); (3) conclude: “the integral converges/diverges, so by the integral test the series converges/diverges”.
Limit comparison: (1) choose \(b_n\) from the dominant terms and say whether \(\sum b_n\) converges; (2) compute \(\lim\frac{a_n}{b_n}=c\) with \(0<c<\infty\); (3) conclude “same behaviour as \(\sum b_n\)”.
Ratio / root test: (1) simplify \(\left|\frac{a_{n+1}}{a_n}\right|\) or \(\sqrt[n]{|a_n|}\); (2) take the limit \(\rho\); (3) compare with \(1\).
Absolute / conditional: (1) test \(\sum|a_n|\); (2) if it diverges, check the three conditions of the alternating series test; (3) conclude “conditionally”.

Watch out

Common mistakes

The integral test needs \(f\) decreasing; show it, e.g. with \(f'(x)<0\) for large \(x\).
The value of \(\int_N^\infty f\,dx\) is not the sum of the series. It only decides convergence.
Ratio test with \(\rho=1\) (any \(p\)-series, any rational function of \(n\)) is inconclusive, not “converges”.
Comparison in the wrong direction proves nothing: \(\frac{1}{n^2+n}\le\frac1n\) and \(\sum\frac1n\) diverges tells you nothing.
Limit comparison needs a finite, positive limit. Check the leading powers carefully: \(\sqrt{n^7}=n^{7/2}\).
“Conditionally convergent” needs both facts: the series converges, and \(\sum|a_n|\) diverges.

Worked examples (from the real quizzes)

Example§10.3

Integral test with integration by parts (Quiz 1, CRNs 10895–10897)

\(\sum_{n=2}^\infty\frac{\ln n}{n^{3/2}}\). \(f(x)=\frac{\ln x}{x^{3/2}}\) is positive and continuous for \(x\ge2\), and decreasing for \(x>e^{2/3}\) because \(f'(x)=\frac{1-\frac32\ln x}{x^{5/2}}<0\) there. By parts: \[\int_2^\infty\frac{\ln x}{x^{3/2}}dx = \left[-\frac{2\ln x}{\sqrt x}-\frac{4}{\sqrt x}\right]_2^\infty = \frac{2\ln2+4}{\sqrt2}<\infty.\] So the series converges.

Example§10.5

Ratio test with factorials (Quiz 1, CRNs 10895–10897)

\(a_n=\frac{(n!)^2}{(2n)!}\): \(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)^2}{(2n+2)(2n+1)}=\dfrac{n+1}{2(2n+1)}\to\dfrac14<1\), so the series converges.

Example§10.5

Root test (Quiz 1, CRNs 10895–10897)

\(\sum\left(\frac{3n+2}{4n+1}\right)^n\): \(\sqrt[n]{a_n}=\frac{3n+2}{4n+1}\to\frac34<1\), so the series converges.

Example§10.4

Limit comparison (Quiz 1, CRN 10890 and CRNs 10895–10897)

\(\sum\frac{n^2+2n}{\sqrt{n^7+n^3}}\): dominant terms give \(\frac{n^2}{n^{7/2}}=\frac1{n^{3/2}}\), so take \(b_n=\frac{1}{n^{3/2}}\) (convergent \(p\)-series, \(p=\frac32\)). Then \(\frac{a_n}{b_n}=\frac{n^{7/2}+2n^{5/2}}{\sqrt{n^7+n^3}}\to1\), which is finite and positive, so the series converges. For \(\sum\frac{\sqrt{n^2+1}}{n^2+3n+1}\) the same idea gives \(b_n=\frac1n\), so it diverges.

Example§10.6

Conditional convergence (Quiz 2, CRNs 10890 and 10895)

\(\sum(-1)^{n+1}\frac{1}{\sqrt n}\): \(\sum\frac{1}{\sqrt n}\) is a \(p\)-series with \(p=\frac12\le1\) (diverges), but \(b_n=\frac1{\sqrt n}\) is positive, decreasing and tends to \(0\), so the alternating series converges. It is conditionally convergent.

Quick checks · true or false?

If the ratio test gives \(\rho=1\), the series converges.

If \(\sum|a_n|\) converges, then \(\sum a_n\) converges.

The integral test tells you the sum: \(\sum_{n=1}^\infty\frac1{n^2}=\int_1^\infty\frac{dx}{x^2}=1\).

\(\sum\frac{(-1)^n}{n}\) converges absolutely.

If \(0\le a_n\le b_n\) and \(\sum a_n\) diverges, then \(\sum b_n\) diverges.

The root test is a natural choice for \(\sum\left(\frac{2n}{3n+1}\right)^n\).

Practice set 3A · Positive-term series

12 problems like Quiz 1 and Quiz 2 (integral, comparison, ratio, root)

Rational terms

new · quiz style

Does \(\displaystyle\sum_{n=1}^\infty\frac{1}{n^2+4n}\) converge or diverge?

Hint

Compare with \(\sum\frac1{n^2}\).

Answer

Converges.

Full solution

\(0<\dfrac{1}{n^2+4n}\le\dfrac1{n^2}\) for all \(n\ge1\).
\(\sum\frac1{n^2}\) is a \(p\)-series with \(p=2>1\): it converges.
By the comparison test the series converges.

Looks like 1/n

new · like Quiz 2 (unlabelled)

Does \(\displaystyle\sum_{n=1}^\infty\frac{n}{n^2+1}\) converge or diverge?

Hint

Dominant terms: \(\frac{n}{n^2}=\frac1n\).

Answer

Diverges.

Full solution

Take \(b_n=\frac1n\) (harmonic series, divergent).
\(\lim\dfrac{a_n}{b_n}=\lim\dfrac{n^2}{n^2+1}=1\), finite and positive.
By the limit comparison test the series diverges.

Integral test with ln

new · like Quiz 1, CRN 10891

Does \(\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\) converge or diverge?

Hint

Integral test with \(u=\ln x\).

Answer

Converges.

Full solution

\(f(x)=\frac{1}{x(\ln x)^3}\) is positive, continuous and decreasing for \(x\ge2\).
With \(u=\ln x\): \(\int_2^\infty\frac{dx}{x(\ln x)^3}=\int_{\ln2}^\infty u^{-3}du=\left[-\frac1{2u^2}\right]_{\ln2}^\infty=\frac{1}{2(\ln2)^2}<\infty\).
By the integral test the series converges.

The integral itself

new · like Quiz 1, CRN 10891

As part of the integral test for \(\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}\), evaluate \(\displaystyle\int_2^\infty\frac{dx}{x(\ln x)^2}\).

Hint

\(u=\ln x\), \(du=\frac{dx}{x}\); the new limits are \(\ln 2\) and \(\infty\).

Answer

\(\dfrac{1}{\ln 2}\)

Full solution

Substitute \(u=\ln x\): \(\displaystyle\int_{\ln2}^{\ln b}\frac{du}{u^2} = \left[-\frac1u\right]_{\ln2}^{\ln b} = \frac1{\ln2}-\frac1{\ln b}\).
As \(b\to\infty\), \(\frac1{\ln b}\to0\), so the integral equals \(\frac{1}{\ln 2}\): finite, and the series converges.

Exponential beats polynomial

new · lecture style

Does \(\displaystyle\sum_{n=1}^\infty n^2e^{-n}\) converge or diverge?

Hint

Ratio test: \(\frac{(n+1)^2e^{-(n+1)}}{n^2e^{-n}}\).

Answer

Converges.

Full solution

\(\dfrac{a_{n+1}}{a_n}=\left(\dfrac{n+1}{n}\right)^2e^{-1}\to\dfrac1e<1\).
By the ratio test the series converges (absolutely).

Factorial on top

new · like Quiz 1, CRNs 10895–10897

Does \(\displaystyle\sum_{n=1}^\infty\frac{n!}{10^n}\) converge or diverge?

Hint

Ratio test: \(\frac{(n+1)!}{n!}=n+1\).

Answer

Diverges.

Full solution

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{10^{n+1}}\cdot\dfrac{10^n}{n!}=\dfrac{n+1}{10}\to\infty\).
\(\rho=\infty>1\): the series diverges. (Also \(a_n\to\infty\), so the \(n\)th-term test works too.)

An nth power

new · like Quiz 1, CRNs 10895–10897

Does \(\displaystyle\sum_{n=1}^\infty\left(\frac{n}{2n+1}\right)^n\) converge or diverge?

Hint

Root test.

Answer

Converges.

Full solution

\(\sqrt[n]{a_n}=\dfrac{n}{2n+1}\to\dfrac12<1\).
By the root test the series converges.

Limit comparison with a root

new · like Quiz 1, CRN 10890

Does \(\displaystyle\sum_{n=1}^\infty\frac{2n+1}{\sqrt{n^5+3}}\) converge or diverge?

Hint

Dominant terms: \(\frac{2n}{n^{5/2}}\).

Answer

Converges.

Full solution

Dominant terms: \(\frac{2n}{n^{5/2}}=\frac2{n^{3/2}}\). Take \(b_n=\frac1{n^{3/2}}\), a convergent \(p\)-series (\(p=\frac32\)).
\(\dfrac{a_n}{b_n}=\dfrac{(2n+1)n^{3/2}}{\sqrt{n^5+3}}=\dfrac{2+1/n}{\sqrt{1+3/n^5}}\to2\), finite and positive.
By the limit comparison test the series converges.

Bounded numerator

new · like the unlabelled Quiz 1

Does \(\displaystyle\sum_{n=1}^\infty\frac{\sin^2 n}{n^2}\) converge or diverge?

Hint

\(0\le\sin^2n\le1\).

Answer

Converges.

Full solution

\(0\le\dfrac{\sin^2n}{n^2}\le\dfrac{1}{n^2}\).
\(\sum\frac1{n^2}\) converges, so the series converges by the comparison test.

ln n over n

new · lecture style

Does \(\displaystyle\sum_{n=1}^\infty\frac{\ln n}{n}\) converge or diverge?

Hint

For \(n\ge3\), \(\ln n>1\).

Answer

Diverges.

Full solution

For \(n\ge3\): \(\dfrac{\ln n}{n}>\dfrac1n\ge0\).
\(\sum\frac1n\) diverges, so by comparison \(\sum\frac{\ln n}{n}\) diverges. (The integral test also works: \(\int\frac{\ln x}{x}dx=\frac12(\ln x)^2\to\infty\).)

The ratio-test limit

new · like Quiz 1, CRNs 10895–10897

For \(\displaystyle\sum_{n=1}^\infty\frac{(2n)!}{(n!)^2\,5^n}\), compute \(\rho=\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\).

Hint

\(\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)\) and \(\frac{(n!)^2}{((n+1)!)^2}=\frac1{(n+1)^2}\).

Answer

\(\rho=\tfrac45\) (so the series converges).

Full solution

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(2n+2)(2n+1)}{(n+1)^2}\cdot\dfrac15=\dfrac{2(2n+1)}{5(n+1)}\).
As \(n\to\infty\) this tends to \(\dfrac{4}{5}<1\): the series converges.

Put the integral-test solution in order

puzzle

Here are the steps of a full-marks solution for “Does \(\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}\) converge?”, shuffled.

Use the arrows to put the steps in order, then check.

Let \(f(x)=\dfrac{1}{x(\ln x)^2}\) for \(x\ge2\). It is positive and continuous there.
\(f\) is decreasing on \([2,\infty)\): both \(x\) and \((\ln x)^2\) increase, so the denominator increases.
Substitute \(u=\ln x\): \(\displaystyle\int_2^b\frac{dx}{x(\ln x)^2}=\int_{\ln2}^{\ln b}\frac{du}{u^2}=\frac1{\ln2}-\frac1{\ln b}\).
Let \(b\to\infty\): the improper integral equals \(\frac1{\ln2}\), a finite number.
By the integral test the series converges. (Its sum is not \(\frac{1}{\ln 2}\).)

Hint

Hypotheses first, then the computation, then the conclusion.

Answer

Hypotheses (positive, continuous) → decreasing → substitution → limit → conclusion.

Full solution

The order mirrors the grading on the solution sheet for Quiz 2 (0.5 pt for the hypotheses, 0.5 pt for the integral, 0.5 pt for the conclusion).

Practice set 3B · Absolute, conditional or divergent?

9 problems like Quiz 2 (CRNs 10890, 10891, 10895, 10897)

Root in the denominator

new · like Quiz 2, CRN 10890

Classify \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n+1}}\).

Hint

\(\sum\frac1{\sqrt{n+1}}\) behaves like \(\sum\frac1{\sqrt n}\).

Answer

Converges conditionally.

Full solution

\(\sum|a_n|=\sum\frac{1}{\sqrt{n+1}}\): limit comparison with \(\frac1{\sqrt n}\) (\(p=\frac12\)) gives limit \(1\), so it diverges.
\(b_n=\frac1{\sqrt{n+1}}\) is positive, decreasing and tends to \(0\): the alternating series test gives convergence.
Conditionally convergent.

Cubic denominator

new · quiz style

Classify \(\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{n}{n^3+1}\).

Hint

\(|a_n|\le\frac{n}{n^3}=\frac1{n^2}\).

Answer

Converges absolutely.

Full solution

\(|a_n|=\dfrac{n}{n^3+1}\le\dfrac{n}{n^3}=\dfrac1{n^2}\), and \(\sum\frac1{n^2}\) converges.
So \(\sum|a_n|\) converges: the series converges absolutely.

Terms do not shrink

new · quiz style

Classify \(\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n}{2n+1}\).

Hint

Always start with the \(n\)th-term test.

Answer

Diverges.

Full solution

\(|a_n|=\frac{n}{2n+1}\to\frac12\ne0\), so \(a_n\not\to0\).
By the \(n\)th-term test the series diverges.

Factorial

new · like Quiz 2, CRN 10897

Classify \(\displaystyle\sum_{n=1}^\infty\frac{(-2)^n}{n!}\).

Hint

Ratio test on \(\frac{2^n}{n!}\).

Answer

Converges absolutely.

Full solution

\(\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{2}{n+1}\to0<1\).
By the ratio test \(\sum|a_n|\) converges: absolutely convergent.

ln n over n, alternating

new · lecture style

Classify \(\displaystyle\sum_{n=2}^\infty(-1)^{n+1}\frac{\ln n}{n}\).

Hint

For the alternating series test, show \(f(x)=\frac{\ln x}{x}\) decreases for \(x>e\).

Answer

Converges conditionally.

Full solution

\(\sum\frac{\ln n}{n}\) diverges (it is bigger than \(\sum\frac1n\) for \(n\ge3\)), so the convergence is not absolute.
\(b_n=\frac{\ln n}{n}>0\), \(b_n\to0\) (L'Hôpital), and \(f'(x)=\frac{1-\ln x}{x^2}<0\) for \(x>e\), so \(b_n\) decreases for \(n\ge3\).
The alternating series test gives convergence: conditionally convergent.

cos(nπ) in disguise

new · like the unlabelled Quiz 1

Classify \(\displaystyle\sum_{n=1}^\infty\frac{\cos(n\pi)}{n^{3/2}}\).

Hint

\(\cos(n\pi)=(-1)^n\).

Answer

Converges absolutely.

Full solution

\(\cos(n\pi)=(-1)^n\), so \(|a_n|=\frac1{n^{3/2}}\).
\(\sum\frac{1}{n^{3/2}}\) is a \(p\)-series with \(p=\frac32>1\): absolutely convergent.

Degree difference one

new · like Quiz 2, CRN 10890

Classify \(\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n+1}{n^2+2}\).

Hint

\(|a_n|\) behaves like \(\frac1n\).

Answer

Converges conditionally.

Full solution

Limit comparison of \(|a_n|=\frac{n+1}{n^2+2}\) with \(\frac1n\): \(\lim\frac{n^2+n}{n^2+2}=1\), so \(\sum|a_n|\) diverges.
\(b_n=\frac{n+1}{n^2+2}\to0\), and it decreases for all \(n\ge1\): for \(f(x)=\frac{x+1}{x^2+2}\), \(f'(x)=\frac{2-2x-x^2}{(x^2+2)^2}<0\) for \(x\ge1\).
By the alternating series test the series converges, so it is conditionally convergent.

Geometric-like with 1/n

new · quiz style

Classify \(\displaystyle\sum_{n=1}^\infty\frac{(-3)^n}{n\,4^n}\).

Hint

Ratio test, or compare \(|a_n|\) with \(\left(\frac34\right)^n\).

Answer

Converges absolutely.

Full solution

\(\dfrac{|a_{n+1}|}{|a_n|}=\dfrac34\cdot\dfrac{n}{n+1}\to\dfrac34<1\).
So \(\sum|a_n|\) converges: absolutely convergent.

How many terms?

new · Thomas §10.6 style

The alternating series \(S=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^3}\) is approximated by \(S_N\). Using the alternating series estimation theorem, what is the smallest \(N\) that guarantees \(|S-S_N|<0.001\)?

Hint

\(|S-S_N|\le b_{N+1}=\frac{1}{(N+1)^3}\).

Answer

\(N=10\).

Full solution

The error is at most the first omitted term: \(|S-S_N|\le\dfrac{1}{(N+1)^3}\).
Need \((N+1)^3>1000\), i.e. \(N+1>10\), so \(N\ge10\).
With \(N=10\): \(b_{11}=\frac1{1331}<0.001\). With \(N=9\): \(b_{10}=\frac1{1000}\) is not less than \(0.001\).

Chapter 4 · Thomas §10.7 · Week 4

Power series

In plain words

The idea

A power series is an “infinite polynomial” \(\sum c_n(x-a)^n\). For each number \(x\) it is an ordinary series, which may converge or diverge. The set of \(x\) where it converges is always an interval centered at \(a\). Its half-width is the radius of convergence \(R\), and the two endpoints have to be checked by hand. This “radius + endpoints” question appears on almost every Quiz 2.

Definition§10.7

Power series, center, radius

A power series about \(x=a\) is \(\sum_{n=0}^\infty c_n(x-a)^n = c_0+c_1(x-a)+c_2(x-a)^2+\cdots\). The number \(a\) is the center.

Theorem§10.7

The Convergence Theorem for Power Series

Exactly one of the following holds:

the series converges only at \(x=a\) (\(R=0\));
there is \(R>0\) such that it converges absolutely for \(|x-a|<R\) and diverges for \(|x-a|>R\). At \(x=a\pm R\) anything can happen;
it converges absolutely for every \(x\) (\(R=\infty\)).

Theorem§10.7

Term-by-term differentiation and integration

If \(f(x)=\sum c_n(x-a)^n\) for \(|x-a|<R\), then on the same open interval \[f'(x)=\sum_{n=1}^\infty n c_n(x-a)^{n-1},\qquad \int f(x)\,dx = C+\sum_{n=0}^\infty c_n\frac{(x-a)^{n+1}}{n+1}.\] The radius stays \(R\); the endpoints may behave differently.

Partial sums \(1+x+\dots+x^n\) of the geometric power series against \(\frac{1}{1-x}\). Inside the shaded interval \((-1,1)\) they close in as \(n\) grows; outside it they move away, however many terms you add.

Recipe

Radius and interval of convergence (4 steps, as on the solution sheets)

Ratio test. Compute \(\left|\dfrac{a_{n+1}}{a_n}\right|\) with the \(x\)'s kept inside, simplify, and let \(n\to\infty\). You get something like \(\dfrac{|x-a|}{b}\) (or \(0\), or \(\infty\)).
Solve \(L<1\). \(\dfrac{|x-a|}{b}<1\iff a-b<x<a+b\). Center \(a\), radius \(R=b\). If you get \(|3x-2|<1\), write it as \(\left|x-\frac23\right|<\frac13\): \(R=\frac13\), not \(1\).
Left endpoint. Substitute \(x=a-R\) into the original series, simplify \((\pm1)^n\), and identify the series: \(p\)-series, alternating series (AST), or \(n\)th-term test.
Right endpoint. Same with \(x=a+R\). Then write the interval with the correct brackets and say where the convergence is absolute and where it is conditional.

Watch out

Common mistakes

Forgetting the endpoints: the ratio test gives \(L=1\) there, so it says nothing.
Reading the radius off \((3x-2)^n\) as \(1\). Factor out the coefficient of \(x\): \(R=\frac13\).
Losing a \((-1)^n\): at \(x=a-R\) the factor \((x-a)^n=(-R)^n\) brings its own \((-1)^n\), which can cancel an existing one: \((-1)^n(-1)^n=1\).
At \(x=a\) every power series converges (all terms after the first are \(0\)).
A sum starting at \(n=0\) with \(\frac1n\) in the term makes no sense (see the errata).

Worked examples (from the real quizzes)

ExampleR = 2

Quiz 2, CRN 10893

\(\sum_{n=1}^\infty\frac{(-1)^nx^n}{n2^n}\).

\(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{|x|}{2}\cdot\dfrac{n}{n+1}\to\dfrac{|x|}2\). Converges absolutely for \(|x|<2\): \(R=2\).
\(x=-2\): \(\sum\frac{(-1)^n(-2)^n}{n2^n}=\sum\frac1n\), harmonic, diverges.
\(x=2\): \(\sum\frac{(-1)^n}{n}\), alternating harmonic, converges (conditionally).
Interval: \((-2,2]\).

ExampleR = 1/3

Quiz 2, CRN 10896, Variant 1

\(\sum\frac{(3x-2)^n}{n}\): the ratio test gives \(|3x-2|<1\iff\frac13<x<1\); since \(3x-2=3\left(x-\frac23\right)\), the center is \(\frac23\) and \(R=\frac13\). At \(x=\frac13\): \(\sum\frac{(-1)^n}{n}\) converges conditionally. At \(x=1\): \(\sum\frac1n\) diverges. Interval \(\left[\frac13,1\right)\).

Exampleneither endpoint

Quiz 2, CRN 10890, Variant A

\(\sum\frac{n(x+1)^n}{4^n}\): ratio \(\frac{n+1}{n}\cdot\frac{|x+1|}{4}\to\frac{|x+1|}{4}\), so \(R=4\) and \(-5<x<3\). At both endpoints the terms are \(n(\pm1)^n\), which do not tend to \(0\): both endpoints are excluded. Interval \((-5,3)\).

Quick checks · true or false?

Every power series converges at its center.

If the radius of convergence is \(R=2\), the series converges at \(x=a\pm2\).

\(\sum c_nx^n\) and \(\sum nc_nx^{n-1}\) have the same radius of convergence.

The radius of convergence of \(\sum (5x)^n\) is \(5\).

Practice set 4 · Radius and interval of convergence

10 problems in the style of Quiz 2

Harmonic at one end

new · like Quiz 2, CRN 10890

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{x^n}{n\,5^n}\).

Hint

Ratio test gives \(\frac{|x|}{5}\). At \(x=5\) you get \(\sum\frac1n\); at \(x=-5\), \(\sum\frac{(-1)^n}{n}\).

Answer

\(R=5\), interval \([-5,5)\).

Full solution

\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x|}{5}\cdot\dfrac{n}{n+1}\to\dfrac{|x|}{5}<1\iff|x|<5\). So \(R=5\).
\(x=-5\): \(\sum\frac{(-5)^n}{n5^n}=\sum\frac{(-1)^n}{n}\) converges by the alternating series test (conditionally).
\(x=5\): \(\sum\frac1n\) diverges.
Interval \([-5,5)\).

Both endpoints included

new · quiz style

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(x-3)^n}{n^2}\).

Hint

At both endpoints \(\sum\left|\frac{(\pm1)^n}{n^2}\right|=\sum\frac1{n^2}\) converges.

Answer

\(R=1\), interval \([2,4]\).

Full solution

Ratio: \(|x-3|\left(\frac{n}{n+1}\right)^2\to|x-3|<1\iff2<x<4\). \(R=1\).
\(x=4\): \(\sum\frac1{n^2}\) converges (\(p=2\)).
\(x=2\): \(\sum\frac{(-1)^n}{n^2}\) converges absolutely.
Interval \([2,4]\), absolute convergence everywhere on it.

Factorial: radius zero

new · Thomas style

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty n!\,(x+1)^n\).

Hint

\(\left|\frac{a_{n+1}}{a_n}\right|=(n+1)|x+1|\). What happens unless \(x=-1\)?

Answer

\(R=0\): it converges only at \(x=-1\). Type the set as {-1}.

Full solution

\(\left|\dfrac{a_{n+1}}{a_n}\right|=(n+1)|x+1|\to\infty\) whenever \(x\ne-1\): divergence.
At \(x=-1\) every term is \(0\): convergence.
So \(R=0\) and the “interval” is the single point \(\{-1\}\).

Factorial below: radius infinite

new · Thomas style

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\).

Hint

\(\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x|}{n+1}\).

Answer

\(R=\infty\), interval \((-\infty,\infty)\). (This is the series of \(e^x\).)

Full solution

\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x|}{n+1}\to0<1\) for every \(x\).
So the series converges absolutely for all real \(x\): \(R=\infty\).

Coefficient inside the power

new · like Quiz 2, CRN 10891

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(2x-1)^n}{\sqrt n}\).

Hint

\(|2x-1|<1\iff\left|x-\frac12\right|<\frac12\).

Answer

\(R=\frac12\), interval \([0,1)\).

Full solution

Ratio: \(|2x-1|\sqrt{\frac{n}{n+1}}\to|2x-1|<1\iff0<x<1\). Center \(\frac12\), \(R=\frac12\).
\(x=0\): \(\sum\frac{(-1)^n}{\sqrt n}\) converges (AST), but not absolutely (\(p=\frac12\)): conditional.
\(x=1\): \(\sum\frac1{\sqrt n}\) diverges (\(p=\frac12\le1\)).
Interval \([0,1)\).

Two signs that cancel

new · like Quiz 2, CRN 10896

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n(x+2)^n}{3^n(n+1)}\).

Hint

At \(x=-5\), \((x+2)^n=(-3)^n\) and the two \((-1)^n\) cancel.

Answer

\(R=3\), interval \((-5,1]\).

Full solution

Ratio: \(\dfrac{|x+2|}{3}\cdot\dfrac{n+1}{n+2}\to\dfrac{|x+2|}{3}<1\iff-5<x<1\). \(R=3\).
\(x=-5\): \(\dfrac{(-1)^n(-3)^n}{3^n(n+1)}=\dfrac{1}{n+1}\): harmonic-like, diverges.
\(x=1\): \(\dfrac{(-1)^n3^n}{3^n(n+1)}=\dfrac{(-1)^n}{n+1}\): converges by the AST (conditionally).
Interval \((-5,1]\).

Polynomial factor in front

new · like Quiz 2, CRN 10890

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{2^n}\).

Hint

At \(x=\pm2\) the terms are \(n^2(\pm1)^n\).

Answer

\(R=2\), interval \((-2,2)\).

Full solution

Ratio: \(\left(\frac{n+1}{n}\right)^2\frac{|x|}{2}\to\frac{|x|}{2}<1\iff|x|<2\).
At \(x=\pm2\): terms \(n^2(\pm1)^n\) do not tend to \(0\), so both endpoints diverge.
Interval \((-2,2)\).

Only even powers

new · Thomas style

Find the radius and the interval of convergence of \(\displaystyle\sum_{n=0}^\infty\frac{(x-1)^{2n}}{4^n}\).

Hint

It is geometric with ratio \(\frac{(x-1)^2}{4}\).

Answer

\(R=2\), interval \((-1,3)\).

Full solution

The series is geometric with \(r=\dfrac{(x-1)^2}{4}\). It converges iff \(\dfrac{(x-1)^2}{4}<1\iff|x-1|<2\).
\(R=2\), center \(1\). At \(x=-1\) and \(x=3\), \(r=1\): every term is \(1\), so both endpoints diverge.
Interval \((-1,3)\).

Differentiate a known series

new · Thomas §10.7 style

Differentiating \(\dfrac1{1-x}=\sum_{n=0}^\infty x^n\) gives \(\dfrac{1}{(1-x)^2}=\sum_{n=0}^\infty c_nx^n\). Find a formula for \(c_n\) in terms of \(n\).

Hint

\(\frac{d}{dx}\sum x^n=\sum_{n\ge1}nx^{n-1}\). Re-index so the power is \(x^n\).

Answer

\(c_n=n+1\).

Full solution

\(\dfrac{1}{(1-x)^2}=\dfrac{d}{dx}\dfrac1{1-x}=\sum_{n=1}^\infty nx^{n-1}=1+2x+3x^2+\cdots\)
Re-index with \(m=n-1\): \(\sum_{m=0}^\infty(m+1)x^m\). So \(c_n=n+1\), valid for \(|x|<1\).

Sum a series with a power series

new · Thomas style

Use \(\sum_{n=1}^\infty nx^{n}=\dfrac{x}{(1-x)^2}\) (for \(|x|<1\)) to find \(\displaystyle\sum_{n=1}^\infty\frac{n}{2^n}\).

Hint

Plug in \(x=\frac12\).

Answer

\(2\)

Full solution

\(\sum\frac{n}{2^n}=\sum n\left(\frac12\right)^n\): this is the formula at \(x=\frac12\), which is inside \((-1,1)\).
\(\dfrac{1/2}{(1-1/2)^2}=\dfrac{1/2}{1/4}=2\).

Chapter 5 · Thomas §10.8–10.10 · Week 4

Taylor and Maclaurin series

In plain words

The idea

Can a function be written as a power series? If it can, the coefficients are forced: \(c_n=\frac{f^{(n)}(a)}{n!}\). The resulting Taylor series agrees with \(f\) at \(a\) in value, slope, curvature and every higher derivative. Cutting it off after degree \(n\) gives the Taylor polynomial \(P_n\), the best degree-\(n\) polynomial approximation near \(a\). In practice you rarely differentiate ten times: you build new series from the few you know.

Definition§10.8

Taylor series, Maclaurin series, Taylor polynomial

If \(f\) has derivatives of all orders at \(a\), its Taylor series at \(a\) is \[\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots\] At \(a=0\) it is called the Maclaurin series. The Taylor polynomial of order \(n\), \(P_n(x)\), stops at the \((x-a)^n\) term.

Theorem§10.9

Taylor's Theorem and the remainder

\(f(x)=P_n(x)+R_n(x)\) with \(R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\) for some \(c\) between \(a\) and \(x\). If \(|f^{(n+1)}|\le M\) on that interval, then \(|R_n(x)|\le M\dfrac{|x-a|^{n+1}}{(n+1)!}\) (Remainder Estimation Theorem). The series equals \(f\) exactly where \(R_n(x)\to0\).

Theorem§10.10

The binomial series

For \(-1<x<1\) and any real \(m\): \((1+x)^m=1+\sum_{k=1}^\infty\binom mk x^k\), where \(\binom mk=\dfrac{m(m-1)\cdots(m-k+1)}{k!}\). For example \(\sqrt{1+x}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\cdots\)

The known series are collected in the Maclaurin table. Memorise the first five rows: \(\frac{1}{1-x}\), \(e^x\), \(\sin x\), \(\cos x\), \(\ln(1+x)\).

Taylor polynomials of \(\ln(1+x)\) about \(0\). They improve on \((-1,1]\) as the degree grows, but beyond \(x=1\) they get worse: the series only represents the function on its interval of convergence.

Recipe

Taylor series by derivatives (the way the Quiz 2 sheets do it)

Compute \(f(a), f'(a), f''(a), f'''(a)\) and look for the pattern of \(f^{(n)}(a)\).
Divide by \(n!\): \(c_n=\dfrac{f^{(n)}(a)}{n!}\). Simplify. Factorials often cancel, e.g. \(\frac{(n+1)!}{n!}=n+1\).
Write \(\sum c_n(x-a)^n\) and the first few terms, e.g. \(1-2(x-1)+3(x-1)^2-\cdots\)

Recipe

Taylor series from known series (usually faster)

Substitute: \(e^{-x^2}=\sum\frac{(-x^2)^n}{n!}\), \(\frac{1}{1+x^2}=\sum(-1)^nx^{2n}\).
Multiply by a power: \(\frac{x^2}{1+x}=x^2\sum(-1)^nx^n=\sum(-1)^nx^{n+2}\).
Rewrite first: \(\frac{2+x}{1-x}=\frac{3}{1-x}-1=2+3x+3x^2+\cdots\)
Differentiate or integrate: \(\frac1{(1-x)^2}=\frac{d}{dx}\frac1{1-x}\), \(\arctan x=\int\frac{dx}{1+x^2}\).
Multiply two series and collect powers: \((1-3x+2x^2)e^x=1-2x-\frac12x^2+\cdots\)
Interval: after substituting \(u=\ldots\), solve the old condition (e.g. \(|u|<1\)) for \(x\). For a sum of two series, take the intersection of the intervals.

Watch out

Common mistakes

Forgetting \(n!\): the coefficient is \(\frac{f^{(n)}(a)}{n!}\), not \(f^{(n)}(a)\).
About \(a\ne0\) the powers are \((x-a)^n\), not \(x^n\).
“First three nonzero terms” means skipping zero coefficients: for \(\sin x\) they are \(x-\frac{x^3}{6}+\frac{x^5}{120}\).
\(e^x=\sum\frac{x^n}{n!}\), not \(\sum\frac{x^n}{n}\) (that is \(-\ln(1-x)\)).
A Taylor series need not equal the function outside its interval of convergence.

Worked examples (from Quiz 2, CRNs 10895–10897)

ExampleMaclaurin

Rewrite first (CRN 10895, Variant 1)

\(f(x)=\frac{2+x}{1-x}=3(1-x)^{-1}-1\). Then \(f^{(n)}(x)=3\,n!\,(1-x)^{-n-1}\), so \(f^{(n)}(0)=3n!\) for \(n\ge1\) and \(f(0)=2\). Series: \(2+3x+3x^2+3x^3+\cdots=2+3\sum_{n=1}^\infty x^n\).

ExampleTaylor

Center 1 (CRN 10896, Variant 1)

\(f(x)=x^{-2}\): \(f^{(n)}(x)=(-1)^n(n+1)!\,x^{-n-2}\), so \(f^{(n)}(1)=(-1)^n(n+1)!\) and \(c_n=\frac{(-1)^n(n+1)!}{n!}=(-1)^n(n+1)\). \[\frac1{x^2}=\sum_{n=0}^\infty(-1)^n(n+1)(x-1)^n=1-2(x-1)+3(x-1)^2-4(x-1)^3+\cdots\]

Examplecombine

First three nonzero terms (CRN 10897, Variants 1 and 2)

\(\cos x-\frac{3}{1-x}\): \(f(0)=-2\), \(f'(0)=-3\), \(f''(0)=-7\), so \(-2-3x-\frac72x^2+\cdots\), converging absolutely exactly for \(-1<x<1\) (the cosine part converges everywhere).

\((1-3x+2x^2)e^x\): \(f(0)=1\), \(f'(0)=-2\), \(f''(0)=-1\), so \(1-2x-\frac12x^2+\cdots\), for all real \(x\).

Quick checks · true or false?

The Taylor series of a polynomial is the polynomial itself.

The Maclaurin series of \(\cos x\) contains only odd powers of \(x\).

\(e^{-x^2}=\sum_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{n!}\) for every \(x\).

The coefficient of \(x^3\) in the Maclaurin series of \(f\) is \(f'''(0)\).

Practice set 5 · Taylor and Maclaurin series

11 problems in the style of Quiz 2 (CRNs 10895–10897)

Substitute into e^x

new · quiz style

Find the Taylor polynomial of degree 3 of \(f(x)=e^{2x}\) about \(x=0\).

Hint

Replace \(x\) by \(2x\) in \(1+x+\frac{x^2}2+\frac{x^3}6\).

Answer

\(1+2x+2x^2+\frac43x^3\)

Full solution

\(e^{u}=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\cdots\) with \(u=2x\).
\(1+2x+\frac{4x^2}{2}+\frac{8x^3}{6}=1+2x+2x^2+\frac43x^3\).

Multiply a known series by x

new · like Quiz 2, CRN 10897

Find the first three nonzero terms of the Maclaurin series of \(f(x)=x\cos 2x\). (Type them as a polynomial.)

Hint

\(\cos u=1-\frac{u^2}{2}+\frac{u^4}{24}-\cdots\) with \(u=2x\), then multiply by \(x\).

Answer

\(x-2x^3+\frac23x^5\)

Full solution

\(\cos2x=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{24}-\cdots=1-2x^2+\frac23x^4-\cdots\)
Multiply by \(x\): \(x-2x^3+\frac23x^5-\cdots\)

Geometric series in disguise

new · like Quiz 2, CRN 10895

Find the first three nonzero terms of the Maclaurin series of \(f(x)=\dfrac{x}{1-x^2}\).

Hint

\(\frac{1}{1-u}=1+u+u^2+\cdots\) with \(u=x^2\).

Answer

\(x+x^3+x^5\)

Full solution

\(\dfrac{1}{1-x^2}=1+x^2+x^4+\cdots\) for \(|x|<1\).
Multiply by \(x\): \(x+x^3+x^5+\cdots=\sum_{n=0}^\infty x^{2n+1}\).

Taylor polynomial of a root

new · like Quiz 2, CRN 10896

Find the Taylor polynomial of degree 3 of \(f(x)=\sqrt x\) about \(x=4\).

Hint

\(f'(x)=\frac12x^{-1/2}\), \(f''(x)=-\frac14x^{-3/2}\), \(f'''(x)=\frac38x^{-5/2}\). Evaluate at \(4\) and divide by \(k!\).

Answer

\(2+\frac{x-4}{4}-\frac{(x-4)^2}{64}+\frac{(x-4)^3}{512}\)

Full solution

\(f(4)=2\), \(f'(4)=\frac14\), \(f''(4)=-\frac14\cdot\frac18=-\frac1{32}\), \(f'''(4)=\frac38\cdot\frac1{32}=\frac{3}{256}\).
Divide by \(0!,1!,2!,3!\): \(2,\ \frac14,\ -\frac1{64},\ \frac{3}{256\cdot6}=\frac1{512}\).
\(P_3(x)=2+\frac14(x-4)-\frac1{64}(x-4)^2+\frac1{512}(x-4)^3\).

ln x about 1

new · Thomas style

Find the Taylor polynomial of degree 3 of \(f(x)=\ln x\) about \(x=1\).

Hint

\(f^{(k)}(x)=(-1)^{k-1}(k-1)!\,x^{-k}\) for \(k\ge1\).

Answer

\((x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}\)

Full solution

\(f(1)=0\), \(f'(1)=1\), \(f''(1)=-1\), \(f'''(1)=2\).
Coefficients \(\frac{f^{(k)}(1)}{k!}\): \(0,\ 1,\ -\frac12,\ \frac13\).
\(P_3=(x-1)-\frac12(x-1)^2+\frac13(x-1)^3\).

One coefficient

new · quiz style

What is the coefficient of \(x^5\) in the Maclaurin series of \(\sin 3x\)?

Hint

The \(u^5\) term of \(\sin u\) is \(\frac{u^5}{5!}\).

Answer

\(\frac{3^5}{5!}=\frac{243}{120}=\frac{81}{40}\)

Full solution

\(\sin u=u-\frac{u^3}{3!}+\frac{u^5}{5!}-\cdots\) with \(u=3x\).
The \(x^5\) term is \(\frac{(3x)^5}{120}=\frac{243}{120}x^5=\frac{81}{40}x^5\).

A limit with series

new · Thomas §10.10 style

Use the Maclaurin series of \(e^x\) to find \(\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}\).

Hint

\(e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots\)

Answer

\(\tfrac12\)

Full solution

\(e^x-1-x=\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\)
Divide by \(x^2\): \(\frac12+\frac{x}{6}+\cdots\to\frac12\).

Binomial series

new · Thomas §10.10 style

Find the first three terms of the binomial series of \((1+x)^{1/3}\).

Hint

\(1+mx+\frac{m(m-1)}{2}x^2\) with \(m=\frac13\).

Answer

\(1+\frac x3-\frac{x^2}9\)

Full solution

\(m=\frac13\): \(\binom m1=\frac13\), \(\binom m2=\frac{\frac13\left(-\frac23\right)}{2}=-\frac19\).
\((1+x)^{1/3}=1+\frac13x-\frac19x^2+\cdots\)

A high derivative for free

new · Thomas style

Let \(f(x)=x^2e^x\). Find \(f^{(10)}(0)\).

Hint

Find the coefficient of \(x^{10}\) in \(x^2\sum\frac{x^n}{n!}\), then use \(c_{10}=\frac{f^{(10)}(0)}{10!}\).

Answer

\(90\)

Full solution

\(x^2e^x=\sum\frac{x^{n+2}}{n!}\). The \(x^{10}\) term comes from \(n=8\): coefficient \(\frac1{8!}\).
\(\frac{f^{(10)}(0)}{10!}=\frac1{8!}\), so \(f^{(10)}(0)=\frac{10!}{8!}=10\cdot9=90\).

Interval for a substituted series

new · like Quiz 2, CRN 10897

The Maclaurin series of \(\ln\!\left(1-\frac x2\right)\) is \(-\sum_{n=1}^\infty\frac{x^n}{n\,2^n}\). On which interval does it converge?

Hint

\(\ln(1+u)\) converges for \(-1<u\le1\); here \(u=-\frac x2\).

Answer

\([-2,2)\)

Full solution

\(\ln(1+u)=\sum\frac{(-1)^{n-1}u^n}{n}\) converges for \(-1<u\le1\).
With \(u=-\frac x2\): \(-1<-\frac x2\le1\iff-2\le x<2\).
Check: at \(x=2\) the series is \(-\sum\frac1n\) (diverges); at \(x=-2\) it is \(-\sum\frac{(-1)^n}{n}\) (converges).

General coefficient about 2

new · like Quiz 2, CRN 10896

The Taylor series of \(f(x)=\dfrac1x\) about \(a=2\) is \(\sum_{n=0}^\infty c_n(x-2)^n\). Find \(c_n\).

Hint

\(\frac1x=\frac{1}{2+(x-2)}=\frac12\cdot\frac{1}{1+\frac{x-2}{2}}\): a geometric series.

Answer

\(c_n=\dfrac{(-1)^n}{2^{n+1}}\)

Full solution

\(\dfrac1x=\dfrac12\cdot\dfrac{1}{1-\left(-\frac{x-2}{2}\right)}=\dfrac12\sum_{n=0}^\infty\left(-\dfrac{x-2}{2}\right)^n\).
So \(c_n=\dfrac{(-1)^n}{2^{n+1}}\), valid for \(|x-2|<2\).
Check with derivatives: \(f^{(n)}(2)=\frac{(-1)^nn!}{2^{n+1}}\), and dividing by \(n!\) gives the same \(c_n\).

Chapter 6 · Thomas §11.1–11.2 · Week 5

Parametric curves and their calculus

In plain words

The idea

Instead of \(y=f(x)\), describe a curve by where a moving point is at time \(t\): \(x=f(t)\), \(y=g(t)\). This handles circles, loops and spirals that are not graphs of functions, and it records the direction of motion. All of calculus still works: slopes come from the chain rule, and length comes from adding up tiny hypotenuses \(\sqrt{dx^2+dy^2}\).

Definition§11.1

Parametric curve

If \(x=f(t)\) and \(y=g(t)\) for \(t\) in an interval \(I\), the points \((x,y)\) form a parametric curve; \(t\) is the parameter and \(I\) the parameter interval. For \(I=[a,b]\), \((f(a),g(a))\) is the initial point and \((f(b),g(b))\) the terminal point.

Key facts§11.1

Standard parametrizations

Circle of radius \(a\) about \((h,k)\): \(x=h+a\cos t,\ y=k+a\sin t,\ 0\le t\le2\pi\) (counterclockwise).
Ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): \(x=a\cos t,\ y=b\sin t\).
Segment from \(P\) to \(Q\): \((x,y)=(1-t)P+tQ,\ 0\le t\le1\).
Any graph \(y=f(x)\): \(x=t,\ y=f(t)\).
Cycloid: \(x=a(t-\sin t),\ y=a(1-\cos t)\).

Theorem§11.2

Slope and concavity

If \(f,g\) are differentiable and \(\frac{dx}{dt}\ne0\): \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}.\] Horizontal tangent: \(\frac{dy}{dt}=0\), \(\frac{dx}{dt}\ne0\). Vertical tangent: \(\frac{dx}{dt}=0\), \(\frac{dy}{dt}\ne0\).

Theorem§11.2

Length, area and surface area

If the curve is traced exactly once for \(a\le t\le b\): \[L=\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt,\qquad A=\int_a^b y\,\frac{dx}{dt}\,dt\ \ (y\ge0),\qquad S=\int_a^b2\pi y\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt.\] The last one is the area of the surface swept out by rotating the curve about the \(x\)-axis.

The cycloid traced by a point on a rolling wheel. Slide \(t\): the orange arrow is the velocity \(\left(\frac{dx}{dt},\frac{dy}{dt}\right)\), always tangent to the curve. At \(t=2\pi\) it is zero: the cusp, where \(\frac{dy}{dx}\) is undefined.

Recipe

Tangent line at t = t₀

Point: \(x_0=f(t_0)\), \(y_0=g(t_0)\).
Slope: \(m=\dfrac{g'(t_0)}{f'(t_0)}\).
Line: \(y-y_0=m(x-x_0)\). If \(f'(t_0)=0\) and \(g'(t_0)\ne0\) it is the vertical line \(x=x_0\).

Recipe

Arc length

Differentiate: \(x'(t)\), \(y'(t)\).
Square, add and factor. Look for a perfect square or a common factor, e.g. \(36t^2+36t^4=36t^2(1+t^2)\).
Take the square root carefully (\(\sqrt{t^2}=t\) only if \(t\ge0\)) and integrate, usually by \(u\)-substitution.

Watch out

Common mistakes

\(\dfrac{d^2y}{dx^2}\ne\dfrac{d^2y/dt^2}{d^2x/dt^2}\). Differentiate \(\frac{dy}{dx}\) with respect to \(t\), then divide by \(\frac{dx}{dt}\).
When eliminating \(t\), keep the restrictions: \(x=t^2\) forces \(x\ge0\).
The arc-length formula counts every pass: if the parameter interval traces the circle twice, you get twice the circumference.

Worked examples

Example§11.1

Eliminating the parameter

\(x=2\cos t,\ y=3\sin t\): \(\left(\frac x2\right)^2+\left(\frac y3\right)^2=\cos^2t+\sin^2t=1\), the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\), traced counterclockwise from \((2,0)\).

Example§11.2

Slope and concavity

\(x=t-t^2,\ y=t-t^3\) at \(t=2\): \(\frac{dy}{dx}=\frac{1-3t^2}{1-2t}=\frac{-11}{-3}=\frac{11}{3}\). Point \((-2,-6)\). Tangent: \(y+6=\frac{11}{3}(x+2)\).

Example§11.2

Arc length with a perfect square

\(x=t^2,\ y=\frac23t^3\), \(0\le t\le\sqrt3\): \(\sqrt{(2t)^2+(2t^2)^2}=2t\sqrt{1+t^2}\), so \(L=\int_0^{\sqrt3}2t\sqrt{1+t^2}\,dt=\left[\tfrac23(1+t^2)^{3/2}\right]_0^{\sqrt3}=\tfrac23(8-1)=\tfrac{14}{3}\).

Quick checks · true or false?

\(\dfrac{d^2y}{dx^2}=\dfrac{d^2y/dt^2}{d^2x/dt^2}\).

The same curve can have many different parametrizations.

\(x=\cos2t,\ y=\sin2t\), \(0\le t\le\pi\) traces the unit circle exactly once.

A parametric curve has a vertical tangent where \(\frac{dy}{dt}=0\).

Practice set 6 · Parametric curves

9 problems (Thomas §11.1–11.2 style)

Eliminate the parameter

new · Thomas style

Find a Cartesian equation of the curve \(x=3+2\cos t,\ y=-1+2\sin t\).

Hint

Solve for \(\cos t\) and \(\sin t\), then use \(\cos^2t+\sin^2t=1\).

Answer

\((x-3)^2+(y+1)^2=4\): the circle with center \((3,-1)\) and radius \(2\).

Full solution

\(\cos t=\frac{x-3}{2}\), \(\sin t=\frac{y+1}{2}\).
\(\left(\frac{x-3}{2}\right)^2+\left(\frac{y+1}{2}\right)^2=1\iff(x-3)^2+(y+1)^2=4\).

Slope at a point

new · quiz style

For \(x=t^3+t,\ y=t^2-2t\), find \(\dfrac{dy}{dx}\) at \(t=2\).

Hint

\(\frac{dy}{dx}=\frac{2t-2}{3t^2+1}\).

Answer

\(\tfrac{2}{13}\)

Full solution

\(\frac{dx}{dt}=3t^2+1=13\), \(\frac{dy}{dt}=2t-2=2\) at \(t=2\).
\(\frac{dy}{dx}=\frac{2}{13}\).

Tangent line to an ellipse

new · Thomas style

Find the tangent line to \(x=2\cos t,\ y=3\sin t\) at \(t=\frac\pi4\).

Hint

Point \(\left(\sqrt2,\frac{3\sqrt2}{2}\right)\); slope \(\frac{3\cos t}{-2\sin t}\).

Answer

\(y-\frac{3\sqrt2}{2}=-\frac32\left(x-\sqrt2\right)\), i.e. \(y=-\frac32x+3\sqrt2\).

Full solution

Point: \(x=2\cdot\frac{\sqrt2}{2}=\sqrt2\), \(y=3\cdot\frac{\sqrt2}{2}\).
\(\frac{dy}{dx}=\frac{3\cos t}{-2\sin t}=-\frac32\) at \(t=\frac\pi4\).
\(y-\frac{3\sqrt2}{2}=-\frac32(x-\sqrt2)\).

Second derivative

new · Thomas style

For \(x=t^2,\ y=t^3\) (\(t>0\)), find \(\dfrac{d^2y}{dx^2}\) as a function of \(t\).

Hint

First \(\frac{dy}{dx}=\frac{3t}{2}\). Then differentiate that with respect to \(t\) and divide by \(\frac{dx}{dt}=2t\).

Answer

\(\dfrac{3}{4t}\)

Full solution

\(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\).
\(\frac{d}{dt}\left(\frac{3t}2\right)=\frac32\); divide by \(\frac{dx}{dt}=2t\): \(\frac{d^2y}{dx^2}=\frac{3}{4t}\).

Horizontal tangents

new · Thomas style

Find all points where \(x=t^2-3,\ y=t^3-3t\) has a horizontal tangent.

Hint

\(\frac{dy}{dt}=3t^2-3=0\). Check that \(\frac{dx}{dt}\ne0\) there.

Answer

\((-2,-2)\) at \(t=1\) and \((-2,2)\) at \(t=-1\).

Full solution

\(\frac{dy}{dt}=3t^2-3=0\iff t=\pm1\); \(\frac{dx}{dt}=2t=\pm2\ne0\).
\(t=1\): \((1-3,\ 1-3)=(-2,-2)\). \(t=-1\): \((-2,\ -1+3)=(-2,2)\).

Length of a spiral

new · Thomas style

Find the length of \(x=e^t\cos t,\ y=e^t\sin t\) for \(0\le t\le\ln2\).

Hint

\(x'^2+y'^2=2e^{2t}\).

Answer

\(\sqrt2\)

Full solution

\(x'=e^t(\cos t-\sin t)\), \(y'=e^t(\sin t+\cos t)\).
\(x'^2+y'^2=e^{2t}\left[(\cos t-\sin t)^2+(\sin t+\cos t)^2\right]=2e^{2t}\).
\(L=\int_0^{\ln2}\sqrt2e^t\,dt=\sqrt2(2-1)=\sqrt2\).

Length with a perfect square

new · Thomas style

Find the length of \(x=1+3t^2,\ y=4+2t^3\) for \(0\le t\le1\).

Hint

\(x'^2+y'^2=36t^2+36t^4=36t^2(1+t^2)\).

Answer

\(4\sqrt2-2\)

Full solution

\(\sqrt{x'^2+y'^2}=6t\sqrt{1+t^2}\) (for \(t\ge0\)).
\(\int_0^16t\sqrt{1+t^2}\,dt=\left[2(1+t^2)^{3/2}\right]_0^1=2(2\sqrt2-1)=4\sqrt2-2\).

Area under a cycloid arch

new · Thomas style

Find the area under one arch of the cycloid \(x=t-\sin t,\ y=1-\cos t\), \(0\le t\le2\pi\).

Hint

\(A=\int y\,\frac{dx}{dt}\,dt=\int_0^{2\pi}(1-\cos t)^2dt\). Use \(\cos^2t=\frac{1+\cos2t}{2}\).

Answer

\(3\pi\)

Full solution

\(A=\int_0^{2\pi}(1-\cos t)(1-\cos t)\,dt=\int_0^{2\pi}\left(1-2\cos t+\cos^2t\right)dt\).
\(=2\pi-0+\pi=3\pi\).

Surface of revolution

new · Thomas §11.2 style

The quarter circle \(x=\cos t,\ y=\sin t\), \(0\le t\le\frac\pi2\), is rotated about the \(x\)-axis. Find the area of the surface it sweeps out.

Hint

The speed is \(1\), so \(S=\int_0^{\pi/2}2\pi\sin t\,dt\).

Answer

\(2\pi\) (half the area of the unit sphere, as expected).

Full solution

\(\sqrt{x'^2+y'^2}=\sqrt{\sin^2t+\cos^2t}=1\).
\(S=\int_0^{\pi/2}2\pi\sin t\,dt=2\pi\left[-\cos t\right]_0^{\pi/2}=2\pi\).

Chapter 7 · Thomas §12.1–12.4 · Week 6

Space, vectors, dot and cross products

In plain words

The idea

A vector is an arrow, a length plus a direction, written by its components \(\langle v_1,v_2,v_3\rangle\). Two products answer two different questions. The dot product \(\mathbf u\cdot\mathbf v\) is a number that measures how much two vectors point the same way (angles, projections, work). The cross product \(\mathbf u\times\mathbf v\) is a vector perpendicular to both, and its length is the area they span (normals, areas, volumes). Quiz 3 in CRNs 10895–10897 was built entirely from these.

12.1 Points, distance, spheres

Definition§12.1

Distance and spheres

\(|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\). The sphere with center \((x_0,y_0,z_0)\) and radius \(a\) is \((x-x_0)^2+(y-y_0)^2+(z-z_0)^2=a^2\). Its interior is “\(<a^2\)”, its exterior “\(>a^2\)”, and the closed ball is “\(\le a^2\)” (boundary included).

Recipe

Center and radius: complete the square

If \(x^2,y^2,z^2\) have a common coefficient (e.g. \(2\)), divide the whole equation by it first.
Group: \((x^2+bx)+(y^2+cy)+(z^2+dz)=e\).
Add \(\left(\frac b2\right)^2,\left(\frac c2\right)^2,\left(\frac d2\right)^2\) to both sides.
Read off center \(\left(-\frac b2,-\frac c2,-\frac d2\right)\) and radius \(\sqrt{\text{right side}}\).

12.2 Vectors

Definition§12.2

Vectors and their algebra

The vector from \(P_1\) to \(P_2\) is \(\overrightarrow{P_1P_2}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\). Length \(|\mathbf v|=\sqrt{v_1^2+v_2^2+v_3^2}\); unit vector in the direction of \(\mathbf v\): \(\frac{\mathbf v}{|\mathbf v|}\). Standard basis \(\mathbf i=\langle1,0,0\rangle,\ \mathbf j=\langle0,1,0\rangle,\ \mathbf k=\langle0,0,1\rangle\), so \(\langle v_1,v_2,v_3\rangle=v_1\mathbf i+v_2\mathbf j+v_3\mathbf k\). Midpoint of \(P_1P_2\): \(\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2,\frac{z_1+z_2}2\right)\).

12.3 The dot product

Definition§12.3

Dot product and angle

\(\mathbf u\cdot\mathbf v=u_1v_1+u_2v_2+u_3v_3=|\mathbf u||\mathbf v|\cos\theta\), so \(\theta=\cos^{-1}\dfrac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}\in[0,\pi]\). \(\mathbf u\) and \(\mathbf v\) are orthogonal iff \(\mathbf u\cdot\mathbf v=0\).

Definition§12.3

Projection and work

\(\operatorname{proj}_{\mathbf v}\mathbf u=\left(\dfrac{\mathbf u\cdot\mathbf v}{|\mathbf v|^2}\right)\mathbf v\) (a vector); the scalar component of \(\mathbf u\) in the direction of \(\mathbf v\) is \(\dfrac{\mathbf u\cdot\mathbf v}{|\mathbf v|}\). A constant force \(\mathbf F\) moving an object along \(\mathbf D=\overrightarrow{PQ}\) does work \(W=\mathbf F\cdot\mathbf D\).

12.4 The cross product

Definition§12.4

Cross product

\[\mathbf u\times\mathbf v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}=(u_2v_3-u_3v_2)\mathbf i-(u_1v_3-u_3v_1)\mathbf j+(u_1v_2-u_2v_1)\mathbf k.\] It is perpendicular to \(\mathbf u\) and \(\mathbf v\) (right-hand rule), \(|\mathbf u\times\mathbf v|=|\mathbf u||\mathbf v|\sin\theta\), and \(\mathbf v\times\mathbf u=-\mathbf u\times\mathbf v\).

Theorem§12.4

Areas and volumes

Area of the parallelogram spanned by \(\mathbf u,\mathbf v\): \(|\mathbf u\times\mathbf v|\). Area of triangle \(PQR\): \(\frac12\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|\). Volume of the parallelepiped spanned by \(\mathbf u,\mathbf v,\mathbf w\): \(|\mathbf u\cdot(\mathbf v\times\mathbf w)|=\left|\det\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix}\right|\).

\(\mathbf u=\langle1,2,-1\rangle\) and \(\mathbf v=\langle3,0,2\rangle\) span the shaded parallelogram. \(\mathbf u\times\mathbf v=\langle4,-5,-6\rangle\) sticks out perpendicular to it, and its length \(\sqrt{77}\) is the area. Drag to rotate; switch on the projection.

Recipe

Cross product without mistakes

Write the \(3\times3\) determinant with \(\mathbf i,\mathbf j,\mathbf k\) on top, then \(\mathbf u\), then \(\mathbf v\).
\(\mathbf i\): cover its row and column and compute \(u_2v_3-u_3v_2\).
\(\mathbf j\): \(-(u_1v_3-u_3v_1)\). Don't forget the minus sign.
\(\mathbf k\): \(u_1v_2-u_2v_1\).
Check: the result dotted with \(\mathbf u\) and with \(\mathbf v\) must give \(0\).

Watch out

Common mistakes

\(\operatorname{proj}_{\mathbf v}\mathbf u\) divides by \(|\mathbf v|^2\), not by \(|\mathbf v|\) (and not by \(|\mathbf u|^2\)).
Triangle area is half of \(|\overrightarrow{PQ}\times\overrightarrow{PR}|\).
Volume uses the absolute value of the triple product.
“Interior of a sphere” is strict (\(<\)); a “closed region” includes its boundary (\(\le\)).
The angle at vertex \(A\) uses the vectors starting at \(A\): \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
A unit normal has two answers, \(\pm\frac{\mathbf n}{|\mathbf n|}\). Both are correct.

Worked examples (Quiz 3, CRNs 10895–10897)

Example§12.3

Angles of a triangle (CRN 10895, Variant 1)

\(A=(-1,0),\ B=(2,1),\ C=(1,-2)\). At \(A\): \(\overrightarrow{AB}=\langle3,1\rangle\), \(\overrightarrow{AC}=\langle2,-2\rangle\), \(\cos A=\frac{6-2}{\sqrt{10}\sqrt8}=\frac1{\sqrt5}\). At \(B\): \(\overrightarrow{BA}=\langle-3,-1\rangle\), \(\overrightarrow{BC}=\langle-1,-3\rangle\), \(\cos B=\frac{6}{10}=\frac35\). At \(C\): \(\cos C=\frac1{\sqrt5}\). So \(A=C=\cos^{-1}\frac1{\sqrt5}\approx63.435^\circ\), \(B=\cos^{-1}\frac35\approx53.130^\circ\); the sum is \(180^\circ\).

Example§12.1

Center and radius (CRN 10896, Variant 1)

\(2x^2+2y^2+2z^2+x+y+z=9\). Divide by \(2\): \(x^2+\frac12x+y^2+\frac12y+z^2+\frac12z=\frac92\). Add \(3\cdot\frac1{16}\): \(\left(x+\frac14\right)^2+\left(y+\frac14\right)^2+\left(z+\frac14\right)^2=\frac{75}{16}\). Center \(\left(-\frac14,-\frac14,-\frac14\right)\), radius \(\frac{5\sqrt3}{4}\).

Example§12.4

Triangle area and unit normal (CRN 10897, Variant 1)

\(P(1,-1,2),\ Q(2,0,-1),\ R(0,2,1)\): \(\overrightarrow{PQ}\times\overrightarrow{PR}=\langle1,1,-3\rangle\times\langle-1,3,-1\rangle=\langle8,4,4\rangle\), length \(\sqrt{96}=4\sqrt6\). Area \(=2\sqrt6\); unit normal \(\pm\frac{1}{\sqrt6}\langle2,1,1\rangle\).

Example§12.3–12.4

Volume and projection (CRNs 10896, 10897)

\(\mathbf u=\mathbf i-\mathbf j+\mathbf k,\ \mathbf v=2\mathbf i+\mathbf j-2\mathbf k,\ \mathbf w=-\mathbf i+2\mathbf j-\mathbf k\): \(\mathbf v\times\mathbf w=\langle3,4,5\rangle\), \(\mathbf u\cdot(\mathbf v\times\mathbf w)=3-4+5=4\), so \(V=4\). And \(\operatorname{proj}_{\mathbf v}\mathbf u\) for \(\mathbf v=\langle10,11,-2\rangle,\ \mathbf u=\langle0,3,4\rangle\): \(\frac{25}{225}\mathbf v=\frac{10}9\mathbf i+\frac{11}9\mathbf j-\frac29\mathbf k\).

Quick checks · true or false?

\(\mathbf u\times\mathbf v=\mathbf v\times\mathbf u\).

For nonzero \(\mathbf u,\mathbf v\): \(\mathbf u\cdot\mathbf v=0\) exactly when \(\mathbf u\perp\mathbf v\).

\(|\mathbf u\times\mathbf v|=|\mathbf u||\mathbf v|\cos\theta\).

\((x-1)^2+(y-1)^2+(z-1)^2\le1\) describes the interior of the sphere of radius 1 about \((1,1,1)\).

\(\mathbf u\cdot(\mathbf u\times\mathbf v)=0\) for all \(\mathbf u,\mathbf v\).

Practice set 7 · Vectors, dot and cross products

13 problems in the style of Quiz 3 (CRNs 10895–10897)

Distance

new · Thomas style

Find the distance between \(P(1,-2,3)\) and \(Q(4,2,-9)\).

Hint

\(\sqrt{3^2+4^2+12^2}\).

Answer

\(13\)

Full solution

\(\overrightarrow{PQ}=\langle3,4,-12\rangle\), so \(|PQ|=\sqrt{9+16+144}=\sqrt{169}=13\).

Center and radius

new · like Quiz 3, CRN 10896

Find the center and radius of the sphere \(x^2+y^2+z^2-4x+6y-2z=2\).

Hint

Add \(4\), \(9\) and \(1\) to both sides.

Answer

Center \((2,-3,1)\), radius \(4\).

Full solution

\((x^2-4x+4)+(y^2+6y+9)+(z^2-2z+1)=2+4+9+1\).
\((x-2)^2+(y+3)^2+(z-1)^2=16\): center \((2,-3,1)\), radius \(4\).

Describe a region

new · like Quiz 3, CRN 10895

Write an inequality describing the closed ball of radius 3 centered at \((0,2,-1)\).

Hint

Closed means the sphere itself is included: use \(\le\).

Answer

\(x^2+(y-2)^2+(z+1)^2\le9\)

Full solution

Squared distance from \((x,y,z)\) to the center: \(x^2+(y-2)^2+(z+1)^2\).
Closed ball: distance at most \(3\), so \(x^2+(y-2)^2+(z+1)^2\le9\).

Unit vector

new · Thomas style

Find the unit vector in the direction of \(\mathbf v=2\mathbf i-\mathbf j+2\mathbf k\).

Hint

\(|\mathbf v|=3\).

Answer

\(\left\langle\frac23,-\frac13,\frac23\right\rangle\)

Full solution

\(|\mathbf v|=\sqrt{4+1+4}=3\), so \(\mathbf u=\frac13\langle2,-1,2\rangle\).

Angle between vectors

new · like Quiz 3, CRN 10895

Find the angle between \(\mathbf u=\langle1,1,0\rangle\) and \(\mathbf v=\langle0,1,1\rangle\) (in radians, or in degrees with deg).

Hint

\(\cos\theta=\frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}=\frac12\).

Answer

\(\frac\pi3\) (that is \(60^\circ\))

Full solution

\(\mathbf u\cdot\mathbf v=1\), \(|\mathbf u|=|\mathbf v|=\sqrt2\).
\(\cos\theta=\frac{1}{2}\), so \(\theta=\frac\pi3\).

Angle in a triangle

new · like Quiz 3, CRN 10895

Triangle \(A(0,0),\ B(4,0),\ C(1,3)\). Find the angle at \(A\).

Hint

Use \(\overrightarrow{AB}=\langle4,0\rangle\) and \(\overrightarrow{AC}=\langle1,3\rangle\).

Answer

\(\cos^{-1}\frac{1}{\sqrt{10}}\approx71.57^\circ\)

Full solution

\(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\), \(|\overrightarrow{AB}|=4\), \(|\overrightarrow{AC}|=\sqrt{10}\).
\(\cos A=\frac{4}{4\sqrt{10}}=\frac1{\sqrt{10}}\), so \(A=\cos^{-1}\frac{1}{\sqrt{10}}\approx71.57^\circ\).

Vector projection

new · like Quiz 3, CRN 10897

Find \(\operatorname{proj}_{\mathbf v}\mathbf u\) for \(\mathbf u=\langle1,2,3\rangle\), \(\mathbf v=\langle1,1,0\rangle\).

Hint

\(\mathbf u\cdot\mathbf v=3\), \(|\mathbf v|^2=2\).

Answer

\(\left\langle\frac32,\frac32,0\right\rangle\)

Full solution

\(\operatorname{proj}_{\mathbf v}\mathbf u=\frac{\mathbf u\cdot\mathbf v}{|\mathbf v|^2}\mathbf v=\frac32\langle1,1,0\rangle=\left\langle\frac32,\frac32,0\right\rangle\).

Scalar component

new · Thomas style

Find the scalar component of \(\mathbf u=\langle3,4,0\rangle\) in the direction of \(\mathbf v=\langle1,2,2\rangle\).

Hint

\(|\mathbf v|=3\).

Answer

\(\frac{11}{3}\)

Full solution

\(\mathbf u\cdot\mathbf v=3+8+0=11\), \(|\mathbf v|=3\), so the scalar component is \(\frac{11}{3}\).

Cross product

new · like Quiz 3, CRN 10896

Compute \(\mathbf u\times\mathbf v\) for \(\mathbf u=\langle1,2,-1\rangle\), \(\mathbf v=\langle3,0,2\rangle\).

Hint

\(\mathbf i(2\cdot2-(-1)\cdot0)-\mathbf j(1\cdot2-(-1)\cdot3)+\mathbf k(1\cdot0-2\cdot3)\).

Answer

\(\langle4,-5,-6\rangle\)

Full solution

\(\mathbf i\): \(2\cdot2-(-1)\cdot0=4\). \(\mathbf j\): \(-(1\cdot2-(-1)\cdot3)=-5\). \(\mathbf k\): \(1\cdot0-2\cdot3=-6\).
Check: \(\langle4,-5,-6\rangle\cdot\langle1,2,-1\rangle=4-10+6=0\) ✓ and \(\cdot\langle3,0,2\rangle=12-12=0\) ✓.

Triangle area

new · like Quiz 3, CRN 10897

Find the area of the triangle with vertices \(A(1,0,0)\), \(B(0,2,0)\), \(C(0,0,3)\).

Hint

\(\overrightarrow{AB}\times\overrightarrow{AC}=\langle6,3,2\rangle\).

Answer

\(\frac72\)

Full solution

\(\overrightarrow{AB}=\langle-1,2,0\rangle\), \(\overrightarrow{AC}=\langle-1,0,3\rangle\), cross product \(\langle6,3,2\rangle\).
Its length is \(\sqrt{36+9+4}=7\); the area is half of that: \(\frac72\).

Volume of a parallelepiped

new · like Quiz 3, CRN 10896

Find the volume of the parallelepiped determined by \(\mathbf u=\langle1,1,0\rangle\), \(\mathbf v=\langle0,1,1\rangle\), \(\mathbf w=\langle1,0,1\rangle\).

Hint

\(\det\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\).

Answer

\(2\)

Full solution

\(\mathbf v\times\mathbf w=\langle1\cdot1-1\cdot0,\ -(0\cdot1-1\cdot1),\ 0\cdot0-1\cdot1\rangle=\langle1,1,-1\rangle\).
\(\mathbf u\cdot(\mathbf v\times\mathbf w)=1+1+0=2\), so \(V=|2|=2\).

Make them perpendicular

new · Thomas style

Find \(c\) so that \(\langle c,2,-1\rangle\) and \(\langle3,c,4\rangle\) are orthogonal.

Hint

Set the dot product equal to \(0\).

Answer

\(c=\frac45\)

Full solution

\(3c+2c-4=0\iff5c=4\iff c=\frac45\).

Work by a constant force

new · Thomas §12.3 style

A constant force \(\mathbf F=2\mathbf i+3\mathbf j+\mathbf k\) moves an object from \((0,0,0)\) to \((4,-1,2)\). Find the work done.

Hint

\(W=\mathbf F\cdot\overrightarrow{PQ}\).

Answer

\(7\)

Full solution

\(\overrightarrow{PQ}=\langle4,-1,2\rangle\), so \(W=8-3+2=7\).

Chapter 8 · Thomas §12.5 · Week 6

Lines and planes in space

In plain words

The idea

A line is a point plus a direction: start at \(P_0\) and walk along \(\mathbf v\). A plane is a point plus a normal: every vector lying in the plane is perpendicular to \(\mathbf n\). Almost every problem in this section comes down to finding the right direction or normal, usually with a cross product, and then plugging in a point.

Definition§12.5

Lines

The line through \(P_0(x_0,y_0,z_0)\) parallel to \(\mathbf v=\langle v_1,v_2,v_3\rangle\) is \(\mathbf r(t)=\mathbf r_0+t\mathbf v\), i.e. \[x=x_0+tv_1,\quad y=y_0+tv_2,\quad z=z_0+tv_3,\qquad -\infty<t<\infty.\] The segment from \(P\) to \(Q\) is \(\mathbf r(t)=(1-t)P+tQ,\ 0\le t\le1\).

Definition§12.5

Planes

The plane through \(P_0\) with normal \(\mathbf n=\langle A,B,C\rangle\): \(\mathbf n\cdot\overrightarrow{P_0P}=0\), i.e. \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\), or \(Ax+By+Cz=D\). The coefficients of \(x,y,z\) are the normal vector.

Theorem§12.5

Distances and angles

Point \(S\) to the line through \(P\) along \(\mathbf v\): \(d=\dfrac{|\overrightarrow{PS}\times\mathbf v|}{|\mathbf v|}\). Point \(S\) to the plane \(Ax+By+Cz=D\): \(d=\dfrac{|Ax_1+By_1+Cz_1-D|}{\sqrt{A^2+B^2+C^2}}\). The angle between two planes is the angle between their normals.

Key facts§12.5

Parallel, intersecting, skew

Two planes are parallel iff their normals are parallel; otherwise they meet in a line with direction \(\mathbf n_1\times\mathbf n_2\).
A line \(\mathbf r_0+t\mathbf v\) is parallel to a plane with normal \(\mathbf n\) iff \(\mathbf v\cdot\mathbf n=0\); it is perpendicular to the plane iff \(\mathbf v\parallel\mathbf n\).
Two lines in space are parallel (directions are multiples), intersecting, or skew (neither).

The plane \(3x+2y+6z=6\) with its normal \(\mathbf n=\langle3,2,6\rangle\) and the line \(x=\frac83+2t,\ y=-2t,\ z=1+t\), which pierces it at \(\left(\frac23,2,0\right)\). Drag to rotate.

Recipe

The five standard constructions

Line through \(P\) and \(Q\): direction \(\overrightarrow{PQ}\), point \(P\).
Plane through three points \(P,Q,R\): \(\mathbf n=\overrightarrow{PQ}\times\overrightarrow{PR}\), then \(\mathbf n\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0\). Check the other two points.
Where a line meets a plane: substitute \(x(t),y(t),z(t)\) into the plane equation, solve for \(t\), plug back in.
Line of intersection of two planes: direction \(\mathbf n_1\times\mathbf n_2\). For a point, set one variable (e.g. \(z=0\)) and solve the two equations.
Parallel or perpendicular objects: parallel planes share \(\mathbf n\); a line perpendicular to a plane has direction \(\mathbf n\).

Watch out

Common mistakes

Mixing up direction and normal: for \(2x-y+z=4\), \(\langle2,-1,1\rangle\) is perpendicular to the plane.
A line has infinitely many correct equations (any point on it, any nonzero multiple of the direction). The answer boxes accept all of them.
Distance formulas need the absolute value.
Not parallel does not mean intersecting: lines in space can be skew.

Worked examples (Thomas §12.5)

Example§12.5

Plane through three points

\(A(0,0,1),\ B(2,0,0),\ C(0,3,0)\): \(\overrightarrow{AB}\times\overrightarrow{AC}=\langle2,0,-1\rangle\times\langle0,3,-1\rangle=\langle3,2,6\rangle\). Through \(A\): \(3x+2y+6(z-1)=0\), i.e. \(3x+2y+6z=6\).

Example§12.5

Where a line meets a plane

\(x=\frac83+2t,\ y=-2t,\ z=1+t\) in \(3x+2y+6z=6\): \(8+6t-4t+6+6t=6\Rightarrow t=-1\). The point is \(\left(\frac23,2,0\right)\).

Example§12.5

Line of intersection

\(3x-6y-2z=15\) and \(2x+y-2z=5\): \(\mathbf n_1\times\mathbf n_2=\langle14,2,15\rangle\). With \(z=0\): \(3x-6y=15,\ 2x+y=5\Rightarrow x=3,\ y=-1\). Line: \(x=3+14t,\ y=-1+2t,\ z=15t\).

Example§12.5

Distance from a point to a plane

\(S(1,1,3)\) to \(3x+2y+6z=6\): \(d=\frac{|3+2+18-6|}{\sqrt{9+4+36}}=\frac{17}{7}\).

Quick checks · true or false?

Two lines in space that are not parallel must intersect.

Planes with normals \(\mathbf n_1\) and \(\mathbf n_2\) are perpendicular iff \(\mathbf n_1\cdot\mathbf n_2=0\).

The line \(\mathbf r(t)=\langle1,2,3\rangle+t\langle1,0,0\rangle\) is parallel to the plane \(x=5\).

\(2x-y+z=4\) and \(-4x+2y-2z=1\) are parallel planes.

Practice set 8 · Lines and planes

11 problems (likely Quiz 4 material)

Line from a point and a direction

new · Thomas style

Find parametric equations of the line through \(P(1,-2,4)\) parallel to \(\mathbf v=2\mathbf i-3\mathbf k\).

Hint

\(\mathbf r(t)=\mathbf r_0+t\mathbf v\). Note \(\mathbf v=\langle2,0,-3\rangle\).

Answer

\(x=1+2t,\ y=-2,\ z=4-3t\)

Full solution

\(\mathbf r(t)=\langle1,-2,4\rangle+t\langle2,0,-3\rangle\), i.e. \(x=1+2t,\ y=-2,\ z=4-3t\).

Line through two points

new · Thomas style

Find parametric equations of the line through \((-2,0,3)\) and \((3,5,-2)\).

Hint

Direction \(\overrightarrow{PQ}=\langle5,5,-5\rangle\) (or \(\langle1,1,-1\rangle\)).

Answer

\(x=-2+t,\ y=t,\ z=3-t\) (any equivalent form)

Full solution

\(\overrightarrow{PQ}=\langle5,5,-5\rangle\parallel\langle1,1,-1\rangle\). Through \((-2,0,3)\): \(x=-2+t,\ y=t,\ z=3-t\).

Parallel plane

new · Thomas style

Find an equation of the plane through \((1,-1,3)\) parallel to \(3x+y+z=7\).

Hint

Same normal \(\langle3,1,1\rangle\); find the new constant.

Answer

\(3x+y+z=5\)

Full solution

Normal \(\langle3,1,1\rangle\). \(D=3(1)+(-1)+3=5\). Plane \(3x+y+z=5\).

Plane through three points

new · quiz style

Find an equation of the plane through \(P(1,1,-1)\), \(Q(2,0,2)\), \(R(0,-2,1)\).

Hint

\(\mathbf n=\overrightarrow{PQ}\times\overrightarrow{PR}=\langle1,-1,3\rangle\times\langle-1,-3,2\rangle\).

Answer

\(7x-5y-4z=6\)

Full solution

\(\langle1,-1,3\rangle\times\langle-1,-3,2\rangle=\langle(-1)(2)-3(-3),\ -(1\cdot2-3(-1)),\ 1(-3)-(-1)(-1)\rangle=\langle7,-5,-4\rangle\).
Through \(P\): \(7(x-1)-5(y-1)-4(z+1)=0\iff7x-5y-4z=6\).
Check \(Q\): \(14-0-8=6\) ✓; \(R\): \(0+10-4=6\) ✓.

Line meets plane

new · Thomas style

Find the point where the line \(x=1-t,\ y=3t,\ z=1+t\) meets the plane \(2x-y+3z=6\).

Hint

Substitute: \(2(1-t)-3t+3(1+t)=6\).

Answer

\(\left(\frac32,-\frac32,\frac12\right)\) at \(t=-\frac12\)

Full solution

\(2-2t-3t+3+3t=6\iff5-2t=6\iff t=-\frac12\).
\(x=\frac32,\ y=-\frac32,\ z=\frac12\).

Distance to a plane

new · Thomas style

Find the distance from \((2,-3,4)\) to the plane \(x+2y+2z=13\).

Hint

\(\frac{|2-6+8-13|}{\sqrt{1+4+4}}\).

Answer

\(3\)

Full solution

\(d=\frac{|2+2(-3)+2(4)-13|}{3}=\frac{|-9|}{3}=3\).

Angle between planes

new · Thomas style

Find the angle between the planes \(x+y=1\) and \(2x+y-2z=2\).

Hint

Normals \(\langle1,1,0\rangle\) and \(\langle2,1,-2\rangle\).

Answer

\(\frac\pi4\) (\(45^\circ\))

Full solution

\(\cos\theta=\frac{2+1+0}{\sqrt2\cdot3}=\frac{1}{\sqrt2}\), so \(\theta=\frac\pi4\).

Line of intersection

new · Thomas style

Find parametric equations of the line of intersection of \(x+y+z=1\) and \(x+y=2\).

Hint

Direction \(\langle1,1,1\rangle\times\langle1,1,0\rangle\). A point: \(x+y=2\) forces \(z=-1\); take \(x=2,\ y=0\).

Answer

\(x=2-t,\ y=t,\ z=-1\)

Full solution

\(\mathbf n_1\times\mathbf n_2=\langle1\cdot0-1\cdot1,\ 1\cdot1-1\cdot0,\ 1\cdot1-1\cdot1\rangle=\langle-1,1,0\rangle\).
A common point: subtracting the equations gives \(z=-1\); with \(y=0\), \(x=2\).
\(x=2-t,\ y=t,\ z=-1\).

Distance to a line

new · Thomas style

Find the distance from \(S(1,1,5)\) to the line \(x=1+t,\ y=3-t,\ z=2t\).

Hint

\(P(1,3,0)\), \(\mathbf v=\langle1,-1,2\rangle\), \(\overrightarrow{PS}=\langle0,-2,5\rangle\).

Answer

\(\sqrt5\)

Full solution

\(\overrightarrow{PS}\times\mathbf v=\langle0,-2,5\rangle\times\langle1,-1,2\rangle=\langle1,5,2\rangle\), length \(\sqrt{30}\).
\(d=\frac{\sqrt{30}}{\sqrt6}=\sqrt5\).

Plane containing a line and a point

new · Thomas style

Find the plane that contains the line \(\mathbf r(t)=\langle1,0,2\rangle+t\langle1,1,1\rangle\) and the point \(Q(0,1,0)\).

Hint

Two vectors in the plane: the direction \(\langle1,1,1\rangle\) and \(\overrightarrow{PQ}\) from \(P(1,0,2)\).

Answer

\(3x-y-2z=-1\)

Full solution

\(\overrightarrow{PQ}=\langle-1,1,-2\rangle\); \(\mathbf n=\langle1,1,1\rangle\times\langle-1,1,-2\rangle=\langle-3,1,2\rangle\).
Through \(P\): \(-3(x-1)+y+2(z-2)=0\iff3x-y-2z=-1\).

Parallel, intersecting or skew?

new · Thomas style

Classify the lines \(L_1:\ \langle1,0,0\rangle+t\langle1,2,3\rangle\) and \(L_2:\ \langle0,1,2\rangle+s\langle2,4,6\rangle\).

Hint

Compare directions first, then check whether a point of \(L_2\) lies on \(L_1\).

Answer

Parallel (and distinct).

Full solution

\(\langle2,4,6\rangle=2\langle1,2,3\rangle\): the directions are parallel.
Is \((0,1,2)\) on \(L_1\)? \(1+t=0\Rightarrow t=-1\), but then \(y=-2\ne1\). No, so the lines are parallel and distinct.

Chapter 9 · Thomas §13.1–13.2 · Week 7

Curves in space and their motion

In plain words

The idea

A vector function \(\mathbf r(t)=\langle f(t),g(t),h(t)\rangle\) is the position of a moving particle. Everything is done one component at a time: differentiate each component to get the velocity, again for the acceleration, and integrate each component to go back. Its length \(|\mathbf v|\) is the speed. Chapter 11's parametric curves were the 2D case of the same idea.

Definition§13.1

Velocity, speed, acceleration, direction

If \(\mathbf r(t)=f(t)\mathbf i+g(t)\mathbf j+h(t)\mathbf k\) then \(\mathbf v(t)=\mathbf r'(t)=f'\mathbf i+g'\mathbf j+h'\mathbf k\) is the velocity, \(|\mathbf v|\) the speed, \(\mathbf a=\mathbf v'=\mathbf r''\) the acceleration, and \(\mathbf T=\frac{\mathbf v}{|\mathbf v|}\) the direction of motion (unit tangent). The tangent line at \(t_0\) is \(\mathbf r(t_0)+s\,\mathbf v(t_0)\).

Theorem§13.1

Differentiation rules

\(\frac{d}{dt}(\mathbf u\cdot\mathbf v)=\mathbf u'\cdot\mathbf v+\mathbf u\cdot\mathbf v'\), \(\frac{d}{dt}(\mathbf u\times\mathbf v)=\mathbf u'\times\mathbf v+\mathbf u\times\mathbf v'\) (keep the order!), \(\frac{d}{dt}\mathbf u(f(t))=f'(t)\mathbf u'(f(t))\). If \(|\mathbf r(t)|\) is constant, then \(\mathbf r\cdot\mathbf r'=0\): the position is perpendicular to the velocity.

Theorem§13.2

Integrals and initial value problems

\(\int\mathbf r(t)\,dt=\left(\int f\right)\mathbf i+\left(\int g\right)\mathbf j+\left(\int h\right)\mathbf k+\mathbf C\), with a constant vector \(\mathbf C\). Given \(\mathbf r'(t)\) and \(\mathbf r(t_0)\), integrate and use \(\mathbf r(t_0)\) to find \(\mathbf C\).

Theorem§13.2

Ideal projectile motion

Launched from the origin with speed \(v_0\) at angle \(\alpha\), with gravity \(g\): \[\mathbf r(t)=(v_0\cos\alpha)\,t\,\mathbf i+\left((v_0\sin\alpha)\,t-\tfrac12gt^2\right)\mathbf j.\] Maximum height \(\frac{(v_0\sin\alpha)^2}{2g}\), flight time \(\frac{2v_0\sin\alpha}{g}\), range \(\frac{v_0^2\sin2\alpha}{g}\).

The helix \(\mathbf r(t)=\langle\cos t,\sin t,\frac t3\rangle\): the velocity (orange) is tangent to the curve and the acceleration (purple) points straight at the axis. Slide \(t\) and drag to rotate.

Recipe

Motion problems, step by step

Differentiate each component: \(\mathbf v=\mathbf r'\), \(\mathbf a=\mathbf r''\). Then plug in \(t_0\).
Speed \(=\sqrt{v_1^2+v_2^2+v_3^2}\): a number, never a vector.
Tangent line at \(t_0\): point \(\mathbf r(t_0)\), direction \(\mathbf v(t_0)\).
Going backwards: integrate each component, add \(\mathbf C=\langle C_1,C_2,C_3\rangle\), and use the initial condition, once for \(\mathbf v\) and once for \(\mathbf r\).

Watch out

Common mistakes

One scalar \(+C\) is not enough: the constant of integration is a vector.
\(\frac{d}{dt}(\mathbf u\times\mathbf v)\) keeps the order of the factors; swapping them flips the sign.
Speed is \(|\mathbf r'(t)|\), not \(|\mathbf r(t)|\).
With degrees, \(\sin30^\circ=\frac12\); keep the calculator-free exact values handy.

Worked examples (Thomas §13.1–13.2)

Example§13.1

The helix

\(\mathbf r(t)=\langle\cos t,\sin t,t\rangle\): \(\mathbf v=\langle-\sin t,\cos t,1\rangle\), speed \(\sqrt{\sin^2t+\cos^2t+1}=\sqrt2\) (constant), \(\mathbf a=\langle-\cos t,-\sin t,0\rangle\). Since \(\mathbf v\cdot\mathbf a=0\), the acceleration is always perpendicular to the velocity, which is why the speed never changes.

Example§13.2

Initial value problem

\(\mathbf r''=-32\mathbf k\), \(\mathbf r'(0)=8\mathbf i+8\mathbf j\), \(\mathbf r(0)=100\mathbf k\). Integrate: \(\mathbf r'=-32t\mathbf k+\mathbf C_1\) with \(\mathbf C_1=8\mathbf i+8\mathbf j\). Again: \(\mathbf r=8t\mathbf i+8t\mathbf j-16t^2\mathbf k+\mathbf C_2\) with \(\mathbf C_2=100\mathbf k\). So \(\mathbf r(t)=8t\mathbf i+8t\mathbf j+(100-16t^2)\mathbf k\).

Example§13.2

A projectile

\(v_0=500\) m/s, \(\alpha=60^\circ\), \(g=9.8\): after \(10\) s, \(\mathbf r(10)=500\cdot\frac12\cdot10\,\mathbf i+\left(500\cdot\frac{\sqrt3}{2}\cdot10-4.9\cdot100\right)\mathbf j\approx2500\,\mathbf i+3840\,\mathbf j\) (metres).

Quick checks · true or false?

Speed is the magnitude of the velocity vector.

If \(|\mathbf r(t)|\) is constant, then \(\mathbf r(t)\perp\mathbf r'(t)\).

The antiderivative of a vector function needs only one scalar constant \(C\).

\(\frac{d}{dt}(\mathbf u\times\mathbf v)=\mathbf u'\times\mathbf v'\).

Practice set 9 · Vector functions and motion

11 problems (week 7 quiz material)

Velocity

new · quiz style

Find the velocity of \(\mathbf r(t)=\langle t^2,e^t,\sin t\rangle\) at \(t=0\).

Hint

\(\mathbf v(t)=\langle2t,e^t,\cos t\rangle\).

Answer

\(\langle0,1,1\rangle\)

Full solution

\(\mathbf v(t)=\langle2t,e^t,\cos t\rangle\), so \(\mathbf v(0)=\langle0,1,1\rangle\).

Constant speed

new · quiz style

Find the speed of \(\mathbf r(t)=\langle3\cos t,3\sin t,4t\rangle\).

Hint

\(|\mathbf v|^2=9\sin^2t+9\cos^2t+16\).

Answer

\(5\) (for every \(t\))

Full solution

\(\mathbf v=\langle-3\sin t,3\cos t,4\rangle\), \(|\mathbf v|=\sqrt{9+16}=5\).

Acceleration

new · quiz style

Find the acceleration of \(\mathbf r(t)=\langle t^3,t^2,t\rangle\) at \(t=1\).

Hint

\(\mathbf a=\mathbf r''=\langle6t,2,0\rangle\).

Answer

\(\langle6,2,0\rangle\)

Full solution

\(\mathbf r'=\langle3t^2,2t,1\rangle\), \(\mathbf r''=\langle6t,2,0\rangle\); at \(t=1\): \(\langle6,2,0\rangle\).

Tangent line

new · Thomas style

Find parametric equations of the tangent line to \(\mathbf r(t)=\langle t,t^2,t^3\rangle\) at \(t=1\).

Hint

Point \(\mathbf r(1)\), direction \(\mathbf r'(1)\).

Answer

\(x=1+s,\ y=1+2s,\ z=1+3s\)

Full solution

\(\mathbf r(1)=\langle1,1,1\rangle\), \(\mathbf r'(t)=\langle1,2t,3t^2\rangle\Rightarrow\mathbf r'(1)=\langle1,2,3\rangle\).
Line: \(\langle1,1,1\rangle+s\langle1,2,3\rangle\).

Definite integral of a vector function

new · Thomas style

Evaluate \(\displaystyle\int_0^1\langle t,\ e^{2t},\ 3t^2\rangle\,dt\).

Hint

Integrate each component separately.

Answer

\(\left\langle\frac12,\frac{e^2-1}{2},1\right\rangle\)

Full solution

\(\int_0^1t\,dt=\frac12\), \(\int_0^1e^{2t}dt=\frac{e^2-1}{2}\), \(\int_0^13t^2dt=1\).

Initial value problem

new · Thomas style

Solve \(\dfrac{d\mathbf r}{dt}=\langle2t,\ 3,\ \cos t\rangle\), \(\mathbf r(0)=\langle1,0,2\rangle\). Give \(\mathbf r(t)\).

Hint

\(\mathbf r=\langle t^2,3t,\sin t\rangle+\mathbf C\), then use \(\mathbf r(0)\).

Answer

\(\mathbf r(t)=\langle t^2+1,\ 3t,\ \sin t+2\rangle\)

Full solution

Integrate: \(\mathbf r=\langle t^2,3t,\sin t\rangle+\mathbf C\).
\(\mathbf r(0)=\mathbf C=\langle1,0,2\rangle\).
\(\mathbf r(t)=\langle t^2+1,3t,\sin t+2\rangle\).

Projectile range

new · Thomas §13.2 style

A ball is launched from the ground at \(20\) m/s at an angle of \(30^\circ\). Take \(g=10\) m/s². How far away does it land? (Exact form.)

Hint

\(y(t)=10t-5t^2=0\) at \(t=2\); then \(x(2)=20\cos30^\circ\cdot2\).

Answer

\(20\sqrt3\approx34.6\) m

Full solution

\(x(t)=20\cos30^\circ\,t=10\sqrt3\,t\), \(y(t)=20\sin30^\circ\,t-5t^2=10t-5t^2\).
\(y=0\) again at \(t=2\). Then \(x(2)=20\sqrt3\). (Same as \(\frac{v_0^2\sin2\alpha}{g}=\frac{400\cdot\frac{\sqrt3}{2}}{10}\).)

Maximum height

new · Thomas §13.2 style

For the same ball (\(20\) m/s at \(30^\circ\), \(g=10\)), find the maximum height.

Hint

\(y'(t)=10-10t=0\) at \(t=1\).

Answer

\(5\) m

Full solution

\(y'(t)=10-10t=0\Rightarrow t=1\); \(y(1)=10-5=5\). (Check: \(\frac{(v_0\sin\alpha)^2}{2g}=\frac{100}{20}=5\).)

Unit tangent vector

new · Thomas style

Find the unit tangent vector \(\mathbf T(t)\) of the helix \(\mathbf r(t)=\langle\cos t,\sin t,t\rangle\).

Hint

\(\mathbf T=\frac{\mathbf v}{|\mathbf v|}\) and \(|\mathbf v|=\sqrt2\).

Answer

\(\mathbf T(t)=\frac{1}{\sqrt2}\langle-\sin t,\cos t,1\rangle\)

Full solution

\(\mathbf v=\langle-\sin t,\cos t,1\rangle\), \(|\mathbf v|=\sqrt2\), so \(\mathbf T=\frac{1}{\sqrt2}\langle-\sin t,\cos t,1\rangle\).

Length of a helix

new · Thomas style

Find the length of one full turn of the helix \(\mathbf r(t)=\langle\cos t,\sin t,t\rangle\), \(0\le t\le2\pi\).

Hint

\(L=\int_0^{2\pi}|\mathbf v|\,dt\) with \(|\mathbf v|=\sqrt2\).

Answer

\(2\sqrt2\,\pi\)

Full solution

\(L=\int_0^{2\pi}\sqrt2\,dt=2\sqrt2\,\pi\).

Velocity perpendicular to acceleration

new · Thomas style

For \(\mathbf r(t)=\langle t^3-3t,\ t^2,\ 0\rangle\), find the positive time \(t\) at which the velocity is perpendicular to the acceleration.

Hint

\(\mathbf v=\langle3t^2-3,2t,0\rangle\), \(\mathbf a=\langle6t,2,0\rangle\). Solve \(\mathbf v\cdot\mathbf a=0\).

Answer

\(t=\frac{\sqrt7}{3}\)

Full solution

\(\mathbf v\cdot\mathbf a=(3t^2-3)6t+4t=18t^3-14t=2t(9t^2-7)\).
Zero at \(t=0\) or \(t^2=\frac79\); the positive solution is \(t=\frac{\sqrt7}{3}\).

Chapter 10 · Thomas §14.1–14.2 · Week 8

Functions of several variables, limits and continuity

In plain words

The idea

A function \(f(x,y)\) assigns a height to every point of a region in the plane. You can picture it in two ways: as a surface \(z=f(x,y)\) in space, or as a contour map of level curves \(f(x,y)=c\), like a hiking map. Limits are harder than in one variable because \((x,y)\) can approach a point from infinitely many directions. One bad path is enough to show that a limit does not exist.

Definition§14.1

Domain, range, level curves

The domain of \(f(x,y)\) is the set of points where the formula makes sense (no division by zero, no negative numbers under even roots, positive arguments of \(\ln\)). The set \(f(x,y)=c\) is a level curve; for \(f(x,y,z)\) the sets \(f=c\) are level surfaces. A region is closed if it contains all its boundary points and bounded if it fits inside a disk.

Definition§14.2

Limit

\(\displaystyle\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L\) if for every \(\varepsilon>0\) there is \(\delta>0\) such that \(|f(x,y)-L|<\varepsilon\) whenever \(0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta\) and \((x,y)\) is in the domain. The value must be the same along every way of approaching.

Theorem§14.2

Limit laws, continuity

Sums, products, quotients (nonzero denominator) and compositions of continuous functions are continuous. Polynomials and rational functions are continuous on their domains, so their limits are found by substitution. \(f\) is continuous at \((x_0,y_0)\) if \(f(x_0,y_0)\) is defined and equals the limit.

Theorem§14.2

Two-path test for nonexistence

If \(f(x,y)\) has different limits along two different paths to \((x_0,y_0)\), then \(\lim_{(x,y)\to(x_0,y_0)}f(x,y)\) does not exist.

Level curves of \(f(x,y)=\frac{2xy}{x^2+y^2}\). Every level line passes through the origin, so arbitrarily close to \((0,0)\) the function takes every value between \(-1\) and \(1\). The limit at the origin cannot exist. Click the picture to read values.

The graph \(z=x^2+y^2\) (scaled by \(\frac12\)). Its level curves \(x^2+y^2=c\) are circles: the horizontal slices of this bowl.

Recipe

Limit of f(x, y), step by step

Substitute. If the formula is continuous at the point (denominator \(\ne0\)), the limit is the value. Done.
Algebra for \(\frac00\): factor and cancel, or multiply by a conjugate. For example \(\frac{x^2-y^2}{x-y}=x+y\).
Suspect it does not exist? Try the paths \(y=mx\). If the answer depends on \(m\), the limit does not exist. If not, also try \(y=mx^2\) (it catches \(\frac{x^2y}{x^4+y^2}\)).
Suspect it exists (at the origin)? Use polar coordinates \(x=r\cos\theta,\ y=r\sin\theta\). If \(|f|\le(\text{something})\cdot r\to0\) for every \(\theta\), the limit is \(0\). Or squeeze: \(\left|\frac{x^2y}{x^2+y^2}\right|\le|y|\).

Watch out

Common mistakes

Getting the same value along every line \(y=mx\) does not prove the limit exists (try \(y=mx^2\)).
In polar form the bound must not depend on \(\theta\): \(r\cos\theta\sin\theta\to0\) is fine, but \(\frac{\cos\theta\sin\theta}{\cos^2\theta}\) alone tells you nothing.
Domain questions ask for a set: describe it with inequalities and say whether the boundary is included (\(\le\) or \(<\)).

Worked examples (Thomas §14.1–14.2)

Example§14.1

Domain

\(f(x,y)=\sqrt{y-x^2}\) needs \(y-x^2\ge0\): the domain is \(y\ge x^2\), the closed region on and above the parabola, which is unbounded.

Example§14.2

Limit by algebra

\(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-xy}{\sqrt x-\sqrt y}=\lim\frac{x(x-y)(\sqrt x+\sqrt y)}{x-y}=\lim x(\sqrt x+\sqrt y)=0\).

Example§14.2

Two-path test

For \(\frac{2xy}{x^2+y^2}\) along \(y=mx\): \(\frac{2mx^2}{x^2(1+m^2)}=\frac{2m}{1+m^2}\), which depends on \(m\) (\(0\) for \(m=0\), \(1\) for \(m=1\)). The limit at \((0,0)\) does not exist.

Quick checks · true or false?

If \(f\to0\) along every line \(y=mx\) through the origin, then \(\lim_{(x,y)\to(0,0)}f=0\).

Polynomials in \(x\) and \(y\) are continuous everywhere.

The level curves of \(f(x,y)=x^2+y^2\) are circles centered at the origin (and the single point \((0,0)\) for \(c=0\)).

The domain of \(f(x,y)=\frac{1}{x-y}\) is the whole plane except the line \(y=x\).

Practice set 10 · Domains, level curves, limits

10 problems (week 8 quiz material)

Domain with a root

new · Thomas style

Describe the domain of \(f(x,y)=\sqrt{9-x^2-y^2}\) by an inequality.

Hint

The radicand must be \(\ge0\).

Answer

\(x^2+y^2\le9\): the closed disk of radius 3.

Full solution

\(9-x^2-y^2\ge0\iff x^2+y^2\le9\). The boundary circle is included (\(\sqrt0=0\) is fine).

Domain with a log

new · Thomas style

Describe the domain of \(f(x,y)=\ln(x+y-2)\) by an inequality.

Hint

The argument of \(\ln\) must be positive.

Answer

\(x+y>2\): the open half-plane above the line \(x+y=2\).

Full solution

\(x+y-2>0\iff x+y>2\). The line itself is excluded.

Level curve through a point

new · Thomas style

Find an equation of the level curve of \(f(x,y)=x^2+4y^2\) that passes through \((2,1)\).

Hint

Find \(c=f(2,1)\) first.

Answer

\(x^2+4y^2=8\), an ellipse.

Full solution

\(c=f(2,1)=4+4=8\). The level curve is \(x^2+4y^2=8\).

Limit by substitution

new · quiz style

Find \(\displaystyle\lim_{(x,y)\to(1,2)}\frac{x^2y+3}{x+y}\).

Hint

The denominator is not \(0\) at \((1,2)\).

Answer

\(\frac53\)

Full solution

Continuous there, so substitute: \(\frac{1\cdot2+3}{1+2}=\frac53\).

Factor and cancel

new · Thomas style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x-y}\) (taken over points with \(x\ne y\)).

Hint

\(x^2-y^2=(x-y)(x+y)\).

Answer

\(0\)

Full solution

\(\frac{x^2-y^2}{x-y}=x+y\) for \(x\ne y\), and \(x+y\to0\).

Does it exist?

new · quiz style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\), or type DNE.

Hint

Try the paths \(y=mx\).

Answer

Does not exist.

Full solution

Along \(y=mx\): \(\frac{mx^2}{x^2+m^2x^2}=\frac{m}{1+m^2}\).
\(m=0\) gives \(0\), \(m=1\) gives \(\frac12\): different limits, so by the two-path test the limit does not exist.

Squeeze it

new · Thomas style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}\), or type DNE.

Hint

\(\frac{x^2}{x^2+y^2}\le1\).

Answer

\(0\)

Full solution

\(\left|\frac{x^2y}{x^2+y^2}\right|=\frac{x^2}{x^2+y^2}|y|\le|y|\to0\).
By the sandwich theorem the limit is \(0\). (Polar check: \(r\cos^2\theta\sin\theta\to0\).)

Difference of fourth powers

new · Thomas style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-y^4}{x^2+y^2}\).

Hint

\(x^4-y^4=(x^2-y^2)(x^2+y^2)\).

Answer

\(0\)

Full solution

The fraction equals \(x^2-y^2\), which tends to \(0\).

A familiar limit in disguise

new · Thomas style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}\).

Hint

Let \(u=x^2+y^2\to0^+\).

Answer

\(1\)

Full solution

With \(u=x^2+y^2\to0\): \(\frac{\sin u}{u}\to1\).

Lines are not enough

new · Thomas style

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^4+y^2}\), or type DNE.

Hint

Along every line \(y=mx\) the limit is \(0\). Now try \(y=kx^2\).

Answer

Does not exist.

Full solution

Along \(y=mx\): \(\frac{mx^3}{x^4+m^2x^2}=\frac{mx}{x^2+m^2}\to0\).
Along \(y=kx^2\): \(\frac{kx^4}{x^4+k^2x^4}=\frac{k}{1+k^2}\), which depends on \(k\).
Two paths with different limits (\(k=0\) gives \(0\), \(k=1\) gives \(\frac12\)), so the limit does not exist.

Chapter 11 · Thomas §14.3–14.4 · Week 9

Partial derivatives and the chain rule

In plain words

The idea

To differentiate \(f(x,y)\) with respect to \(x\), pretend \(y\) is a constant number and use the ordinary Calculus I rules. The result \(f_x\) is the slope of the surface in the \(x\)-direction, and \(f_y\) is the slope in the \(y\)-direction. The chain rule then says how fast \(f\) changes when \(x\) and \(y\) themselves depend on \(t\): add up one contribution per path.

Definition§14.3

Partial derivatives

\[f_x(x_0,y_0)=\frac{\partial f}{\partial x}\Big|_{(x_0,y_0)}=\lim_{h\to0}\frac{f(x_0+h,y_0)-f(x_0,y_0)}{h},\qquad f_y(x_0,y_0)=\lim_{h\to0}\frac{f(x_0,y_0+h)-f(x_0,y_0)}{h}.\] Second-order partials: \(f_{xx}=\frac{\partial^2f}{\partial x^2}\), \(f_{xy}=(f_x)_y=\frac{\partial^2f}{\partial y\,\partial x}\), and so on.

Theorem§14.3

Mixed derivative theorem (Clairaut)

If \(f,f_x,f_y,f_{xy},f_{yx}\) are continuous on an open region containing \((a,b)\), then \(f_{xy}(a,b)=f_{yx}(a,b)\).

Theorem§14.3

Differentiability

If \(f_x\) and \(f_y\) are continuous on an open region, then \(f\) is differentiable there, and differentiable implies continuous. Just having partial derivatives at a point does not imply continuity.

Theorem§14.4

Chain rules

If \(w=f(x,y)\) with \(x=x(t),\ y=y(t)\): \(\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}\) (one more term for each extra variable).
If \(x=x(r,s),\ y=y(r,s)\): \(\dfrac{\partial w}{\partial r}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial r}\), and similarly for \(s\).

Theorem§14.4

Implicit differentiation

If \(F(x,y)=0\) defines \(y\) as a function of \(x\) and \(F_y\ne0\): \(\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}\). If \(F(x,y,z)=0\) defines \(z(x,y)\) and \(F_z\ne0\): \(\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}\), \(\dfrac{\partial z}{\partial y}=-\dfrac{F_y}{F_z}\).

Recipe

Partial derivatives without mistakes

Circle the variable you differentiate in; every other letter is a constant.
Apply the usual rules: product, quotient, and the chain rule on inner expressions such as \(e^{xy}\) or \(\sin(3x+2y)\).
For a value at a point, differentiate first, then substitute.
Chain rule: draw the tree \(w\to x,y\to t\); multiply along each branch, add over the branches, and substitute \(x(t),y(t)\) at the end.

Watch out

Common mistakes

\(\frac{\partial}{\partial x}e^{xy}=y\,e^{xy}\): the inner derivative \(y\) is easy to forget.
\(\frac{\partial}{\partial x}(xy^2)=y^2\), not \(2xy\).
In the implicit formula, the minus sign belongs there: \(\frac{dy}{dx}=-\frac{F_x}{F_y}\).
\(f_{xy}=f_{yx}\) needs continuity of the second partials (true for all functions in this course, but it is a theorem, not a definition).

Worked examples (Thomas §14.3–14.4)

Example§14.3

Values at a point

\(f=x^2+3xy+y-1\): \(f_x=2x+3y\), \(f_y=3x+1\). At \((4,-5)\): \(f_x=-7\), \(f_y=13\).

Example§14.3

Chain rule inside

\(f=y\sin xy\): \(f_x=y\cdot y\cos xy=y^2\cos xy\) (\(y\) is a constant factor; the inner derivative of \(xy\) in \(x\) is \(y\)).

Example§14.4

Chain rule with one parameter

\(w=xy\), \(x=\cos t\), \(y=\sin t\): \(\frac{dw}{dt}=y(-\sin t)+x\cos t=-\sin^2t+\cos^2t=\cos2t\).

Example§14.4

Implicit derivative

\(F(x,y)=y^2-x^2-\sin xy=0\): \(\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{-2x-y\cos xy}{2y-x\cos xy}=\frac{2x+y\cos xy}{2y-x\cos xy}\).

Quick checks · true or false?

\(f_{xy}=f_{yx}\) for every function \(f\).

If \(f_x\) and \(f_y\) exist at a point, then \(f\) is continuous there.

\(\dfrac{\partial}{\partial x}\left(y^2\right)=0\).

If \(w=f(x,y)\), \(x=t^2\), \(y=t^3\), then \(\frac{dw}{dt}=2t\,f_x+3t^2f_y\).

Practice set 11 · Partial derivatives and the chain rule

12 problems (week 9 quiz material)

First partial in x

new · quiz style

Let \(f(x,y)=x^3y^2-4xy+y^3\). Find \(f_x\).

Hint

\(y\) is a constant: \(\frac{\partial}{\partial x}(x^3y^2)=3x^2y^2\).

Answer

\(f_x=3x^2y^2-4y\)

Full solution

\(\frac{\partial}{\partial x}(x^3y^2)=3x^2y^2\), \(\frac{\partial}{\partial x}(-4xy)=-4y\), \(\frac{\partial}{\partial x}y^3=0\).

First partial in y

new · quiz style

For the same \(f(x,y)=x^3y^2-4xy+y^3\), find \(f_y\).

Hint

Now \(x\) is the constant.

Answer

\(f_y=2x^3y-4x+3y^2\)

Full solution

\(\frac{\partial}{\partial y}(x^3y^2)=2x^3y\), \(\frac{\partial}{\partial y}(-4xy)=-4x\), \(\frac{\partial}{\partial y}y^3=3y^2\).

Value at a point

new · Thomas style

Let \(f(x,y)=x^2y^3-2x\). Find \(f_x(1,-1)\).

Hint

\(f_x=2xy^3-2\).

Answer

\(-4\)

Full solution

\(f_x=2xy^3-2\); at \((1,-1)\): \(2(1)(-1)-2=-4\).

Product rule in a partial

new · Thomas style

Let \(f(x,y)=e^{xy}\ln y\) (\(y>0\)). Find \(f_y\).

Hint

Product rule: \((e^{xy})_y=xe^{xy}\) and \((\ln y)_y=\frac1y\).

Answer

\(f_y=xe^{xy}\ln y+\dfrac{e^{xy}}{y}\)

Full solution

\(f_y=(xe^{xy})\ln y+e^{xy}\cdot\frac1y\).

Mixed partial

new · Thomas style

Let \(f(x,y)=x^2\sin y\). Find \(f_{xy}\).

Hint

\(f_x=2x\sin y\), then differentiate in \(y\).

Answer

\(f_{xy}=2x\cos y\)

Full solution

\(f_x=2x\sin y\), \(f_{xy}=2x\cos y\). (Check: \(f_y=x^2\cos y\), \(f_{yx}=2x\cos y\).)

Inverse tangent

new · Thomas style

Let \(f(x,y)=\tan^{-1}\!\left(\dfrac yx\right)\) for \(x>0\). Find \(f_x\).

Hint

\(\frac{d}{du}\tan^{-1}u=\frac{1}{1+u^2}\), with \(u=\frac yx\) and \(u_x=-\frac{y}{x^2}\).

Answer

\(f_x=-\dfrac{y}{x^2+y^2}\)

Full solution

\(f_x=\dfrac{1}{1+y^2/x^2}\cdot\left(-\dfrac{y}{x^2}\right)=-\dfrac{y}{x^2+y^2}\).

Chain rule, one parameter

new · Thomas style

\(w=x^2+y^2\) with \(x=t^2\), \(y=2t\). Find \(\dfrac{dw}{dt}\) in terms of \(t\).

Hint

\(w_x\frac{dx}{dt}+w_y\frac{dy}{dt}=2x(2t)+2y(2)\).

Answer

\(4t^3+8t\)

Full solution

\(\frac{dw}{dt}=2x\cdot2t+2y\cdot2=2t^2\cdot2t+4t\cdot2=4t^3+8t\).

Chain rule, two parameters

new · Thomas style

\(w=x^2y\) with \(x=s+t\), \(y=s-t\). Find \(\dfrac{\partial w}{\partial s}\) at \((s,t)=(1,0)\).

Hint

\(w_x x_s+w_y y_s=2xy\cdot1+x^2\cdot1\). At \((1,0)\): \(x=y=1\).

Answer

\(3\)

Full solution

\(\frac{\partial w}{\partial s}=2xy+x^2\); at \(s=1,t=0\): \(x=1,y=1\), so \(2+1=3\).

Polar substitution

new · Thomas style

\(z=xy\) with \(x=r\cos\theta\), \(y=r\sin\theta\). Find \(\dfrac{\partial z}{\partial\theta}\) in terms of \(r\) and \(\theta\).

Hint

\(z=r^2\cos\theta\sin\theta=\frac{r^2}{2}\sin2\theta\).

Answer

\(r^2\cos2\theta\)

Full solution

\(z=\frac{r^2}{2}\sin2\theta\), so \(\frac{\partial z}{\partial\theta}=r^2\cos2\theta\). (Chain rule: \(y\cdot(-r\sin\theta)+x\cdot r\cos\theta=r^2(\cos^2\theta-\sin^2\theta)\).)

Implicit dy/dx

new · Thomas style

Find \(\dfrac{dy}{dx}\) for the curve \(x^3+y^3-6xy=0\), in terms of \(x\) and \(y\).

Hint

\(\frac{dy}{dx}=-\frac{F_x}{F_y}\) with \(F_x=3x^2-6y\), \(F_y=3y^2-6x\).

Answer

\(\dfrac{2y-x^2}{y^2-2x}\)

Full solution

\(-\dfrac{3x^2-6y}{3y^2-6x}=\dfrac{6y-3x^2}{3y^2-6x}=\dfrac{2y-x^2}{y^2-2x}\).

Implicit ∂z/∂x

new · Thomas style

\(x^2+y^2+z^2=1\) defines \(z\) as a function of \(x\) and \(y\) (with \(z\ne0\)). Find \(\dfrac{\partial z}{\partial x}\).

Hint

\(-\frac{F_x}{F_z}\).

Answer

\(-\dfrac xz\)

Full solution

\(F_x=2x\), \(F_z=2z\), so \(\frac{\partial z}{\partial x}=-\frac{2x}{2z}=-\frac xz\).

Laplace's equation

new · Thomas style

For \(f(x,y)=e^x\sin y\), compute \(f_{xx}+f_{yy}\).

Hint

\(f_{xx}=e^x\sin y\), \(f_{yy}=-e^x\sin y\).

Answer

\(0\) (so \(f\) is harmonic)

Full solution

\(f_{xx}=e^x\sin y\) and \(f_{yy}=-e^x\sin y\); the sum is \(0\).

Chapter 12 · Thomas §14.5–14.6 · Week 10

Gradient, directional derivatives, tangent planes

In plain words

The idea

The gradient \(\nabla f=\langle f_x,f_y\rangle\) packs both partial derivatives into one vector. It points uphill as steeply as possible, and its length is that steepest slope. The rate of change in any other unit direction \(\mathbf u\) is just the dot product \(\nabla f\cdot\mathbf u\). The gradient is also perpendicular to level curves and level surfaces, which is why it gives normal vectors to tangent planes.

Definition§14.5

Directional derivative and gradient

For a unit vector \(\mathbf u=\langle u_1,u_2\rangle\): \(\left(D_{\mathbf u}f\right)_{P_0}=\lim_{s\to0}\dfrac{f(x_0+su_1,y_0+su_2)-f(x_0,y_0)}{s}\). The gradient is \(\nabla f=f_x\mathbf i+f_y\mathbf j\) (\(+f_z\mathbf k\) in 3D).

Theorem§14.5

Directional derivative = gradient · unit vector

If \(f\) is differentiable at \(P_0\): \(\left(D_{\mathbf u}f\right)_{P_0}=\nabla f(P_0)\cdot\mathbf u=|\nabla f|\cos\theta\). Hence:

\(f\) increases fastest in the direction of \(\nabla f\), at rate \(|\nabla f|\);
it decreases fastest in the direction of \(-\nabla f\), at rate \(-|\nabla f|\);
the rate is \(0\) in directions perpendicular to \(\nabla f\);
\(\nabla f\) is normal to the level curve (level surface) through the point.

Theorem§14.6

Tangent planes and normal lines

Level surface \(F(x,y,z)=c\) at \(P_0\): tangent plane \(F_x(P_0)(x-x_0)+F_y(P_0)(y-y_0)+F_z(P_0)(z-z_0)=0\); normal line \(\mathbf r(t)=P_0+t\,\nabla F(P_0)\).
Graph \(z=f(x,y)\) at \((a,b)\): \(z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\).

Definition§14.6

Linearization and differential

\(L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\) is the linearization (the tangent plane as a function), \(f\approx L\) near \((a,b)\). The total differential \(df=f_x\,dx+f_y\,dy\) estimates the change \(\Delta f\).

Level curves of \(f=x^2+2y^2\) (ellipses) with gradient arrows: every arrow crosses the level curves at a right angle and points to higher values. Click anywhere to read \(f\) and \(\nabla f\).

The surface \(z=4-x^2-y^2\) with its tangent plane at \((0.8,0.6)\): near the point of contact, the plane and the surface are almost indistinguishable. That is linearization.

Recipe

Directional derivative, step by step

Compute \(\nabla f\) and evaluate it at \(P_0\).
Normalize the direction: \(\mathbf u=\frac{\mathbf v}{|\mathbf v|}\). Forgetting this is the most common lost point.
\(D_{\mathbf u}f=\nabla f(P_0)\cdot\mathbf u\).
Max rate \(=|\nabla f(P_0)|\), in the direction \(\frac{\nabla f}{|\nabla f|}\).

Recipe

Tangent plane, step by step

Write the surface as \(F(x,y,z)=c\) (e.g. \(z=x^2+y^2\) becomes \(x^2+y^2-z=0\)).
\(\mathbf n=\nabla F(P_0)\).
Plane: \(\mathbf n\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0\). For a graph you can use \(z=f(a,b)+f_x(x-a)+f_y(y-b)\) directly.

Watch out

Common mistakes

\(D_{\mathbf u}f\) needs a unit vector: “in the direction of \(3\mathbf i+4\mathbf j\)” means \(\mathbf u=\langle\frac35,\frac45\rangle\).
The gradient is a vector; the directional derivative and the max rate are numbers.
The normal line of a level surface uses \(\nabla F\), not the gradient of \(f(x,y)\) without the \(z\)-component.

Worked examples (Thomas §14.5–14.6)

Example§14.5

Directional derivative

\(f=xe^y+\cos xy\) at \((2,0)\) toward \(\mathbf v=3\mathbf i-4\mathbf j\): \(\nabla f=\langle e^y-y\sin xy,\ xe^y-x\sin xy\rangle=\langle1,2\rangle\) at \((2,0)\); \(\mathbf u=\langle\frac35,-\frac45\rangle\); \(D_{\mathbf u}f=\frac35-\frac85=-1\).

Example§14.6

Tangent plane to a level surface

\(x^2+y^2+z-9=0\) at \((1,2,4)\): \(\nabla F=\langle2x,2y,1\rangle=\langle2,4,1\rangle\). Plane \(2(x-1)+4(y-2)+(z-4)=0\iff2x+4y+z=14\).

Example§14.6

Linearization

\(f=x^2-xy+\frac12y^2+3\) at \((3,2)\): \(f=8\), \(f_x=2x-y=4\), \(f_y=-x+y=-1\). \(L=8+4(x-3)-(y-2)=4x-y-2\).

Quick checks · true or false?

A directional derivative can be larger than \(|\nabla f|\).

\(\nabla f(P)\) is perpendicular to the level curve of \(f\) through \(P\).

If \(\nabla f(P)=\mathbf 0\), every directional derivative at \(P\) is \(0\).

\(D_{\mathbf v}f=\nabla f\cdot\mathbf v\) for any vector \(\mathbf v\), unit or not.

Practice set 12 · Gradient, tangent planes, linearization

11 problems (week 10 quiz material)

Gradient in 3D

new · quiz style

Find \(\nabla f\) at \((1,-1,2)\) for \(f(x,y,z)=x^2y+yz^3\).

Hint

\(\nabla f=\langle2xy,\ x^2+z^3,\ 3yz^2\rangle\).

Answer

\(\langle-2,9,-12\rangle\)

Full solution

\(\nabla f=\langle2xy,\ x^2+z^3,\ 3yz^2\rangle\). At \((1,-1,2)\): \(\langle-2,\ 1+8,\ -12\rangle\).

Directional derivative

new · quiz style

Find the derivative of \(f(x,y)=x^2+y^2\) at \((1,2)\) in the direction of \(\mathbf v=3\mathbf i+4\mathbf j\).

Hint

\(\nabla f(1,2)=\langle2,4\rangle\), \(\mathbf u=\langle\frac35,\frac45\rangle\).

Answer

\(\frac{22}{5}\)

Full solution

\(\nabla f=\langle2x,2y\rangle=\langle2,4\rangle\); \(\mathbf u=\frac15\langle3,4\rangle\).
\(D_{\mathbf u}f=\frac{6+16}{5}=\frac{22}{5}\).

Directional derivative in 3D

new · Thomas style

Find the derivative of \(f=xyz\) at \((1,1,1)\) in the direction of \(\langle1,2,2\rangle\).

Hint

\(\nabla f=\langle yz,xz,xy\rangle=\langle1,1,1\rangle\); \(|\langle1,2,2\rangle|=3\).

Answer

\(\frac53\)

Full solution

\(D_{\mathbf u}f=\langle1,1,1\rangle\cdot\frac13\langle1,2,2\rangle=\frac53\).

Maximum rate of change

new · Thomas style

What is the maximum rate of change of \(f(x,y)=xe^y\) at \((2,0)\)?

Hint

It is \(|\nabla f(2,0)|\).

Answer

\(\sqrt5\), in the direction \(\langle1,2\rangle\).

Full solution

\(\nabla f=\langle e^y,xe^y\rangle=\langle1,2\rangle\) at \((2,0)\); \(|\nabla f|=\sqrt5\).

Steepest descent

new · Thomas style

In which direction does \(f(x,y)=x^2-y^2\) decrease most rapidly at \((1,1)\)? (Any vector pointing that way.)

Hint

\(-\nabla f\).

Answer

Along \(-\nabla f(1,1)=\langle-2,2\rangle\), e.g. the unit vector \(\frac{1}{\sqrt2}\langle-1,1\rangle\).

Full solution

\(\nabla f=\langle2x,-2y\rangle=\langle2,-2\rangle\). Fastest decrease: \(-\nabla f=\langle-2,2\rangle\).

Tangent plane to a sphere

new · quiz style

Find the tangent plane to \(x^2+y^2+z^2=9\) at \((1,2,2)\).

Hint

\(\nabla F=\langle2x,2y,2z\rangle=\langle2,4,4\rangle\).

Answer

\(x+2y+2z=9\)

Full solution

\(2(x-1)+4(y-2)+4(z-2)=0\iff x+2y+2z=9\).

Normal line

new · Thomas style

Find the normal line to \(x^2+y^2+z^2=9\) at \((1,2,2)\).

Hint

Through the point, along \(\nabla F\).

Answer

\(x=1+t,\ y=2+2t,\ z=2+2t\) (it passes through the center of the sphere)

Full solution

Direction \(\nabla F=\langle2,4,4\rangle\parallel\langle1,2,2\rangle\): \(\mathbf r(t)=\langle1,2,2\rangle+t\langle1,2,2\rangle\).

Tangent plane to a graph

new · quiz style

Find the tangent plane to \(z=x^2+y^2\) at the point where \((x,y)=(1,2)\).

Hint

\(f(1,2)=5\), \(f_x=2\), \(f_y=4\).

Answer

\(z=5+2(x-1)+4(y-2)\), i.e. \(z=2x+4y-5\).

Full solution

\(z=f(1,2)+f_x(1,2)(x-1)+f_y(1,2)(y-2)=5+2(x-1)+4(y-2)\).

Linearization

new · Thomas style

Find the linearization of \(f(x,y)=\sqrt{x^2+y^2}\) at \((3,4)\).

Hint

\(f(3,4)=5\), \(f_x=\frac{x}{\sqrt{x^2+y^2}}=\frac35\), \(f_y=\frac45\).

Answer

\(L=5+\frac35(x-3)+\frac45(y-4)=\frac{3x+4y}{5}\)

Full solution

\(L=5+\frac35(x-3)+\frac45(y-4)=\frac35x+\frac45y+5-\frac95-\frac{16}{5}=\frac{3x+4y}{5}\).

Total differential

new · Thomas style

Use the differential to estimate the change in \(f(x,y)=x^2y\) when \((x,y)\) moves from \((2,3)\) to \((2.01,2.98)\).

Hint

\(df=2xy\,dx+x^2\,dy\) with \(dx=0.01\), \(dy=-0.02\).

Answer

\(df=12(0.01)+4(-0.02)=0.04\)

Full solution

At \((2,3)\): \(f_x=2xy=12\), \(f_y=x^2=4\).
\(df=12(0.01)+4(-0.02)=0.12-0.08=0.04\).

Tangent line to a level curve

new · Thomas style

Find the tangent line to the level curve \(x^2+2y^2=3\) at \((1,1)\).

Hint

\(\nabla f(1,1)=\langle2,4\rangle\) is normal to the curve.

Answer

\(x+2y=3\)

Full solution

\(2(x-1)+4(y-1)=0\iff x+2y=3\).

Chapter 13 · Thomas §14.7 · Week 11

Extreme values and saddle points

In plain words

The idea

At the top of a hill or the bottom of a bowl the surface is flat, so both partial derivatives are zero. Solving \(f_x=0,\ f_y=0\) gives the candidate points. A flat point can also be a saddle (up in one direction, down in another), and the second-derivative test, using \(D=f_{xx}f_{yy}-f_{xy}^2\), tells the three cases apart. On a closed bounded region, also check the boundary.

Definition§14.7

Local extrema, critical points, saddle points

\(f(a,b)\) is a local maximum if \(f(a,b)\ge f(x,y)\) for all \((x,y)\) near \((a,b)\) (local minimum: \(\le\)). An interior point where \(f_x=f_y=0\) or where a partial derivative does not exist is a critical point. A critical point is a saddle point if every disk around it contains points where \(f\) is larger and points where \(f\) is smaller.

Theorem§14.7

First derivative test

If \(f\) has a local extremum at an interior point \((a,b)\) and the first partials exist there, then \(f_x(a,b)=0\) and \(f_y(a,b)=0\).

Theorem§14.7

Second derivative test

Suppose \(f_x(a,b)=f_y(a,b)=0\) and the second partials are continuous near \((a,b)\). Let \(D=f_{xx}f_{yy}-f_{xy}^2\) at \((a,b)\).

\(D>0\) and \(f_{xx}<0\): local maximum;
\(D>0\) and \(f_{xx}>0\): local minimum;
\(D<0\): saddle point;
\(D=0\): the test is inconclusive.

Theorem§14.7

Absolute extrema on a closed bounded region

A continuous \(f\) on a closed bounded region \(R\) attains an absolute max and min. To find them: (1) list the interior critical points; (2) on each boundary piece, reduce to one variable and find its critical points and endpoints (corners); (3) evaluate \(f\) at all candidates and pick the largest and smallest values.

Level curves of \(f=3y^2-2y^3-3x^2+6xy\). Around the maximum \((2,2)\) the curves close up into small ovals; at the saddle \((0,0)\) two level curves cross. That crossing is how a saddle looks on a contour map.

\(z=x^2-y^2\): a minimum along the \(x\)-direction and a maximum along the \(y\)-direction, which together make a saddle at the origin (\(D=2\cdot(-2)-0=-4<0\)).

Recipe

Find and classify all critical points

Compute \(f_x\) and \(f_y\) and solve the system \(f_x=0,\ f_y=0\). Factor instead of dividing, so you don't lose solutions like \(x=0\).
Compute \(f_{xx}\), \(f_{yy}\), \(f_{xy}\) and \(D=f_{xx}f_{yy}-f_{xy}^2\).
Evaluate \(D\) (and if needed \(f_{xx}\)) at each critical point; make a small table.
Conclude for every point: local max, local min, saddle, or inconclusive. Give the function value if it is asked for.

Watch out

Common mistakes

Dividing an equation by \(x\) or \(y\) (losing the solutions \(x=0\), \(y=0\)).
Checking the sign of \(f_{xx}\) when \(D<0\): it doesn't matter, a negative \(D\) is always a saddle.
Reading \(D=0\) as “saddle” or “no extremum”: it means the test gives no information (\(x^4+y^4\) has a minimum and \(D=0\)).
Absolute extrema: forgetting the boundary and the corners.

Worked examples (Thomas §14.7)

Example§14.7

A local maximum

\(f=xy-x^2-y^2-2x-2y+4\): \(f_x=y-2x-2=0,\ f_y=x-2y-2=0\Rightarrow(-2,-2)\). \(f_{xx}=-2,\ f_{yy}=-2,\ f_{xy}=1\), \(D=4-1=3>0\), \(f_{xx}<0\): local maximum, \(f(-2,-2)=8\).

Example§14.7

A saddle and a maximum

\(f=3y^2-2y^3-3x^2+6xy\): \(f_x=-6x+6y=0\Rightarrow y=x\); \(f_y=6y-6y^2+6x=0\Rightarrow12x-6x^2=0\Rightarrow x=0,2\). \(f_{xx}=-6,\ f_{yy}=6-12y,\ f_{xy}=6\), so \(D=-72+72y\). At \((0,0)\): \(D=-72\), a saddle. At \((2,2)\): \(D=72>0\), \(f_{xx}<0\), a local maximum with \(f(2,2)=8\).

Example§14.7

Absolute extrema on a triangle

\(f=2+2x+4y-x^2-y^2\) on the triangle bounded by \(x=0,\ y=0,\ y=9-x\). Interior: \((1,2)\), \(f=7\). Edge \(y=0\): \(2+2x-x^2\), giving \(f(1,0)=3\) and corners \(f(0,0)=2\), \(f(9,0)=-61\). Edge \(x=0\): \(f(0,2)=6\), \(f(0,9)=-43\). Edge \(y=9-x\): \(-43+16x-2x^2\), giving \(f(4,5)=-11\). Absolute max \(7\) at \((1,2)\), absolute min \(-61\) at \((9,0)\).

Quick checks · true or false?

Every critical point is a local maximum or a local minimum.

If \(D(a,b)=0\) at a critical point, the second derivative test is inconclusive.

A continuous function on a closed bounded region attains an absolute maximum.

If \(D<0\) and \(f_{xx}>0\), the point is a local minimum.

Practice set 13 · Critical points and extreme values

10 problems (week 11 quiz material)

A quadratic with one minimum

new · quiz style

Find the local extremum of \(f(x,y)=x^2+xy+y^2+3x-3y+4\) and its value.

Hint

\(f_x=2x+y+3=0\), \(f_y=x+2y-3=0\).

Answer

Local minimum at \((-3,3)\), value \(-5\).

Full solution

From \(f_y=0\): \(x=3-2y\). Then \(2(3-2y)+y+3=0\Rightarrow y=3,\ x=-3\).
\(f_{xx}=2,\ f_{yy}=2,\ f_{xy}=1\), \(D=3>0\), \(f_{xx}>0\): local minimum.
\(f(-3,3)=9-9+9-9-9+4=-5\).

Minimum and saddle

new · quiz style

Find and classify the critical points of \(f(x,y)=x^3-12x+y^2\).

Hint

\(f_x=3x^2-12\), \(f_y=2y\). Then \(D=6x\cdot2\).

Answer

Local minimum \((2,0)\); saddle point \((-2,0)\).

Full solution

\(3x^2-12=0\Rightarrow x=\pm2\); \(2y=0\Rightarrow y=0\).
\(D=f_{xx}f_{yy}-f_{xy}^2=(6x)(2)-0=12x\).
\((2,0)\): \(D=24>0\), \(f_{xx}=12>0\): minimum. \((-2,0)\): \(D=-24<0\): saddle.

Four critical points

new · Thomas style

Find and classify the critical points of \(f(x,y)=xy(1-x-y)\).

Hint

\(f_x=y(1-2x-y)\), \(f_y=x(1-x-2y)\). Set each factor to zero in all combinations.

Answer

Local maximum at \(\left(\frac13,\frac13\right)\); saddle points \((0,0),(1,0),(0,1)\).

Full solution

Solutions: \((0,0)\); \(y=0,\ x=1\); \(x=0,\ y=1\); and \(2x+y=1,\ x+2y=1\Rightarrow\left(\frac13,\frac13\right)\).
\(f_{xx}=-2y\), \(f_{yy}=-2x\), \(f_{xy}=1-2x-2y\).
At \((0,0),(1,0),(0,1)\): \(D=-1<0\), saddles. At \(\left(\frac13,\frac13\right)\): \(D=\frac49-\frac19=\frac13>0\), \(f_{xx}=-\frac23<0\): maximum.

Two minima

new · Thomas style

Find and classify the critical points of \(f(x,y)=x^4+y^4-4xy+1\).

Hint

\(4x^3=4y\) and \(4y^3=4x\) give \(x=x^9\).

Answer

Minima at \((1,1)\) and \((-1,-1)\); saddle at \((0,0)\).

Full solution

\(y=x^3\) and \(x=y^3=x^9\Rightarrow x(x^8-1)=0\Rightarrow x=0,\pm1\); points \((0,0),(1,1),(-1,-1)\).
\(D=(12x^2)(12y^2)-16=144x^2y^2-16\).
\((0,0)\): \(D=-16\), saddle. \((\pm1,\pm1)\): \(D=128>0\), \(f_{xx}=12>0\), minima.

Classify one point

new · quiz style

Classify the critical point \((0,0)\) of \(f(x,y)=x^2-3xy+y^2\).

Hint

\(D=2\cdot2-(-3)^2\).

Answer

Saddle point (\(D=-5<0\)).

Full solution

\(f_{xx}=2,\ f_{yy}=2,\ f_{xy}=-3\Rightarrow D=4-9=-5<0\): saddle point.

When the test fails

new · Thomas style

What does the second derivative test say about \((0,0)\) for \(f(x,y)=x^4+y^4\)?

Hint

All second partials vanish at the origin.

Answer

The test is inconclusive (\(D=0\)). Directly, \(f\ge0=f(0,0)\), so it is in fact a minimum.

Full solution

\(f_{xx}=12x^2,\ f_{yy}=12y^2,\ f_{xy}=0\) are all \(0\) at the origin, so \(D=0\): no conclusion from the test.
Looking at \(f\) itself: \(x^4+y^4\ge0\), so \((0,0)\) is an absolute minimum.

Two maxima

new · Thomas style

The function \(f(x,y)=4xy-x^4-y^4\) has two local maxima. What is the maximum value?

Hint

Critical points: \(y=x^3\), \(x=y^3\). Check \((1,1)\) and \((-1,-1)\).

Answer

\(2\), attained at \((1,1)\) and \((-1,-1)\).

Full solution

Critical points \((0,0),(1,1),(-1,-1)\). \(D=144x^2y^2-16\).
At \((\pm1,\pm1)\): \(D=128>0\), \(f_{xx}=-12<0\): maxima, with \(f=4-1-1=2\).

Absolute extrema on a rectangle

new · Thomas style

Find the absolute maximum and minimum values of \(f(x,y)=x^2-2x+y^2\) on the rectangle \(0\le x\le2,\ -1\le y\le1\).

Hint

\(f=(x-1)^2+y^2-1\). Interior critical point \((1,0)\); then check the edges and the corners.

Answer

Max \(1\) (at the four corners), min \(-1\) (at \((1,0)\)).

Full solution

Interior: \(f_x=2x-2=0,\ f_y=2y=0\Rightarrow(1,0)\), \(f=-1\).
Edges: \(y=\pm1\): \(x^2-2x+1\) on \([0,2]\), min \(0\) at \(x=1\), max \(1\) at \(x=0,2\). \(x=0\) or \(x=2\): \(y^2\) on \([-1,1]\), values between \(0\) and \(1\).
Largest value \(1\) (corners), smallest \(-1\) at \((1,0)\).

Maximize a product

new · Thomas style

Three positive numbers add up to \(9\). What is the largest possible value of their product?

Hint

Maximize \(P=xy(9-x-y)\) over \(x,y>0\).

Answer

\(27\) (all three equal to \(3\))

Full solution

\(P_x=y(9-2x-y)=0\), \(P_y=x(9-x-2y)=0\) with \(x,y>0\Rightarrow2x+y=9,\ x+2y=9\Rightarrow x=y=3\).
\(D>0\), \(P_{xx}=-2y<0\): a maximum, \(P=3\cdot3\cdot3=27\).

Closest point on a plane

new · Thomas style

Find the point on the plane \(x+y+z=3\) closest to the origin.

Hint

Minimize the squared distance \(x^2+y^2+(3-x-y)^2\).

Answer

\((1,1,1)\)

Full solution

\(g(x,y)=x^2+y^2+(3-x-y)^2\); \(g_x=2x-2(3-x-y)=0\), \(g_y=2y-2(3-x-y)=0\Rightarrow x=y=1\).
Then \(z=3-1-1=1\). (Check: the closest point lies along the normal \(\langle1,1,1\rangle\).)

Chapter 14 · Thomas §15.1–15.3 · Weeks 12–13

Double integrals and area

In plain words

The idea

\(\iint_R f\,dA\) adds up \(f\) times tiny areas over a region \(R\). When \(f\ge0\) it is the volume under the surface, and when \(f=1\) it is the area of \(R\). You never compute it from the definition: Fubini's theorem turns it into two ordinary integrals, one inside the other. All the thinking goes into the limits of integration, which you read from a sketch of the region.

Theorem§15.1

Fubini's theorem (rectangles)

If \(f\) is continuous on \(R=[a,b]\times[c,d]\), then \(\displaystyle\iint_Rf\,dA=\int_a^b\!\!\int_c^df(x,y)\,dy\,dx=\int_c^d\!\!\int_a^bf(x,y)\,dx\,dy\). The order does not matter on a rectangle, and if \(f(x,y)=g(x)h(y)\) the integral splits into \(\left(\int g\right)\left(\int h\right)\).

Theorem§15.2

Fubini's theorem (general regions)

Vertical slices (type I): if \(R:\ a\le x\le b,\ g_1(x)\le y\le g_2(x)\), then \(\iint_Rf\,dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f\,dy\,dx\).
Horizontal slices (type II): if \(R:\ c\le y\le d,\ h_1(y)\le x\le h_2(y)\), then \(\iint_Rf\,dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f\,dx\,dy\).

Definition§15.3

Area and average value

\(\text{Area}(R)=\iint_R1\,dA\). The average value of \(f\) on \(R\) is \(\dfrac{1}{\text{Area}(R)}\iint_Rf\,dA\).

The region between \(y=x^2\) (bottom) and \(y=x+2\) (top) for \(-1\le x\le2\). Slide the vertical slice: at every \(x\), \(y\) runs from the bottom curve to the top curve. Those are the inner limits. Area \(=\int_{-1}^2\int_{x^2}^{x+2}dy\,dx=\frac92\).

The same idea with horizontal slices: between \(x=y^2\) (left) and \(x=2y\) (right) for \(0\le y\le2\).

Recipe

Setting up the limits (sketch first)

Sketch the region and mark where the curves intersect (solve the equations).
Vertical slices (\(dy\,dx\)): draw a vertical line through \(R\). It enters at \(y=g_1(x)\) (inner lower limit) and leaves at \(y=g_2(x)\) (inner upper limit). Then let \(x\) run over all slices: \(a\le x\le b\).
Horizontal slices (\(dx\,dy\)): the same with a horizontal line; inner limits are functions of \(y\), outer limits are numbers.
Integrate the inner integral, treating the outer variable as a constant; substitute the limits; then do the outer integral.
Reversing the order: never just swap the limits. Redraw the region and describe it with slices in the other direction.

Watch out

Common mistakes

The outer limits must be numbers; only the inner limits may contain the outer variable.
\(\int_0^1\int_0^xf\,dy\,dx\ne\int_0^1\int_0^yf\,dx\,dy\): the region changes. Redraw it.
Some inner integrals are impossible in one order (\(e^{y^2}\), \(\frac{\sin x}{x}\), \(\cos(y^2)\)): reversing the order is the intended trick.
For an area use the integrand \(1\), not the curves.

Worked examples (Thomas §15.1–15.3)

Example§15.1

Over a rectangle

\(\iint_R(100-6x^2y)\,dA\), \(R:0\le x\le2,\ -1\le y\le1\): inner \(\int_{-1}^1(100-6x^2y)\,dy=\left[100y-3x^2y^2\right]_{-1}^1=200\); outer \(\int_0^2200\,dx=400\).

Example§15.2

Choosing the order

\(\iint_R\frac{\sin x}{x}\,dA\) over the triangle bounded by \(y=0,\ y=x,\ x=1\). The order \(dx\,dy\) would need \(\int\frac{\sin x}{x}dx\) (impossible), so use vertical slices: \(\int_0^1\int_0^x\frac{\sin x}{x}\,dy\,dx=\int_0^1\sin x\,dx=1-\cos1\).

Example§15.3

Area between curves

Between \(y=x^2\) and \(y=x+2\): they meet at \(x=-1,2\). \(A=\int_{-1}^2\int_{x^2}^{x+2}dy\,dx=\int_{-1}^2(x+2-x^2)\,dx=\frac92\).

Quick checks · true or false?

For continuous \(f\) on a rectangle, \(\int_a^b\int_c^df\,dy\,dx=\int_c^d\int_a^bf\,dx\,dy\).

In \(\int_a^b\int_{g_1(x)}^{g_2(x)}f\,dy\,dx\), the inner limits may depend on \(x\).

\(\iint_R1\,dA\) is the area of \(R\).

\(\int_0^1\int_0^xf\,dy\,dx=\int_0^1\int_0^yf\,dx\,dy\).

Practice set 14 · Double integrals and area

12 problems (weeks 12–13 quiz material)

Over a rectangle

new · quiz style

Evaluate \(\displaystyle\int_0^2\int_0^1(x+y)\,dy\,dx\).

Hint

Inner: \(\int_0^1(x+y)\,dy=x+\frac12\).

Answer

\(3\)

Full solution

Inner: \(\left[xy+\frac{y^2}{2}\right]_0^1=x+\frac12\).
Outer: \(\int_0^2\left(x+\frac12\right)dx=2+1=3\).

A product splits

new · quiz style

Evaluate \(\displaystyle\iint_R xy^2\,dA\) for \(R=[1,2]\times[0,3]\).

Hint

\(\left(\int_1^2x\,dx\right)\left(\int_0^3y^2dy\right)\).

Answer

\(\frac{27}{2}\)

Full solution

\(\int_1^2x\,dx=\frac32\), \(\int_0^3y^2\,dy=9\). Product \(\frac{27}{2}\).

Choose the easy order

new · Thomas style

Evaluate \(\displaystyle\iint_R xe^{xy}\,dA\) for \(R=[0,1]\times[0,1]\).

Hint

Integrate in \(y\) first: \(\int_0^1xe^{xy}dy=e^x-1\).

Answer

\(e-2\)

Full solution

\(\int_0^1xe^{xy}\,dy=\left[e^{xy}\right]_{y=0}^{1}=e^x-1\).
\(\int_0^1(e^x-1)\,dx=e-1-1=e-2\).

Variable inner limit

new · quiz style

Evaluate \(\displaystyle\int_0^1\int_0^x(x^2+y)\,dy\,dx\).

Hint

Inner: \(\left[x^2y+\frac{y^2}{2}\right]_0^x=x^3+\frac{x^2}{2}\).

Answer

\(\frac{5}{12}\)

Full solution

Inner: \(x^3+\frac{x^2}{2}\).
Outer: \(\frac14+\frac16=\frac5{12}\).

Over a triangle

new · quiz style

Evaluate \(\displaystyle\iint_R xy\,dA\), where \(R\) is the triangle with vertices \((0,0)\), \((1,0)\), \((1,1)\).

Hint

\(R:\ 0\le x\le1,\ 0\le y\le x\).

Answer

\(\frac18\)

Full solution

\(\int_0^1\int_0^xxy\,dy\,dx=\int_0^1\frac{x^3}{2}\,dx=\frac18\).

Reverse the order

new · Thomas style

Evaluate \(\displaystyle\int_0^1\int_x^1e^{y^2}\,dy\,dx\) by reversing the order of integration.

Hint

The region is \(0\le x\le y\le1\): for each \(y\), \(x\) runs from \(0\) to \(y\).

Answer

\(\frac{e-1}{2}\)

Full solution

Region: \(0\le x\le1,\ x\le y\le1\), the triangle above \(y=x\). Horizontally: \(0\le y\le1,\ 0\le x\le y\).
\(\int_0^1\int_0^ye^{y^2}\,dx\,dy=\int_0^1ye^{y^2}\,dy=\left[\frac12e^{y^2}\right]_0^1=\frac{e-1}{2}\).

Reverse to make it possible

new · Thomas style

Evaluate \(\displaystyle\int_0^2\int_{x^2}^4x\cos(y^2)\,dy\,dx\).

Hint

The region is \(0\le x\le\sqrt y,\ 0\le y\le4\).

Answer

\(\frac{\sin16}{4}\)

Full solution

Reversed: \(\int_0^4\int_0^{\sqrt y}x\cos(y^2)\,dx\,dy=\int_0^4\frac{y}{2}\cos(y^2)\,dy\).
With \(u=y^2\): \(\frac14\int_0^{16}\cos u\,du=\frac{\sin16}{4}\).

Area between a line and a parabola

new · Thomas style

Use a double integral to find the area between \(y=x\) and \(y=x^2\).

Hint

They meet at \(x=0\) and \(x=1\); \(x\) is on top.

Answer

\(\frac16\)

Full solution

\(A=\int_0^1\int_{x^2}^xdy\,dx=\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16\).

Area with horizontal slices

new · Thomas style

Find the area of the region between \(x=y^2\) and \(x=2y\).

Hint

They meet at \(y=0\) and \(y=2\). Horizontal slices: \(y^2\le x\le2y\).

Answer

\(\frac43\)

Full solution

\(A=\int_0^2\int_{y^2}^{2y}dx\,dy=\int_0^2(2y-y^2)\,dy=4-\frac83=\frac43\).

Average value

new · Thomas §15.3 style

Find the average value of \(f(x,y)=xy\) on the square \([0,2]\times[0,2]\).

Hint

Divide the integral by the area \(4\).

Answer

\(1\)

Full solution

\(\iint xy\,dA=\left(\int_0^2x\,dx\right)^2=4\). Average \(=\frac44=1\).

Volume over a triangle

new · Thomas style

Find the volume under the plane \(z=4-x-y\) over the triangle \(x\ge0,\ y\ge0,\ x+y\le2\).

Hint

\(\int_0^2\int_0^{2-x}(4-x-y)\,dy\,dx\).

Answer

\(\frac{16}{3}\)

Full solution

Inner: \(\left[(4-x)y-\frac{y^2}{2}\right]_0^{2-x}=\frac{(2-x)(6-x)}{2}\).
Outer: \(\frac12\int_0^2(12-8x+x^2)\,dx=\frac12\left(24-16+\frac83\right)=\frac{16}{3}\).

Read off a limit

new · Thomas style

The region \(R\) is bounded by \(y=x^2\) and \(y=4\). With horizontal slices, \(\iint_Rf\,dA=\int_0^4\int_{h_1(y)}^{h_2(y)}f\,dx\,dy\). What is \(h_2(y)\)?

Hint

Solve \(y=x^2\) for \(x\); the right-hand boundary is the positive root.

Answer

\(h_2(y)=\sqrt y\) (and \(h_1(y)=-\sqrt y\))

Full solution

A horizontal line at height \(y\) meets the parabola at \(x=\pm\sqrt y\): it enters at \(-\sqrt y\) and leaves at \(\sqrt y\).

Chapter 15 · Thomas §16.1–16.3 · Weeks 14–15

Line integrals, vector fields, conservative fields

In plain words

The idea

A line integral adds something up along a curve. For a function \(f\) (like the density of a wire) you add \(f\,ds\). For a vector field \(\mathbf F\) (like a force) you add only the part of \(\mathbf F\) that pushes along the curve, \(\mathbf F\cdot d\mathbf r\): that is work. The recipe is always the same: parametrize, substitute, integrate in \(t\). For conservative fields \(\mathbf F=\nabla f\) there is a shortcut, \(f(B)-f(A)\), which is the fundamental theorem of calculus in a new form.

Definition§16.1

Line integral of a function

If \(C\) is \(\mathbf r(t)\), \(a\le t\le b\), then \(\displaystyle\int_Cf\,ds=\int_a^bf(\mathbf r(t))\,|\mathbf r'(t)|\,dt\). It does not depend on the direction of travel. Mass of a wire with density \(\delta\): \(M=\int_C\delta\,ds\).

Definition§16.2

Work, circulation, flux

Work of \(\mathbf F=\langle M,N,P\rangle\) along \(C\): \(\displaystyle W=\int_C\mathbf F\cdot\mathbf T\,ds=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt=\int_CM\,dx+N\,dy+P\,dz\). When \(C\) is closed this is called the circulation. Flux of \(\mathbf F=\langle M,N\rangle\) across a simple closed plane curve traversed counterclockwise: \(\displaystyle\oint_CM\,dy-N\,dx\).

Theorem§16.3

Fundamental theorem of line integrals

If \(\mathbf F=\nabla f\) on a domain containing \(C\) from \(A\) to \(B\), then \(\displaystyle\int_C\mathbf F\cdot d\mathbf r=f(B)-f(A)\). So the integral is path independent, and \(\oint_C\mathbf F\cdot d\mathbf r=0\) around every closed curve.

Theorem§16.3

Component test for conservative fields

On an open simply connected domain with continuous partials, \(\mathbf F=\langle M,N,P\rangle\) is conservative iff \(\dfrac{\partial P}{\partial y}=\dfrac{\partial N}{\partial z},\ \dfrac{\partial M}{\partial z}=\dfrac{\partial P}{\partial x},\ \dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}\). In the plane: \(\dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}\).

The field \(\mathbf F=\langle x-y,\ x\rangle\) and the unit circle traversed counterclockwise. The field mostly pushes along the circle (circulation \(2\pi\)) and partly outward across it (flux \(\pi\)). It is not conservative: \(N_x=1\ne M_y=-1\).

Recipe

Computing a line integral (any kind)

Parametrize \(C\): segment \(P\to Q\) is \(P+t(Q-P),\ 0\le t\le1\); circle of radius \(a\) is \((a\cos t,a\sin t),\ 0\le t\le2\pi\); graph \(y=g(x)\) is \((t,g(t))\).
Substitute \(x(t),y(t),z(t)\) into \(f\) or \(\mathbf F\).
Multiply by \(|\mathbf r'(t)|\) (for \(ds\)) or dot with \(\mathbf r'(t)\) (for work).
Integrate from \(t=a\) to \(t=b\). For a path made of several pieces, add the pieces.

Recipe

Conservative? Find the potential

Test: \(M_y=N_x\) (and the two other equalities in 3D). If one fails, the field is not conservative.
\(f=\int M\,dx+g(y,z)\): the “constant” of integration may depend on the other variables.
Differentiate in \(y\) and compare with \(N\) to find \(g\); repeat with \(z\) in 3D.
Check \(\nabla f=\mathbf F\). Then work from \(A\) to \(B\) is \(f(B)-f(A)\).

Watch out

Common mistakes

\(ds=|\mathbf r'(t)|\,dt\), not \(dt\). For work you need \(\mathbf r'(t)\,dt\), the vector, not its length.
Reversing the direction of \(C\) changes the sign of the work but not of \(\int_Cf\,ds\).
The component test needs a simply connected domain: \(\mathbf F=\frac{\langle-y,x\rangle}{x^2+y^2}\) passes the test but has circulation \(2\pi\) around the origin.
When finding a potential, the “constant” is a function of the remaining variables.

Worked examples (Thomas §16.1–16.3)

Example§16.1

A scalar line integral

\(\int_C(x-3y^2+z)\,ds\) along the segment from \((0,0,0)\) to \((1,1,1)\): \(\mathbf r=\langle t,t,t\rangle\), \(|\mathbf r'|=\sqrt3\); \(\int_0^1(2t-3t^2)\sqrt3\,dt=\sqrt3(1-1)=0\).

Example§16.2

Work along a curve

\(\mathbf F=(y-x^2)\mathbf i+(z-y^2)\mathbf j+(x-z^2)\mathbf k\) along \(\mathbf r=\langle t,t^2,t^3\rangle\), \(0\le t\le1\): \(\mathbf F(\mathbf r)\cdot\mathbf r'=(t^3-t^4)2t+(t-t^6)3t^2\), so \(W=\int_0^1(2t^4-2t^5+3t^3-3t^8)\,dt=\frac{29}{60}\).

Example§16.3

Finding a potential

\(\mathbf F=(e^x\cos y+yz)\mathbf i+(xz-e^x\sin y)\mathbf j+(xy+z)\mathbf k\). The component test passes. \(f=\int M\,dx=e^x\cos y+xyz+g(y,z)\); \(f_y=-e^x\sin y+xz+g_y=N\Rightarrow g_y=0\); \(f_z=xy+g_z=xy+z\Rightarrow g=\frac{z^2}{2}\). So \(f=e^x\cos y+xyz+\frac{z^2}{2}+C\).

Quick checks · true or false?

The work done by any field around a closed curve is zero.

If \(\mathbf F=\nabla f\), then \(\int_C\mathbf F\cdot d\mathbf r\) depends only on the endpoints of \(C\).

\(\int_Cf\,ds\) changes sign when \(C\) is traversed in the opposite direction.

\(\mathbf F=\langle-y,x\rangle\) is conservative.

Practice set 15 · Line integrals and conservative fields

13 problems (weeks 14–15 quiz material)

Along a segment

new · quiz style

Evaluate \(\int_C(x+y)\,ds\), where \(C\) is the segment from \((0,0)\) to \((3,4)\).

Hint

\(\mathbf r=\langle3t,4t\rangle\), \(|\mathbf r'|=5\).

Answer

\(\frac{35}{2}\)

Full solution

\(\int_0^1(3t+4t)\cdot5\,dt=35\int_0^1t\,dt=\frac{35}{2}\).

Along a quarter circle

new · quiz style

Evaluate \(\int_Cxy\,ds\), where \(C\) is the quarter of the unit circle from \((1,0)\) to \((0,1)\).

Hint

\(\mathbf r=\langle\cos t,\sin t\rangle\), \(0\le t\le\frac\pi2\), \(|\mathbf r'|=1\).

Answer

\(\frac12\)

Full solution

\(\int_0^{\pi/2}\cos t\sin t\,dt=\left[\frac{\sin^2t}{2}\right]_0^{\pi/2}=\frac12\).

Work along a parabola

new · quiz style

Find the work done by \(\mathbf F=\langle y,x\rangle\) along \(\mathbf r(t)=\langle t,t^2\rangle\), \(0\le t\le1\).

Hint

\(\mathbf F(\mathbf r)\cdot\mathbf r'=t^2\cdot1+t\cdot2t\).

Answer

\(1\)

Full solution

\(\int_0^1(t^2+2t^2)\,dt=1\). (Also: \(\mathbf F=\nabla(xy)\), so \(W=1\cdot1-0=1\).)

Circulation of a rotation field

new · Thomas style

Find the circulation of \(\mathbf F=\langle-y,x\rangle\) around the unit circle, counterclockwise.

Hint

\(\mathbf F(\mathbf r)\cdot\mathbf r'=\sin^2t+\cos^2t=1\).

Answer

\(2\pi\)

Full solution

\(\mathbf F=\langle-\sin t,\cos t\rangle\), \(\mathbf r'=\langle-\sin t,\cos t\rangle\): dot product \(1\). \(\int_0^{2\pi}1\,dt=2\pi\).

Flux of a radial field

new · Thomas style

Find the outward flux of \(\mathbf F=\langle x,y\rangle\) across the unit circle.

Hint

\(\oint M\,dy-N\,dx=\int_0^{2\pi}(\cos t\cdot\cos t-\sin t\cdot(-\sin t))\,dt\).

Answer

\(2\pi\)

Full solution

\(M\,dy-N\,dx=\cos^2t\,dt+\sin^2t\,dt=dt\). Flux \(=2\pi\).

Circulation and flux together

new · Thomas style

For \(\mathbf F=\langle x-y,\ x\rangle\) and the unit circle (counterclockwise), find the circulation and the outward flux.

Hint

Circulation integrand \(1-\sin t\cos t\); flux integrand \(\cos^2t\).

Answer

Circulation \(2\pi\), flux \(\pi\).

Full solution

\(\mathbf F\cdot\mathbf r'=(\cos t-\sin t)(-\sin t)+\cos t\cos t=1-\sin t\cos t\), integral \(2\pi\).
\(M\,dy-N\,dx=(\cos t-\sin t)\cos t+\cos t\sin t=\cos^2t\), integral \(\pi\).

Conservative or not? (1)

new · quiz style

Is \(\mathbf F=\langle2xy+3,\ x^2-4y^3\rangle\) conservative?

Hint

Compare \(M_y\) and \(N_x\).

Answer

Yes: \(M_y=2x=N_x\).

Full solution

\(M_y=2x\), \(N_x=2x\); equal on the whole plane, which is simply connected, so \(\mathbf F\) is conservative.

Potential function

new · quiz style

Find a potential \(f\) with \(\nabla f=\langle2xy+3,\ x^2-4y^3\rangle\).

Hint

\(f=\int(2xy+3)\,dx=x^2y+3x+g(y)\); then match \(f_y\) with \(N\).

Answer

\(f=x^2y+3x-y^4\) (+ any constant)

Full solution

\(f=x^2y+3x+g(y)\).
\(f_y=x^2+g'(y)=x^2-4y^3\Rightarrow g'(y)=-4y^3\Rightarrow g=-y^4\).
\(f=x^2y+3x-y^4+C\).

Work by the shortcut

new · quiz style

Find the work done by \(\mathbf F=\langle2xy+3,\ x^2-4y^3\rangle\) along any path from \((0,0)\) to \((1,2)\).

Hint

\(f(1,2)-f(0,0)\) with \(f=x^2y+3x-y^4\).

Answer

\(-11\)

Full solution

\(f(1,2)=2+3-16=-11\), \(f(0,0)=0\). Work \(=-11\).

A potential in 3D

new · Thomas style

Find a potential for \(\mathbf F=\langle yz,\ xz,\ xy+2z\rangle\).

Hint

\(\int yz\,dx=xyz+g(y,z)\).

Answer

\(f=xyz+z^2\) (+ any constant)

Full solution

\(f=xyz+g(y,z)\); \(f_y=xz+g_y=xz\Rightarrow g=h(z)\).
\(f_z=xy+h'(z)=xy+2z\Rightarrow h=z^2\). So \(f=xyz+z^2\).

Conservative or not? (2)

new · quiz style

Is \(\mathbf F=\langle x-y,\ x+y\rangle\) conservative?

Hint

\(M_y=-1\), \(N_x=1\).

Answer

No.

Full solution

\(M_y=-1\ne N_x=1\), so the component test fails: not conservative.

An exact differential

new · Thomas style

Evaluate \(\displaystyle\int_{(1,1)}^{(3,-2)}y\,dx+x\,dy\) along any path.

Hint

\(y\,dx+x\,dy=d(xy)\).

Answer

\(-7\)

Full solution

\(f=xy\) is a potential, so the integral is \(3(-2)-1\cdot1=-7\).

Mass of a wire

new · Thomas §16.1 style

A wire lies along the helix \(\mathbf r(t)=\langle\cos t,\sin t,t\rangle\), \(0\le t\le2\pi\), with density \(\delta(x,y,z)=z\). Find its mass.

Hint

\(ds=\sqrt2\,dt\) and \(\delta=t\).

Answer

\(2\sqrt2\,\pi^2\)

Full solution

\(M=\int_0^{2\pi}t\sqrt2\,dt=\sqrt2\cdot\frac{(2\pi)^2}{2}=2\sqrt2\pi^2\).

Tools

Tools for checking your own work

Exams and quizzes are calculator-free, so use these tools to check work you did on paper, not to replace it. Each tool explains what it computed, and most draw a picture. Inputs use the same typing rules as the answer boxes; press Enter or the button.

Antiderivative checker

Is \(F\) an antiderivative of \(f\)? Checks \(F'=f\) at 40 sample points, using exact Taylor-mode differentiation (no rounding from finite differences).

f(x) = F(x) = variable sample on

Definite and improper integrals

Adaptive Simpson's rule. For an infinite limit or a vertical asymptote at an endpoint it integrates over shrinking or doubling pieces and reports whether the integral converges or diverges. This is the integral test step. Split at interior asymptotes yourself.

f(x) = from to

Series explorer

Partial sums \(S_N\) up to \(N=10^5\), the terms \(a_n\), and the trends of \(\left|\frac{a_{n+1}}{a_n}\right|\) (ratio test) and \(\sqrt[n]{|a_n|}\) (root test). Numbers are evidence, not a proof: a quiz still wants the test written out.

aₙ = from n =

Taylor polynomial grapher

Exact Taylor coefficients (computed with power-series arithmetic, not numerical derivatives) and a plot with a degree slider.

f(x) = center a = degree n =

Parametric and polar grapher

Draws the curve with its direction and a movable velocity vector, and computes the arc length \(\int\sqrt{x'^2+y'^2}\,dt\) (and for polar curves \(\frac12\int r^2\,d\theta\)).

type x(t) = y(t) = r(θ) = from to

Vector calculator

Dot and cross products, angle, projection, areas, and the triple product (volume) if you give a third vector. The 3D picture can be rotated.

u = v = w (optional) =

Gradient, tangent plane and critical point explorer

For \(f(x,y)\) or \(f(x,y,z)\) at a point: the value, \(\nabla f\), the directional derivative, the tangent plane, and the second-derivative test numbers \(f_{xx},f_{yy},f_{xy},D\). For two variables it also draws the contour map.

f = point direction

Double integral calculator

Iterated integral with variable inner limits, plus the region and a movable slice so you can check your limits.

f(x, y) = order outer from outer to inner from inner to

Line integral calculator

Work \(\int_C\mathbf F\cdot d\mathbf r\), flux (plane curves), length of \(C\), and the component test for conservative fields. Leave \(P\) and \(z(t)\) empty for a plane field.

M = N = P = x(t) = y(t) = z(t) = t from to

Endless practice: a fresh problem every click

New numbers every time, for every quiz type in the syllabus: sequence limits, geometric and telescoping sums, absolute/conditional convergence, intervals of convergence, Taylor polynomials, parametric slopes, dot and cross products, lines and planes, velocity and acceleration, partial derivatives, gradients and tangent planes, critical points, double integrals, potentials. Each problem comes with a hint, the answer and a step-by-step solution, and every generator's keys are tested against independent computations.

Mock quiz: two random problems against the clock

Pick the topics you expect, start the timer, and solve on paper. Hints and solutions stay hidden until you finish; then you get a score out of 3, like the real quizzes.

Past quizzes · Fall 2026

Every past quiz, with solutions

All 13 quiz papers from the folder, newest first, from eight sections (CRNs 10890–10897). Every task has answer boxes, the solution from the sheet rewritten step by step, and a link to the original PDF. Where a sheet contains a mistake, the problem says so and the Errata list it.

In plain words

How to use the papers

Pick a variant, press Start timer, and solve it on paper in the real time limit. Hints and solutions stay hidden until you stop. Type your final answers, press Finish for a score out of 3, then read the solutions. Do the other variant of the same paper the next day.

What the quizzes test

Every paper so far is built from a small set of problem types. The table shows which types appear where, and is the basis for the predicted next quizzes.

Problem type	Appeared in	Learn it in
Limit of a sequence (conjugate, \(e\)-limit, sandwich, growth)	Quiz 1: CRNs 10890, 10891, 10892, 10893, 10895–97, unlabelled	10.1
Converge or diverge? (integral, comparison, ratio, root test)	Quiz 1: 10890, 10891, 10893, 10895–97, unlabelled · Quiz 2: unlabelled	10.3–10.6
Sum of a geometric or telescoping series	Quiz 1: 10893 · Quiz 2: 10892	10.2
Absolute, conditional, or divergent	Quiz 2: 10890, 10891, 10895, 10897	10.5–10.6
Radius and interval of convergence	Quiz 2: 10890, 10891, 10893, 10896	10.7
Taylor / Maclaurin series, first nonzero terms	Quiz 2: 10895, 10896, 10897	10.8–10.10
Angles, projections, cross products, volumes, spheres	Quiz 3: 10895, 10896, 10897	12.1–12.4

The CRN-to-paper match comes from the file names and headers. The Quiz 1 file for CRNs 10895–10897 contains three papers (A/B variants each) without saying which CRN got which, so they are labelled Paper 1–3. Two sheets do not show a CRN at all; they are labelled “unlabelled”.

Quiz 2 · CRN not stated on the sheet

Full solution

Conditions (0.5 pt). \(f(x)=\frac{x}{x^2+4}\) is positive and continuous on \([1,\infty)\). \(f'(x)=\frac{4-x^2}{(x^2+4)^2}<0\) for \(x>2\), so \(f\) is eventually decreasing, which is enough.
Integral (0.5 pt). With \(u=x^2+4\), \(du=2x\,dx\): \(\displaystyle\int_1^t\frac{x\,dx}{x^2+4}=\frac12\int_5^{t^2+4}\frac{du}{u}=\frac12\ln\frac{t^2+4}{5}\to\infty\).
Conclusion (0.5 pt). The integral diverges, so the series diverges by the integral test.

Note on the sheet. The handwritten solution substitutes \(u=x^2+4\) but keeps the old limits \(1\) and \(t\), and writes the result as \(\frac12\ln|t|\). The correct antiderivative evaluation is \(\frac12\ln\frac{t^2+4}{5}\) (see Errata). The conclusion, divergence, is still right.

Question 2

sheet p. 4

Determine whether \(\displaystyle\sum_{n=1}^\infty\sqrt{\frac{n+4}{n^4+4}}\) converges or diverges.

Hint

Dominant terms: \(\sqrt{\frac{n}{n^4}}=\frac{1}{n^{3/2}}\).

Answer

Converges (limit comparison with \(\frac1{n^{3/2}}\), limit \(1\)).

Full solution

Comparison series (0.5 pt). \(b_n=\frac{1}{n^{3/2}}\), a convergent \(p\)-series (\(p=\frac32>1\)).
Limit (0.5 pt). \(\dfrac{a_n}{b_n}=n^{3/2}\sqrt{\dfrac{n+4}{n^4+4}}=\sqrt{\dfrac{n^3(n+4)}{n^4+4}}=\sqrt{\dfrac{1+4/n}{1+4/n^4}}\to1\).
Conclusion (0.5 pt). \(0<1<\infty\), so the series converges like \(\sum\frac1{n^{3/2}}\).

The sheet writes the limit as \(\frac{\lim\sqrt{1/n^3}}{\lim 1/n^{3/2}}\), a quotient of two limits that are both \(0\). The intended computation is the limit of the ratio, as above.

Quiz 3 · CRNs 10895, 10896, 10897

12.1–12.4

Solutions dated 15 Oct 20266 variants × 2 questions × 1.5 ptsAngles, spheres, regions, cross products, projectionsOpen the PDF ↗

CRN 10895 · Variant 1

Question 1: angles of a triangle

sheet p. 1

Find the measures of the angles of the triangle with vertices \(A=(-1,0)\), \(B=(2,1)\), \(C=(1,-2)\). (Radians, or degrees with deg.)

Hint

At each vertex use the two side vectors starting there, and \(\cos\theta=\frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}\).

Answer

\(A=C=\cos^{-1}\frac1{\sqrt5}\approx63.435^\circ\), \(B=\cos^{-1}\frac35\approx53.130^\circ\).

Full solution

At \(A\): \(\overrightarrow{AB}=\langle3,1\rangle\), \(\overrightarrow{AC}=\langle2,-2\rangle\); \(\cos A=\frac{6-2}{\sqrt{10}\sqrt8}=\frac{4}{4\sqrt5}=\frac1{\sqrt5}\).
At \(B\): \(\overrightarrow{BA}=\langle-3,-1\rangle\), \(\overrightarrow{BC}=\langle-1,-3\rangle\); \(\cos B=\frac{3+3}{\sqrt{10}\sqrt{10}}=\frac35\).
At \(C\): \(\overrightarrow{CA}=\langle-2,2\rangle\), \(\overrightarrow{CB}=\langle1,3\rangle\); \(\cos C=\frac{-2+6}{\sqrt8\sqrt{10}}=\frac1{\sqrt5}\).
\(A=C\approx63.435^\circ\), \(B\approx53.130^\circ\); the sum is \(180^\circ\) as a check.

Question 2: interior and exterior

sheet p. 1

Write inequalities describing (a) the interior and (b) the exterior of the sphere of radius 1 centered at \((1,1,1)\).

Hint

Squared distance to the center compared with \(1\). The sphere itself belongs to neither.

Answer

(a) \((x-1)^2+(y-1)^2+(z-1)^2<1\); (b) \((x-1)^2+(y-1)^2+(z-1)^2>1\).

Full solution

The squared distance from \((x,y,z)\) to \((1,1,1)\) is \(d^2=(x-1)^2+(y-1)^2+(z-1)^2\).
Interior: distance strictly less than 1, \(d^2<1\). Exterior: strictly greater, \(d^2>1\). Equality is the sphere itself, excluded from both.

CRN 10895 · Variant 2

Question 1: angle between diagonals

sheet p. 2

Find the angles between the diagonals of the rectangle with vertices \(A=(1,0)\), \(B=(0,3)\), \(C=(3,4)\), \(D=(4,1)\).

Hint

The diagonals are \(AC\) and \(BD\).

Answer

\(90^\circ\) (all four angles): the diagonals are perpendicular.

Full solution

\(\overrightarrow{AC}=\langle2,4\rangle\), \(\overrightarrow{BD}=\langle4,-2\rangle\).
\(\overrightarrow{AC}\cdot\overrightarrow{BD}=8-8=0\), so \(\cos\theta=0\) and \(\theta=90^\circ\). (This “rectangle” is in fact a square.)

Question 2: a closed shell

sheet p. 2

Write inequalities describing the closed region bounded by the spheres of radii 1 and 2 centered at the origin.

Hint

Closed means both spheres are included.

Answer

\(1\le x^2+y^2+z^2\le4\)

Full solution

With \(r=\sqrt{x^2+y^2+z^2}\): outside the radius-1 sphere and inside the radius-2 sphere, boundaries included, so \(1\le r\le2\).
Squaring (all quantities are nonnegative): \(1\le x^2+y^2+z^2\le4\).

CRN 10896 · Variant 1

Question 1: volume of a parallelepiped

sheet p. 3

Find the volume of the parallelepiped determined by \(\mathbf u=\mathbf i-\mathbf j+\mathbf k\), \(\mathbf v=2\mathbf i+\mathbf j-2\mathbf k\), \(\mathbf w=-\mathbf i+2\mathbf j-\mathbf k\).

Hint

\(V=|\mathbf u\cdot(\mathbf v\times\mathbf w)|\).

Answer

\(4\)

Full solution

\(\mathbf v\times\mathbf w=(1\cdot(-1)-(-2)\cdot2)\mathbf i-(2(-1)-(-2)(-1))\mathbf j+(2\cdot2-1\cdot(-1))\mathbf k=3\mathbf i+4\mathbf j+5\mathbf k\).
\(\mathbf u\cdot(\mathbf v\times\mathbf w)=3-4+5=4\), so \(V=|4|=4\).

Question 2: center and radius

sheet p. 3

Find the center and radius of the sphere \(2x^2+2y^2+2z^2+x+y+z=9\).

Hint

Divide by 2 first; then add \(\frac1{16}\) three times.

Answer

Center \(\left(-\frac14,-\frac14,-\frac14\right)\), radius \(\frac{5\sqrt3}{4}\).

Full solution

Divide by 2: \(x^2+\frac12x+y^2+\frac12y+z^2+\frac12z=\frac92\).
Complete the squares: \(\left(x+\frac14\right)^2+\left(y+\frac14\right)^2+\left(z+\frac14\right)^2=\frac92+3\cdot\frac1{16}=\frac{75}{16}\).
Center \(\left(-\frac14,-\frac14,-\frac14\right)\), radius \(\sqrt{\frac{75}{16}}=\frac{5\sqrt3}{4}\).

CRN 10896 · Variant 2

Question 1: volume of a parallelepiped

sheet p. 4

Find the volume of the parallelepiped determined by \(\mathbf u=\mathbf i+\mathbf j-2\mathbf k\), \(\mathbf v=-\mathbf i-\mathbf k\), \(\mathbf w=2\mathbf i+4\mathbf j-2\mathbf k\).

Hint

\(\mathbf v\times\mathbf w=\langle4,-4,-4\rangle\).

Answer

\(8\)

Full solution

\(\mathbf v\times\mathbf w=(0\cdot(-2)-(-1)\cdot4)\mathbf i-((-1)(-2)-(-1)(2))\mathbf j+((-1)(4)-0)\mathbf k=4\mathbf i-4\mathbf j-4\mathbf k\).
\(\mathbf u\cdot(\mathbf v\times\mathbf w)=4-4+8=8\), so \(V=8\).

Question 2: center and radius

sheet p. 4

Find the center and radius of the sphere \(3x^2+3y^2+3z^2+2y-2z=9\).

Hint

Divide by 3: \(x^2+y^2+\frac23y+z^2-\frac23z=3\).

Answer

Center \(\left(0,-\frac13,\frac13\right)\), radius \(\frac{\sqrt{29}}{3}\).

Full solution

Divide by 3 and complete squares: \(x^2+\left(y+\frac13\right)^2+\left(z-\frac13\right)^2=3+\frac19+\frac19=\frac{29}{9}\).
Center \(\left(0,-\frac13,\frac13\right)\), radius \(\frac{\sqrt{29}}{3}\).

CRN 10897 · Variant 1

Question 1: area and unit normal

sheet p. 5

Find the area of triangle \(PQR\) and a unit vector perpendicular to plane \(PQR\), where \(P=(1,-1,2)\), \(Q=(2,0,-1)\), \(R=(0,2,1)\).

Hint

\(\overrightarrow{PQ}\times\overrightarrow{PR}=\langle8,4,4\rangle\).

Answer

Area \(2\sqrt6\); \(\mathbf n=\pm\frac1{\sqrt6}(2\mathbf i+\mathbf j+\mathbf k)\).

Full solution

\(\overrightarrow{PQ}=\langle1,1,-3\rangle\), \(\overrightarrow{PR}=\langle-1,3,-1\rangle\), cross product \(\langle8,4,4\rangle\).
Area \(=\frac12\sqrt{64+16+16}=\frac12\sqrt{96}=2\sqrt6\).
Unit normal: \(\frac{\langle8,4,4\rangle}{\sqrt{96}}=\frac{\langle2,1,1\rangle}{\sqrt6}\); either sign is acceptable.

Question 2: vector projection

sheet p. 5

Find \(\operatorname{proj}_{\mathbf v}\mathbf u\) for \(\mathbf v=10\mathbf i+11\mathbf j-2\mathbf k\), \(\mathbf u=3\mathbf j+4\mathbf k\).

Hint

\(\mathbf u\cdot\mathbf v=25\), \(|\mathbf v|^2=225\).

Answer

\(\frac{10}{9}\mathbf i+\frac{11}{9}\mathbf j-\frac29\mathbf k\)

Full solution

\(\mathbf u\cdot\mathbf v=0+33-8=25\), \(|\mathbf v|^2=100+121+4=225\).
\(\operatorname{proj}_{\mathbf v}\mathbf u=\frac{25}{225}\langle10,11,-2\rangle=\frac19\langle10,11,-2\rangle\).

CRN 10897 · Variant 2

Question 1: area and unit normal

sheet p. 6

Find the area of triangle \(PQR\) and a unit vector perpendicular to plane \(PQR\), where \(P=(1,1,1)\), \(Q=(2,1,3)\), \(R=(3,-1,1)\).

Hint

\(\overrightarrow{PQ}\times\overrightarrow{PR}=\langle4,4,-2\rangle\), of length \(6\).

Answer

Area \(3\); \(\mathbf n=\pm\frac13(2\mathbf i+2\mathbf j-\mathbf k)\).

Full solution

\(\overrightarrow{PQ}=\langle1,0,2\rangle\), \(\overrightarrow{PR}=\langle2,-2,0\rangle\), cross product \(\langle4,4,-2\rangle\), length \(\sqrt{36}=6\).
Area \(=\frac12\cdot6=3\); \(\mathbf n=\pm\frac16\langle4,4,-2\rangle=\pm\frac13\langle2,2,-1\rangle\).

Question 2: vector projection

sheet p. 6

Find \(\operatorname{proj}_{\mathbf v}\mathbf u\) for \(\mathbf v=2\mathbf i+10\mathbf j-11\mathbf k\), \(\mathbf u=2\mathbf i+2\mathbf j+\mathbf k\).

Hint

\(\mathbf u\cdot\mathbf v=13\), \(\mathbf v\cdot\mathbf v=225\).

Answer

\(\frac{26}{225}\mathbf i+\frac{26}{45}\mathbf j-\frac{143}{225}\mathbf k\)

Full solution

\(\mathbf u\cdot\mathbf v=4+20-11=13\), \(\mathbf v\cdot\mathbf v=4+100+121=225\).
\(\operatorname{proj}_{\mathbf v}\mathbf u=\frac{13}{225}\langle2,10,-11\rangle=\left\langle\frac{26}{225},\frac{26}{45},-\frac{143}{225}\right\rangle\).

Quiz 2 · CRN 10890

10.5–10.7

Solutions file dated 13 Oct 20262 variants × 2 questionsAbsolute/conditional, radius and intervalOpen the PDF ↗

Variant A

Question 1

sheet p. 1

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{\sqrt n}\) converges absolutely, converges conditionally, or diverges.

Hint

Alternating series test for convergence; \(p\)-series for \(\sum|a_n|\).

Answer

Converges conditionally.

Full solution

Alternating form. \(\sum(-1)^{n+1}b_n\) with \(b_n=\frac1{\sqrt n}>0\).
Decreasing. \(n+1>n\Rightarrow\sqrt{n+1}>\sqrt n\Rightarrow\frac{1}{\sqrt{n+1}}<\frac1{\sqrt n}\).
Limit zero. \(\frac1{\sqrt n}\to0\). By the alternating series test the series converges.
Absolute? \(\sum\frac1{n^{1/2}}\) is a \(p\)-series with \(p=\frac12\le1\): diverges. So the convergence is conditional.

Question 2

sheet p. 2

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{n(x+1)^n}{4^n}\).

Hint

Ratio test gives \(\frac{|x+1|}{4}\). At the endpoints the terms are \(n(\pm1)^n\).

Answer

\(R=4\); interval \((-5,3)\).

Full solution

Ratio test (center \(-1\)): \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+1}{n}\cdot\frac{|x+1|}{4}\to\frac{|x+1|}{4}\).
\(\frac{|x+1|}4<1\iff-5<x<3\): \(R=4\).
\(x=-5\): \(\sum n(-1)^n\); the terms do not tend to \(0\): excluded.
\(x=3\): \(\sum n\) diverges: excluded. Interval \((-5,3)\).

Variant B

Question 1

sheet p. 3

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{n}{n^2+1}\) converges absolutely, converges conditionally, or diverges.

Hint

Show \(b_{n}-b_{n+1}>0\); compare \(\frac{n}{n^2+1}\) with \(\frac1{2n}\).

Answer

Converges conditionally.

Full solution

\(b_n=\frac{n}{n^2+1}>0\).
Decreasing: \(b_n-b_{n+1}=\dfrac{n(n^2+2n+2)-(n+1)(n^2+1)}{(n^2+1)(n^2+2n+2)}=\dfrac{n^2+n-1}{(n^2+1)(n^2+2n+2)}>0\) for \(n\ge1\).
Limit: \(\frac{n}{n^2+1}=\frac{1/n}{1+1/n^2}\to0\). The AST gives convergence.
Not absolute: \(n^2+1\le2n^2\Rightarrow\frac{n}{n^2+1}\ge\frac1{2n}\), and \(\sum\frac1{2n}\) diverges. So conditionally convergent.

Question 2

sheet p. 4

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(x-2)^n}{n3^n}\).

Hint

At \(x=-1\): \(\sum\frac{(-1)^n}{n}\); at \(x=5\): \(\sum\frac1n\).

Answer

\(R=3\); interval \([-1,5)\).

Full solution

Ratio test: \(\frac{|x-2|}{3}\cdot\frac{n}{n+1}\to\frac{|x-2|}{3}<1\iff-1<x<5\); \(R=3\).
\(x=-1\): \(\sum\frac{(-3)^n}{n3^n}=\sum\frac{(-1)^n}{n}\) converges by the AST (conditionally): included.
\(x=5\): \(\sum\frac1n\) diverges: excluded.
Interval \([-1,5)\).

Quiz 2 · CRN 10891

10.5–10.7

Solutions file dated 13 Oct 20262 variants × 2 questionsAbsolute/conditional, radius and intervalOpen the PDF ↗

Variant A

Question 1

sheet p. 1

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^n\frac{\arctan(2n)}{n^2+4}\) converges absolutely, converges conditionally, or diverges.

Hint

\(0<\arctan t<\frac\pi2\) for \(t>0\).

Answer

Converges absolutely.

Full solution

Absolute-value series: \(\left|(-1)^n\frac{\arctan2n}{n^2+4}\right|=\frac{\arctan2n}{n^2+4}\) (positive since \(2n>0\)).
Bound: \(0\le\frac{\arctan2n}{n^2+4}\le\frac{\pi/2}{n^2+4}\le\frac{\pi}{2n^2}\).
\(\sum\frac{\pi}{2n^2}\) converges (\(p=2\)), so by comparison \(\sum|a_n|\) converges: absolutely convergent. No alternating series test is needed.

Question 2

sheet p. 2

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(x+1)^n}{\sqrt n}\).

Hint

At \(x=-2\): \(\sum\frac{(-1)^n}{\sqrt n}\) (AST); at \(x=0\): \(\sum\frac1{\sqrt n}\).

Answer

\(R=1\); interval \([-2,0)\).

Full solution

Ratio test (center \(-1\)): \(|x+1|\sqrt{\frac{n}{n+1}}\to|x+1|<1\iff-2<x<0\); \(R=1\).
\(x=-2\): \(\sum\frac{(-1)^n}{\sqrt n}\): \(b_n=\frac1{\sqrt n}\) decreases to \(0\), so it converges, conditionally (\(p=\frac12\)). Included.
\(x=0\): \(\sum\frac1{\sqrt n}\) diverges. Excluded.
Interval \([-2,0)\).

Variant B

Question 1

sheet p. 3

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^n\frac{(n+2)^n}{(4n)^n}\) converges absolutely, converges conditionally, or diverges.

Hint

Root test on \(|a_n|=\left(\frac{n+2}{4n}\right)^n\).

Answer

Converges absolutely.

Full solution

\(|a_n|=\left(\frac{n+2}{4n}\right)^n\).
Root test: \(\sqrt[n]{|a_n|}=\frac{n+2}{4n}=\frac14+\frac1{2n}\to\frac14<1\).
So \(\sum|a_n|\) converges: absolutely convergent.

Question 2

sheet p. 4

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(3x+1)^n}{n}\).

Hint

\(|3x+1|<1\iff\left|x+\frac13\right|<\frac13\).

Answer

\(R=\frac13\); interval \(\left[-\frac23,0\right)\).

Full solution

Ratio test: \(|3x+1|\frac{n}{n+1}\to|3x+1|<1\iff-\frac23<x<0\). Since \(3x+1=3\left(x+\frac13\right)\): center \(-\frac13\), \(R=\frac13\).
\(x=-\frac23\): \(3x+1=-1\), so \(\sum\frac{(-1)^n}{n}\) converges by the AST (conditionally). Included.
\(x=0\): \(\sum\frac1n\), harmonic, diverges. Excluded.

Quiz 2 · CRN 10893

10.7

Dated 11 Oct 202610 minutes · 1 exercise · 3 ptsRadius and interval of convergenceOpen the PDF ↗

Exercise 1 (3 pts)

Radius and interval of convergence

sheet p. 1–2

Find the radius of convergence and the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^nx^n}{n2^n}\).

Watch out

The sheet prints n = 0

The quiz sheet writes the sum from \(n=0\), but the \(n=0\) term \(\frac{(-1)^0x^0}{0\cdot2^0}\) divides by zero. The solution itself (correctly) works with \(n=1\) when it substitutes the endpoints, so the series is meant to start at \(n=1\). Listed in the Errata.

Hint

Ratio test: \(\frac{|x|}{2}\cdot\frac{n}{n+1}\). At \(x=-2\) the two \((-1)^n\) cancel.

Answer

\(R=2\); interval \((-2,2]\).

Full solution

\(\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{x^{n+1}}{x^n}\cdot\dfrac{2^n}{2^{n+1}}\cdot\dfrac{n}{n+1}\right|=\dfrac{|x|}{2}\cdot\dfrac{n}{n+1}\to\dfrac{|x|}{2}\).
Absolute convergence when \(\frac{|x|}{2}<1\iff-2<x<2\): \(R=2\).
\(x=-2\): \(\sum\frac{(-1)^n(-1)^n2^n}{n2^n}=\sum\frac{(-1)^{2n}}{n}=\sum\frac1n\), harmonic, diverges.
\(x=2\): \(\sum\frac{(-1)^n}{n}\), alternating harmonic: \(b_n=\frac1n>0\), \(\frac1{n+1}\le\frac1n\), \(\frac1n\to0\), so it converges.
\(R=2\), interval \((-2,2]\).

Quiz 2 · CRNs 10895, 10896, 10897

10.5–10.8

Solutions dated 9 Oct 20266 variants × 2 questions × 1.5 ptsAbsolute/conditional, intervals, Taylor and Maclaurin seriesOpen the PDF ↗

CRN 10895 · Variant 1

Question 1

sheet p. 1

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{1+n}{n^2}\) converges absolutely, converges conditionally, or diverges.

Hint

\(\frac{n+1}{n^2}=\frac1n+\frac1{n^2}\ge\frac1n\).

Answer

Converges conditionally.

Full solution

\(\sum\left|a_n\right|=\sum\left(\frac1n+\frac1{n^2}\right)\ge\sum\frac1n\), which diverges: not absolutely convergent.
\(b_n=\frac1n+\frac1{n^2}>0\) decreases (both parts decrease) and \(b_n\to0\): the alternating series test gives convergence.
Converges, but not absolutely: conditionally convergent.

Question 2

sheet p. 1

Find the Maclaurin series for \(f(x)=\dfrac{2+x}{1-x}\). (Type the terms up to \(x^3\).)

Hint

\(\frac{2+x}{1-x}=\frac{3}{1-x}-1\).

Answer

\(2+3x+3x^2+3x^3+\cdots=2+3\sum_{n=1}^\infty x^n\)

Full solution

Rewrite: \(f(x)=3(1-x)^{-1}-1\), so \(f(0)=2\).
For \(n\ge1\): \(f^{(n)}(x)=3\,n!\,(1-x)^{-n-1}\), so \(f^{(n)}(0)=3n!\) and \(c_n=\frac{3n!}{n!}=3\).
Series: \(2+3x+3x^2+3x^3+\cdots=2+3\sum_{n\ge1}x^n\) (valid for \(|x|<1\)).

CRN 10895 · Variant 2

Question 1

sheet p. 2

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n+3}\) converges absolutely, converges conditionally, or diverges.

Hint

\(\sum\frac1{n+3}=\sum_{k\ge4}\frac1k\).

Answer

Converges conditionally.

Full solution

\(\sum\frac1{n+3}=\sum_{k=4}^\infty\frac1k\): the harmonic series without its first three terms, still divergent.
\(b_n=\frac1{n+3}\): positive, \(b_{n+1}=\frac1{n+4}<b_n\), and \(b_n\to0\). The AST gives convergence.
Conditionally convergent.

Question 2

sheet p. 2

Find the Maclaurin series for \(f(x)=\dfrac{x^2}{x+1}\). (Type the terms up to \(x^5\).)

Hint

\(\frac{1}{1+x}=\sum(-1)^nx^n\); multiply by \(x^2\).

Answer

\(\sum_{n=0}^\infty(-1)^nx^{n+2}=x^2-x^3+x^4-x^5+\cdots\)

Full solution

Geometric series with \(r=-x\): \(\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^nx^n\) for \(|x|<1\).
Multiply by \(x^2\): \(\frac{x^2}{1+x}=\sum_{n=0}^\infty(-1)^nx^{n+2}=x^2-x^3+x^4-x^5+\cdots\)

CRN 10896 · Variant 1

Question 1

sheet p. 3

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(3x-2)^n}{n}\), identifying where convergence is absolute or conditional.

Hint

\(3x-2=3\left(x-\frac23\right)\).

Answer

\(R=\frac13\); \(\left[\frac13,1\right)\): absolute on \(\left(\frac13,1\right)\), conditional at \(x=\frac13\).

Full solution

Ratio test: \(\left|\frac{a_{n+1}}{a_n}\right|=|3x-2|\frac{n}{n+1}\to|3x-2|<1\iff\frac13<x<1\). Center \(\frac23\), \(R=\frac13\).
\(x=\frac13\): \(3x-2=-1\), \(\sum\frac{(-1)^n}{n}\) converges conditionally (AST, harmonic absolute series).
\(x=1\): \(\sum\frac1n\) diverges.

Question 2

sheet p. 3

Find the Taylor series generated by \(f(x)=\dfrac1{x^2}\) at \(a=1\): give the general coefficient \(c_n\) of \((x-1)^n\) and the polynomial up to \((x-1)^3\).

Hint

\(f^{(n)}(x)=(-1)^n(n+1)!\,x^{-n-2}\).

Answer

\(c_n=(-1)^n(n+1)\): \(1-2(x-1)+3(x-1)^2-4(x-1)^3+\cdots\)

Full solution

\(f'=-2x^{-3}\), \(f''=6x^{-4}\), \(f'''=-24x^{-5}\); in general \(f^{(n)}(x)=(-1)^n(n+1)!\,x^{-n-2}\).
\(f^{(n)}(1)=(-1)^n(n+1)!\), so \(c_n=\frac{(-1)^n(n+1)!}{n!}=(-1)^n(n+1)\).

CRN 10896 · Variant 2

Question 1

sheet p. 4

Find the radius and interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n(x+2)^n}{n}\).

Hint

At \(x=-3\) the two \((-1)^n\) cancel.

Answer

\(R=1\); interval \((-3,-1]\), conditional at \(x=-1\).

Full solution

Ratio test: \(|x+2|\frac{n}{n+1}\to|x+2|<1\iff-3<x<-1\); center \(-2\), \(R=1\).
\(x=-3\): \(\frac{(-1)^n(-1)^n}{n}=\frac1n\), harmonic, diverges.
\(x=-1\): \(\sum\frac{(-1)^n}{n}\) converges conditionally.

Question 2

sheet p. 4

Find the Taylor series generated by \(f(x)=\dfrac{1}{(1-x)^3}\) at \(a=0\): give the coefficient \(c_n\) of \(x^n\).

Hint

\(f^{(n)}(x)=\frac{(n+2)!}{2!}(1-x)^{-n-3}\).

Answer

\(c_n=\frac{(n+1)(n+2)}{2}\): \(1+3x+6x^2+10x^3+\cdots\)

Full solution

\(f'=3(1-x)^{-4}\), \(f''=3\cdot4(1-x)^{-5}\), \(f'''=3\cdot4\cdot5(1-x)^{-6}\); in general \(f^{(n)}(0)=\frac{(n+2)!}{2}\).
\(c_n=\frac{(n+2)!}{2\,n!}=\frac{(n+1)(n+2)}{2}\).

CRN 10897 · Variant 1

Question 1

sheet p. 5

Determine whether \(\displaystyle\sum_{n=1}^\infty(-1)^nn^2\left(\frac13\right)^n\) converges absolutely, converges conditionally, or diverges.

Hint

Ratio test on \(\frac{n^2}{3^n}\).

Answer

Converges absolutely.

Full solution

\(|a_n|=\frac{n^2}{3^n}\). Ratio: \(\frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=\frac13\left(1+\frac1n\right)^2\to\frac13<1\).
The series of absolute values converges: absolutely convergent.

Question 2

sheet p. 5

Find the first three nonzero terms of the Maclaurin series of \(f(x)=\cos x-\dfrac{3}{1-x}\) and the values of \(x\) for which the series converges absolutely.

Hint

\(f(0)=-2\), \(f'(0)=-3\), \(f''(0)=-7\). The cosine part converges everywhere; the geometric part on \((-1,1)\).

Answer

\(-2-3x-\frac72x^2+\cdots\); absolutely for \(-1<x<1\).

Full solution

\(f=\cos x-3(1-x)^{-1}\), \(f'=-\sin x-3(1-x)^{-2}\), \(f''=-\cos x-6(1-x)^{-3}\).
At \(0\): \(f(0)=1-3=-2\), \(f'(0)=-3\), \(f''(0)=-1-6=-7\). So \(-2-3x-\frac72x^2+\cdots\)
\(\cos x\): absolutely for all \(x\); \(\frac{1}{1-x}\): absolutely on \((-1,1)\). Together: exactly \(-1<x<1\).

CRN 10897 · Variant 2

Question 1

sheet p. 6

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{(-20)^n}{n!}\) converges absolutely, converges conditionally, or diverges.

Hint

Ratio test on \(\frac{20^n}{n!}\).

Answer

Converges absolutely.

Full solution

\(b_n=\frac{20^n}{n!}\): \(\frac{b_{n+1}}{b_n}=\frac{20}{n+1}\to0<1\).
So \(\sum|a_n|\) converges: absolutely convergent (the terms grow at first, up to \(n=20\), but that does not matter).

Question 2

sheet p. 6

Find the first three nonzero terms of the Maclaurin series of \(f(x)=(1-3x+2x^2)e^x\) and the values of \(x\) for which it converges absolutely.

Hint

With \(P=1-3x+2x^2\): \(f'=(P+P')e^x\), \(f''=(P+2P'+P'')e^x\).

Answer

\(1-2x-\frac12x^2+\cdots\), for all real \(x\).

Full solution

\(P(0)=1\), \(P'(0)=-3\), \(P''(0)=4\). So \(f(0)=1\), \(f'(0)=1-3=-2\), \(f''(0)=1-6+4=-1\).
\(1-2x-\frac12x^2+\cdots\) (check: \((1-3x+2x^2)(1+x+\frac{x^2}2)=1-2x-\frac12x^2+\cdots\)).
\(e^x\) converges absolutely everywhere, and multiplying by a polynomial keeps that: all \(x\in\mathbb R\).

Quiz 1 · CRN 10893

Diverges.

Full solution

\(p\)-series with \(p=0.2=\frac15<1\): diverges.

Quiz 1 · CRNs 10895, 10896, 10897

10.1–10.5

Solutions dated 30 Sep 20263 papers × variants A/B × 2 questionsConjugates, integral, ratio, root, limit comparisonOpen the PDF ↗

The file holds three papers without CRN labels (one per section); they are listed in file order as Papers 1–3.

Paper 1 · Variant A

Question 1

sheet p. 1, 7

Determine whether \(a_n=n\left(\sqrt{n^2+4}-n\right)\) converges or diverges. If it converges, find its limit, justifying every algebraic step.

Hint

Rationalize: multiply and divide by \(\sqrt{n^2+4}+n\).

Answer

Converges to \(2\).

Full solution

\(a_n=n\cdot\dfrac{(n^2+4)-n^2}{\sqrt{n^2+4}+n}=\dfrac{4n}{\sqrt{n^2+4}+n}\).
Divide by \(n\): \(a_n=\dfrac{4}{\sqrt{1+4/n^2}+1}\to\dfrac42=2\).

Question 2

sheet p. 1, 7

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{\ln n}{n^{3/2}}\) converges or diverges. Choose an appropriate test and justify why it applies.

Hint

Integral test; integrate by parts with \(u=\ln x\), \(dv=x^{-3/2}dx\).

Answer

Converges; the integral is \(\frac{2\ln2+4}{\sqrt2}\).

Full solution

\(f(x)=\frac{\ln x}{x^{3/2}}\) is positive and continuous for \(x\ge2\) and eventually decreasing: \(f'(x)=\frac{1-\frac32\ln x}{x^{5/2}}<0\) for \(x>e^{2/3}\approx1.95\).
By parts: \(\int\frac{\ln x}{x^{3/2}}dx=-\frac{2\ln x}{\sqrt x}-\frac{4}{\sqrt x}+C\).
\(\int_2^\infty=\left[-\frac{2\ln x}{\sqrt x}-\frac{4}{\sqrt x}\right]_2^\infty=0+\frac{2\ln2+4}{\sqrt2}<\infty\) (using \(\frac{\ln x}{\sqrt x}\to0\)). The series converges.

Paper 1 · Variant B

Question 1

sheet p. 2, 8

Determine whether \(a_n=n\left(\sqrt{n^2+9}-n\right)\) converges or diverges. If it converges, find its limit.

Hint

Same conjugate trick: \(a_n=\frac{9n}{\sqrt{n^2+9}+n}\).

Answer

Converges to \(\frac92\).

Full solution

\(a_n=n\cdot\dfrac{9}{\sqrt{n^2+9}+n}=\dfrac{9}{\sqrt{1+9/n^2}+1}\to\dfrac92\).

Question 2

sheet p. 2, 8

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{\ln n}{n^{4/3}}\) converges or diverges.

Hint

By parts: \(\int\ln x\,x^{-4/3}dx=-3x^{-1/3}\ln x-9x^{-1/3}+C\).

Answer

Converges; the integral is \(\frac{3\ln2+9}{2^{1/3}}\).

Full solution

\(f(x)=\frac{\ln x}{x^{4/3}}\): positive, continuous, eventually decreasing.
\(\int_2^\infty\frac{\ln x}{x^{4/3}}dx=\left[-3x^{-1/3}\ln x-9x^{-1/3}\right]_2^\infty=\frac{3\ln2+9}{2^{1/3}}<\infty\).
By the integral test the series converges.

Paper 2 · Variant A

Question 1

sheet p. 3, 9

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}\) converges or diverges.

Hint

Integral test with \(u=\ln x\): \(\int u^{-1/2}du\).

Answer

Diverges.

Full solution

\(f(x)=\frac{1}{x\sqrt{\ln x}}\) is positive, continuous and decreasing for \(x\ge2\): the integral test applies.
\(u=\ln x\): \(\int_2^\infty\frac{dx}{x\sqrt{\ln x}}=\int_{\ln2}^\infty u^{-1/2}du=\left[2\sqrt u\right]_{\ln2}^\infty=\infty\).
The series diverges.

Question 2

sheet p. 3, 9

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{(n!)^2}{(2n)!}\) converges or diverges, showing the essential limit calculation.

Hint

\(\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(2n+2)(2n+1)}\).

Answer

Converges (\(\rho=\frac14\)).

Full solution

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)^2}{(2n+2)(2n+1)}=\dfrac{n+1}{2(2n+1)}\to\dfrac14<1\).
By the ratio test the series converges.

Paper 2 · Variant B

Question 1

sheet p. 4, 10

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^{2/3}}\) converges or diverges. Verify the hypotheses of your test.

Hint

Integral test: \(\int u^{-2/3}du=3u^{1/3}\).

Answer

Diverges.

Full solution

\(f(x)=\frac{1}{x(\ln x)^{2/3}}\) is positive, continuous and decreasing for \(x>1\).
With \(u=\ln x\): \(\int_{\ln2}^\infty u^{-2/3}du=\left[3u^{1/3}\right]_{\ln2}^\infty=\infty\).
The series diverges.

Question 2

sheet p. 4, 10

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{(n!)^2}{(2n+1)!}\) converges or diverges.

Hint

\(\frac{(2n+1)!}{(2n+3)!}=\frac{1}{(2n+3)(2n+2)}\).

Answer

Converges (\(\rho=\frac14\)).

Full solution

\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)^2}{(2n+3)(2n+2)}=\dfrac{n+1}{2(2n+3)}\to\dfrac14<1\).
Converges by the ratio test.

Paper 3 · Variant A

Question 1

sheet p. 5, 11

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{\sqrt{n^2+1}}{n^2+3n+1}\) converges or diverges, choosing a suitable comparison series.

Hint

Dominant terms: \(\frac{n}{n^2}=\frac1n\).

Answer

Diverges (limit comparison with \(\frac1n\), limit \(1\)).

Full solution

Compare with \(b_n=\frac1n\): \(\lim\dfrac{\sqrt{n^2+1}/(n^2+3n+1)}{1/n}=\lim\dfrac{\sqrt{1+1/n^2}}{1+3/n+1/n^2}=1\).
\(0<1<\infty\) and \(\sum\frac1n\) diverges, so the series diverges.

Question 2

sheet p. 5, 11

Determine whether \(\displaystyle\sum_{n=1}^\infty\left(\frac{3n+2}{4n+1}\right)^n\) converges or diverges.

Hint

Root test.

Answer

Converges (\(\rho=\frac34\)).

Full solution

\(\sqrt[n]{a_n}=\frac{3n+2}{4n+1}\to\frac34<1\): converges by the root test.

Paper 3 · Variant B

Question 1

sheet p. 6, 12

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{\sqrt{n^2+4}}{n^2+n+2}\) converges or diverges.

Hint

Compare with \(\frac1n\).

Answer

Diverges.

Full solution

\(\lim\dfrac{a_n}{1/n}=\lim\dfrac{\sqrt{1+4/n^2}}{1+1/n+2/n^2}=1\).
The harmonic series diverges, so the given series diverges.

Question 2

sheet p. 6, 12

Determine whether \(\displaystyle\sum_{n=1}^\infty\left(\frac{2n+3}{3n+4}\right)^n\) converges or diverges.

Hint

Root test.

Answer

Converges (\(\rho=\frac23\)).

Full solution

\(\sqrt[n]{a_n}=\frac{2n+3}{3n+4}\to\frac23<1\): converges by the root test.

Quiz 2 · CRN 10892

10.2

Dated 28 Sep 202610 minutes · Exercise 1 (3 pts)Telescoping and geometric sumsOpen the PDF ↗

Exercise 1: find the sum of each series

(a) A telescoping series

sheet p. 1

Find \(\displaystyle\sum_{n=1}^\infty\left(\tan^{-1}(n)-\tan^{-1}(n+1)\right)\).

Hint

Write out \(S_N\); only two terms survive.

Answer

\(-\frac\pi4\)

Full solution

\(S_N=\left(\tan^{-1}1-\tan^{-1}2\right)+\left(\tan^{-1}2-\tan^{-1}3\right)+\cdots+\left(\tan^{-1}N-\tan^{-1}(N+1)\right)=\tan^{-1}1-\tan^{-1}(N+1)\).
\(S=\lim S_N=\frac\pi4-\frac\pi2=-\frac\pi4\).

(b) A sum of two series

sheet p. 1–2

Find \(\displaystyle\sum_{n=1}^\infty\left[\frac{3^{n+1}}{4^n}+\frac{5^n}{2^{n+1}}\right]\), or type diverges.

Hint

The first part converges to \(9\), but look at the ratio of the second part.

Answer

Diverges.

Full solution

First part: \(\sum\frac{3^{n+1}}{4^n}=3\sum\left(\frac34\right)^n=3\cdot\frac{3/4}{1-3/4}=9\).
Second part: \(\sum\frac{5^n}{2^{n+1}}=\frac12\sum\left(\frac52\right)^n\): geometric with \(\left|\frac52\right|>1\), divergent.
Convergent + divergent is divergent: the original series diverges.

Quiz 1 · CRN 10892

\(\frac1e\)

Full solution

\(\lim\left(\frac{n}{n+1}\right)^n=\lim\left(1+\frac1n\right)^{-n}=e^{-1}\).

Quiz 1 · CRN 10890

10.1–10.4

Question papers + solution sheet (Set 2)2 variants · Q1 (1.5 pts, two parts) + Q2 (1.5 pts)Sandwich, growth rates, comparisonOpen the PDF ↗

Variant A

Question 1(a)

sheet p. 3

Does \(a_n=\dfrac{3^n}{n!}\) converge or diverge? If it converges, find the limit.

Hint

Factorials beat exponentials. For a sandwich, write \(\frac{3^n}{n!}=\frac{3^3}{3!}\cdot\frac34\cdot\frac35\cdots\frac3n\).

Answer

Converges to \(0\).

Full solution

For \(n\ge3\): \(\dfrac{3^n}{n!}=\dfrac{3^3}{3!}\cdot\dfrac34\cdot\dfrac35\cdots\dfrac{3}{n-1}\cdot\dfrac3n\le\dfrac{27}{6}\cdot1\cdots1\cdot\dfrac3n=\dfrac{27}{2n}\), since every middle factor is \(\le1\).
So \(0<\frac{3^n}{n!}\le\frac{27}{2n}\to0\), and by the sandwich theorem \(\frac{3^n}{n!}\to0\). (0.75 pt)

Note on the sheet. The sheet bounds \(\frac{3^n}{n!}\le\frac9n\) “for \(n\ge3\)”. That inequality fails for \(n=3,4,5\) (at \(n=3\): \(\frac{27}{6}=4.5>\frac93=3\)) and only holds from \(n=6\) on. The sandwich theorem only needs the bound eventually, so the conclusion stands, but the stated range is wrong (see Errata).

Question 1(b)

sheet p. 3

Does \(a_n=\dfrac{\sin(4n)}{n}\) converge or diverge? If it converges, find the limit.

Hint

\(-1\le\sin4n\le1\).

Answer

Converges to \(0\).

Full solution

\(-\frac1n\le\frac{\sin4n}{n}\le\frac1n\) and both bounds tend to \(0\): sandwich gives \(0\). (0.75 pt)

Question 2

sheet p. 3

Determine whether \(\displaystyle\sum_{n=1}^\infty\frac{n^2+2n}{\sqrt{n^7+n^3}}\) converges or diverges.

Hint

Dominant terms \(\frac{n^2}{n^{7/2}}=\frac{1}{n^{3/2}}\).

Answer

Converges (limit comparison with \(p=\frac32\), limit \(1\)).

Full solution

Order (0.5): numerator \(\sim n^2\), denominator \(\sim n^{7/2}\), so \(a_n\sim\frac1{n^{3/2}}\).
Test (0.5): limit comparison with \(\sum\frac1{n^{3/2}}\), convergent since \(p=\frac32>1\).
Limit (0.5): \(\lim\frac{n^{7/2}+2n^{5/2}}{\sqrt{n^7+n^3}}=1\), finite and positive, so the series converges.

Answer

Converges.

Full solution

Bound: \(n\le\frac{5^n}{2}\) for all \(n\ge1\) (true at \(n=1\), and the right side grows much faster). So \(5^n-n\ge\frac{5^n}{2}>0\).
Hence \(0<\frac{1}{5^n-n}\le\frac{2}{5^n}=2\left(\frac15\right)^n\).
\(\sum2\left(\frac15\right)^n\) is a convergent geometric series (\(r=\frac15\)), so the series converges by the comparison test. (Limit comparison with \(\left(\frac15\right)^n\) also works: \(\lim\frac{5^n}{5^n-n}=1\).)

Quiz 1 · CRN 10891

10.1–10.3

Question papers + solution sheet2 variants · Q1 (1.5 pts) + Q2 (1.5 pts)Conjugate, log limit, integral testOpen the PDF ↗

Variant A

Question 1

sheet p. 4

Does \(a_n=\sqrt{n^2+4n}-n\) converge or diverge? If it converges, find its limit.

Hint

Multiply and divide by the conjugate \(\sqrt{n^2+4n}+n\).

Answer

Converges to \(2\).

Full solution

Rationalize (0.5): \(a_n=\dfrac{(n^2+4n)-n^2}{\sqrt{n^2+4n}+n}=\dfrac{4n}{\sqrt{n^2+4n}+n}\).
Divide by \(n\) (0.5): \(\dfrac{4}{\sqrt{1+4/n}+1}\).
Limit (0.5): \(\frac{4}{1+1}=2\): converges to \(2\).

Question 2

sheet p. 4

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{1}{n\ln n}\) converges or diverges.

Hint

\(u=\ln x\) gives \(\int\frac{du}{u}=\ln|u|\).

Answer

Diverges (the integral is \(\infty\)).

Full solution

Set-up (0.5): \(f(x)=\frac1{x\ln x}\) is continuous, positive and decreasing for \(x\ge2\); consider \(\lim_{t\to\infty}\int_2^t\frac{dx}{x\ln x}\).
Antiderivative (0.5): \(u=\ln x\): \(\int\frac{du}{u}=\ln|\ln x|\).
Limit (0.5): \(\lim_{t\to\infty}\left(\ln(\ln t)-\ln(\ln2)\right)=\infty\): the series diverges.

Variant B

Question 1

sheet p. 3

Does \(a_n=\ln(3n+5)-\ln(n+2)\) converge or diverge? If it converges, find its limit.

Hint

\(\ln A-\ln B=\ln\frac AB\).

Answer

Converges to \(\ln3\).

Full solution

Combine (0.5): \(a_n=\ln\dfrac{3n+5}{n+2}\).
Inner limit (0.5): \(\dfrac{3+5/n}{1+2/n}\to3\).
Conclude (0.5): \(\ln\) is continuous, so \(a_n\to\ln3\).

Question 2

sheet p. 3

Determine whether \(\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}\) converges or diverges.

Hint

The antiderivative is \(-\frac1{\ln x}\).

Answer

Converges (the integral is \(\frac1{\ln2}\)).

Full solution

Set-up (0.5): \(f(x)=\frac{1}{x(\ln x)^2}\) continuous, positive, decreasing for \(x\ge2\).
Antiderivative (0.5): \(u=\ln x\): \(\int u^{-2}du=-\frac1u=-\frac1{\ln x}\).
Limit (0.5): \(\left[-\frac1{\ln x}\right]_2^t\to0+\frac1{\ln2}=\frac1{\ln2}\), finite: converges.

Quiz 1 · CRN not stated on the sheet

10.1–10.2

MATH 1222 header, “6 minutes”2 variants × 2 questionsGeometric sequences, sandwich, nth-term testOpen the PDF ↗

Variant 1

Question 1

sheet p. 1

Find the limit of the sequence \(a_n=\dfrac{1}{(0.9)^n}\) (type inf if it diverges to \(\infty\)).

Hint

\(\frac{1}{0.9^n}=\left(\frac{10}{9}\right)^n\).

Answer

\(+\infty\): the sequence diverges.

Full solution

\(a_n=\left(\frac{10}{9}\right)^n\) and \(\frac{10}9>1\).
So \(a_n\) grows without bound: \(\lim a_n=+\infty\).

Question 2

sheet p. 1

Is \(\displaystyle\sum_{n=0}^\infty\frac{e^n}{e^n+n}\) convergent or divergent?

Hint

Divide top and bottom by \(e^n\).

Answer

Divergent.

Full solution

\(a_n=\dfrac{e^n}{e^n+n}=\dfrac{1}{1+n/e^n}\), and \(\frac{n}{e^n}\to0\).
So \(a_n\to1\ne0\): by the \(n\)th-term test the series diverges.

Variant 2

Question 1

sheet p. 2

Find the limit of the sequence \(a_n=\dfrac{\sin^2n}{2^n}\).

Hint

\(0\le\sin^2n\le1\).

Answer

\(0\)

Full solution

\(0\le\frac{\sin^2n}{2^n}\le\frac1{2^n}\), and \(\frac1{2^n}\to0\).
By the sandwich theorem, \(\lim a_n=0\).

Question 2

sheet p. 2

Is \(\displaystyle\sum_{n=0}^\infty\cos n\pi\) convergent or divergent?

Hint

\(\cos n\pi=(-1)^n\).

Answer

Divergent.

Full solution

\(a_n=\cos n\pi=(-1)^n\) alternates between \(1\) and \(-1\), so \(\lim a_n\) does not exist.
By the \(n\)th-term test the series diverges.

Predicted quizzes

The next quizzes, predicted

New quizzes for every remaining topic, built the way the real ones are: two problems, 1.5 points each, about 10 minutes, one standard problem type each, in two variants with different numbers.

In plain words

How these were predicted

1. The syllabus fixes the order: series (weeks 1–4), parametric curves (5), vectors, lines and planes (6), vector functions (7), several variables (8–11), double integrals (12–13), line integrals (14–15). 2. The 13 past papers show the format: every task so far is a textbook-standard type (see the table), and variants differ only in the numbers. 3. Pace: a quiz usually covers the previous lecture or two. So each predicted quiz below takes the two most standard problem types of its section, as they appear in Thomas' exercise lists. These are predictions, not leaked papers. If you can do all of them, you are ready for anything of this shape.

Where each section stood on 19 October 2026 (from the dates on the sheets):

CRNs 10895–10897: Quiz 3 (15 Oct) was 12.1–12.4 → next most likely lines and planes, then vector functions.
CRN 10893: Quiz 2 (11 Oct) was power series → next most likely Taylor series, then parametric curves or vectors.
CRNs 10890, 10891: Quiz 2 (≈13 Oct) was alternating series and power series → next most likely Taylor series, then vectors.
CRN 10892 and the unlabelled section: latest sheets cover series sums and tests → Taylor series and vectors.

Everyone should also prepare the midterm practice (Chapters 10–13) and, later, the final practice.

Predicted quiz · Taylor and Maclaurin series

10.8–10.10

Variant 1

Question 1 (1.5 pts)

predicted

Find the first four nonzero terms of the Maclaurin series of \(f(x)=xe^{-x}\).

Hint

\(e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots\), then multiply by \(x\).

Answer

\(x-x^2+\frac{x^3}{2}-\frac{x^4}{6}\)

Full solution

\(e^{-x}=\sum\frac{(-x)^n}{n!}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots\)
Multiply by \(x\): \(x-x^2+\frac{x^3}{2}-\frac{x^4}{6}+\cdots=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!}\).

Question 2 (1.5 pts)

predicted

The Taylor series of \(f(x)=\ln x\) about \(a=2\) is \(\ln2+\sum_{n=1}^\infty c_n(x-2)^n\). Find \(c_n\) for \(n\ge1\).

Hint

\(f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^n}\) for \(n\ge1\).

Answer

\(c_n=\dfrac{(-1)^{n+1}}{n\,2^n}\)

Full solution

\(f'=x^{-1}\), \(f''=-x^{-2}\), \(f'''=2x^{-3}\), …, \(f^{(n)}(x)=(-1)^{n+1}(n-1)!\,x^{-n}\).
\(c_n=\frac{f^{(n)}(2)}{n!}=\frac{(-1)^{n+1}(n-1)!}{2^n\,n!}=\frac{(-1)^{n+1}}{n2^n}\).
Equivalently \(\ln x=\ln2+\ln\left(1+\frac{x-2}{2}\right)\) and use the \(\ln(1+u)\) series.

Variant 2

Question 1 (1.5 pts)

predicted

Find the first three nonzero terms of the Maclaurin series of \(f(x)=\cos(x^2)\).

Hint

Substitute \(u=x^2\) into \(\cos u=1-\frac{u^2}{2}+\frac{u^4}{24}-\cdots\)

Answer

\(1-\frac{x^4}{2}+\frac{x^8}{24}\)

Full solution

\(\cos(x^2)=1-\frac{(x^2)^2}{2!}+\frac{(x^2)^4}{4!}-\cdots=1-\frac{x^4}{2}+\frac{x^8}{24}-\cdots\)

Question 2 (1.5 pts)

predicted

Find the first three nonzero terms of the Maclaurin series of \(f(x)=\dfrac{1}{1+4x^2}\) and the interval on which the series converges.

Hint

Geometric series with \(r=-4x^2\); it needs \(|4x^2|<1\).

Answer

\(1-4x^2+16x^4-\cdots\) on \(\left(-\frac12,\frac12\right)\).

Full solution

\(\frac1{1-r}=\sum r^n\) with \(r=-4x^2\): \(\sum(-4)^nx^{2n}=1-4x^2+16x^4-\cdots\)
Converges iff \(4x^2<1\iff|x|<\frac12\). At \(x=\pm\frac12\), \(r=-1\): the terms \((-1)^n\) don't tend to \(0\), so both endpoints are excluded.

Predicted quiz · Parametric curves

11.1–11.2

Variant 1

Question 1 (1.5 pts)

predicted

For \(x=t^2-1,\ y=t^3-t\), find \(\frac{dy}{dx}\) at \(t=2\) and the tangent line there.

Hint

\(\frac{dy}{dx}=\frac{3t^2-1}{2t}\); the point is \((3,6)\).

Answer

\(\frac{dy}{dx}=\frac{11}{4}\); \(y-6=\frac{11}{4}(x-3)\).

Full solution

At \(t=2\): \(x=3\), \(y=6\), \(\frac{dx}{dt}=4\), \(\frac{dy}{dt}=11\).
Slope \(\frac{11}{4}\); line \(y-6=\frac{11}{4}(x-3)\).

Question 2 (1.5 pts)

predicted

Find the length of the curve \(x=\frac13t^3,\ y=\frac12t^2\), \(0\le t\le\sqrt3\).

Hint

\(x'^2+y'^2=t^4+t^2=t^2(t^2+1)\).

Answer

\(\frac73\)

Full solution

\(\sqrt{t^4+t^2}=t\sqrt{t^2+1}\) for \(t\ge0\).
\(\int_0^{\sqrt3}t\sqrt{t^2+1}\,dt=\left[\frac13(t^2+1)^{3/2}\right]_0^{\sqrt3}=\frac13(8-1)=\frac73\).

Variant 2

Question 1 (1.5 pts)

predicted

Eliminate the parameter: \(x=e^t,\ y=e^{2t}+1\). Give a Cartesian equation.

Hint

\(e^{2t}=(e^t)^2\).

Answer

\(y=x^2+1\) with \(x>0\) (the right half of the parabola).

Full solution

\(e^{2t}=x^2\), so \(y=x^2+1\). Since \(x=e^t>0\), only the part with \(x>0\) is traced.

Question 2 (1.5 pts)

predicted

Find the points where \(x=t^3-3t,\ y=t^2\) has a vertical tangent.

Hint

Vertical: \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}\ne0\).

Answer

\((-2,1)\) at \(t=1\) and \((2,1)\) at \(t=-1\).

Full solution

\(\frac{dx}{dt}=3t^2-3=0\iff t=\pm1\); \(\frac{dy}{dt}=2t\ne0\) there.
\(t=1\): \((-2,1)\); \(t=-1\): \((2,1)\).

Predicted quiz · Vectors, dot and cross products

12.1–12.4

CRNs 10895–10897 already had this quiz (see Quiz 3 above); for the other sections it is the most likely next topic after Taylor series.

Variant 1

Question 1 (1.5 pts)

predicted

Find the angle between \(\mathbf u=2\mathbf i+\mathbf j-2\mathbf k\) and \(\mathbf v=\mathbf i+\mathbf j\).

Hint

\(\mathbf u\cdot\mathbf v=3\), \(|\mathbf u|=3\), \(|\mathbf v|=\sqrt2\).

Answer

\(\frac\pi4\) (\(45^\circ\))

Full solution

\(\cos\theta=\frac{3}{3\sqrt2}=\frac1{\sqrt2}\Rightarrow\theta=\frac\pi4\).

Question 2 (1.5 pts)

predicted

Find the area of triangle \(ABC\) with \(A(1,0,1)\), \(B(2,1,0)\), \(C(0,2,3)\), and a unit vector perpendicular to its plane.

Hint

\(\overrightarrow{AB}\times\overrightarrow{AC}=\langle1,1,-1\rangle\times\langle-1,2,2\rangle\).

Answer

Area \(\frac{\sqrt{26}}{2}\); \(\mathbf n=\pm\frac{1}{\sqrt{26}}\langle4,-1,3\rangle\).

Full solution

\(\langle1,1,-1\rangle\times\langle-1,2,2\rangle=\langle1\cdot2-(-1)\cdot2,\ (-1)(-1)-1\cdot2,\ 1\cdot2-1\cdot(-1)\rangle=\langle4,-1,3\rangle\).
Length \(\sqrt{26}\); area \(\frac{\sqrt{26}}2\); unit normal \(\pm\frac{\langle4,-1,3\rangle}{\sqrt{26}}\).

Variant 2

Question 1 (1.5 pts)

predicted

Find the center and radius of the sphere \(x^2+y^2+z^2+6x-2z+1=0\).

Hint

\((x+3)^2+y^2+(z-1)^2=-1+9+1\).

Answer

Center \((-3,0,1)\), radius \(3\).

Full solution

\((x^2+6x+9)+y^2+(z^2-2z+1)=-1+9+1=9\).
Center \((-3,0,1)\), radius \(3\).

Question 2 (1.5 pts)

predicted

Find \(\operatorname{proj}_{\mathbf v}\mathbf u\) for \(\mathbf u=4\mathbf i-\mathbf j+2\mathbf k\) and \(\mathbf v=\mathbf i+2\mathbf j+2\mathbf k\).

Hint

\(\mathbf u\cdot\mathbf v=6\), \(|\mathbf v|^2=9\).

Answer

\(\frac23\mathbf i+\frac43\mathbf j+\frac43\mathbf k\)

Full solution

\(\frac{\mathbf u\cdot\mathbf v}{|\mathbf v|^2}\mathbf v=\frac69\langle1,2,2\rangle=\left\langle\frac23,\frac43,\frac43\right\rangle\).

Predicted quiz · Lines and planes

12.5

Variant 1

Question 1 (1.5 pts)

predicted

Find an equation of the plane through \(P(1,2,-1)\) with normal vector \(\mathbf n=3\mathbf i-\mathbf j+2\mathbf k\).

Hint

\(3(x-1)-(y-2)+2(z+1)=0\).

Answer

\(3x-y+2z=-1\)

Full solution

\(3(x-1)-(y-2)+2(z+1)=0\iff3x-y+2z=3-2-2=-1\).

Question 2 (1.5 pts)

predicted

Find parametric equations of the line through \((2,0,-3)\) perpendicular to the plane \(x-4y+2z=7\).

Hint

Direction = the plane's normal.

Answer

\(x=2+t,\ y=-4t,\ z=-3+2t\)

Full solution

The normal \(\langle1,-4,2\rangle\) is the direction: \(\mathbf r(t)=\langle2,0,-3\rangle+t\langle1,-4,2\rangle\).

Variant 2

Question 1 (1.5 pts)

predicted

Find the plane through \(A(1,0,0)\), \(B(0,2,0)\), \(C(0,0,3)\).

Hint

\(\overrightarrow{AB}\times\overrightarrow{AC}=\langle-1,2,0\rangle\times\langle-1,0,3\rangle\).

Answer

\(6x+3y+2z=6\) (equivalently \(\frac x1+\frac y2+\frac z3=1\))

Full solution

Normal \(\langle6,3,2\rangle\); through \(A\): \(6(x-1)+3y+2z=0\iff6x+3y+2z=6\).

Question 2 (1.5 pts)

predicted

Find the distance from \((1,-2,3)\) to the plane \(2x-y+2z=1\).

Hint

\(\frac{|2+2+6-1|}{\sqrt{4+1+4}}\).

Answer

\(3\)

Full solution

\(d=\frac{|2(1)-(-2)+2(3)-1|}{3}=\frac{9}{3}=3\).

Predicted quiz · Vector functions and motion

13.1–13.2

Variant 1

Question 1 (1.5 pts)

predicted

For \(\mathbf r(t)=\langle e^t,\ e^{-t},\ \sqrt2\,t\rangle\), find the velocity and the speed at \(t=0\).

Hint

\(\mathbf v=\langle e^t,-e^{-t},\sqrt2\rangle\).

Answer

\(\langle1,-1,\sqrt2\rangle\); speed \(2\).

Full solution

\(\mathbf v(0)=\langle1,-1,\sqrt2\rangle\), \(|\mathbf v(0)|=\sqrt{1+1+2}=2\).

Question 2 (1.5 pts)

predicted

Solve \(\mathbf r''(t)=\langle0,0,-10\rangle\), \(\mathbf r'(0)=\langle3,4,5\rangle\), \(\mathbf r(0)=\langle0,0,0\rangle\).

Hint

Integrate twice; use \(\mathbf r'(0)\) for the first constant and \(\mathbf r(0)\) for the second.

Answer

\(\mathbf r(t)=\langle3t,\ 4t,\ 5t-5t^2\rangle\)

Full solution

\(\mathbf r'(t)=\langle3,4,5-10t\rangle\).
\(\mathbf r(t)=\langle3t,4t,5t-5t^2\rangle\) (the constant is \(\mathbf 0\)).

Variant 2

Question 1 (1.5 pts)

predicted

Find the tangent line to \(\mathbf r(t)=\langle2\cos t,\ 2\sin t,\ t\rangle\) at \(t=\frac\pi2\).

Hint

Point \(\mathbf r(\frac\pi2)=\langle0,2,\frac\pi2\rangle\), direction \(\mathbf r'(\frac\pi2)\).

Answer

\(x=-2s,\ y=2,\ z=\frac\pi2+s\)

Full solution

\(\mathbf r'(t)=\langle-2\sin t,2\cos t,1\rangle\Rightarrow\mathbf r'(\frac\pi2)=\langle-2,0,1\rangle\).
Line: \(\langle0,2,\frac\pi2\rangle+s\langle-2,0,1\rangle\).

Question 2 (1.5 pts)

predicted

Evaluate \(\displaystyle\int_0^{\pi/2}\langle\cos t,\ \sin t,\ 2t\rangle\,dt\).

Hint

Componentwise.

Answer

\(\left\langle1,1,\frac{\pi^2}{4}\right\rangle\)

Full solution

\(\int_0^{\pi/2}\cos t\,dt=1\), \(\int_0^{\pi/2}\sin t\,dt=1\), \(\int_0^{\pi/2}2t\,dt=\frac{\pi^2}{4}\).

Predicted quiz · Functions of several variables, limits

14.1–14.2

Variant 1

Question 1 (1.5 pts)

predicted

Describe the domain of \(f(x,y)=\sqrt{x-y^2}\) with an inequality.

Hint

The radicand must be \(\ge0\).

Answer

\(x\ge y^2\): the closed region on and to the right of the parabola \(x=y^2\).

Full solution

\(x-y^2\ge0\iff x\ge y^2\).

Question 2 (1.5 pts)

predicted

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}\), or type DNE.

Hint

Paths \(y=mx\).

Answer

Does not exist.

Full solution

Along \(y=mx\): \(\frac{1-m^2}{1+m^2}\), e.g. \(1\) for \(m=0\) and \(0\) for \(m=1\). Two different values: the limit does not exist.

Variant 2

Question 1 (1.5 pts)

predicted

Find the level curve of \(f(x,y)=y-x^2\) through \((1,3)\).

Hint

\(c=f(1,3)\).

Answer

\(y=x^2+2\)

Full solution

\(c=3-1=2\), so the level curve is \(y-x^2=2\), a parabola.

Question 2 (1.5 pts)

predicted

Find \(\displaystyle\lim_{(x,y)\to(0,0)}\frac{3x^2y}{x^2+y^2}\), or type DNE.

Hint

\(\frac{x^2}{x^2+y^2}\le1\).

Answer

\(0\)

Full solution

\(\left|\frac{3x^2y}{x^2+y^2}\right|\le3|y|\to0\), so by the sandwich theorem the limit is \(0\).

Predicted quiz · Partial derivatives and the chain rule

14.3–14.4

Variant 1

Question 1 (1.5 pts)

predicted

Find \(f_x\) and \(f_y\) for \(f(x,y)=x^2e^{3y}\).

Hint

Chain rule on \(e^{3y}\).

Answer

\(f_x=2xe^{3y}\), \(f_y=3x^2e^{3y}\)

Full solution

\(f_x=2x\,e^{3y}\); \(f_y=x^2\cdot3e^{3y}\).

Question 2 (1.5 pts)

predicted

\(w=x^2y+y^2\), \(x=\cos t\), \(y=\sin t\). Find \(\frac{dw}{dt}\) at \(t=0\).

Hint

At \(t=0\): \(x=1,\ y=0,\ x'=0,\ y'=1\).

Answer

\(1\)

Full solution

\(\frac{dw}{dt}=(2xy)x'+(x^2+2y)y'\). At \(t=0\): \(0\cdot0+1\cdot1=1\).

Variant 2

Question 1 (1.5 pts)

predicted

For \(f(x,y)=\ln(x^2+y^2)\), find \(f_{xx}\).

Hint

\(f_x=\frac{2x}{x^2+y^2}\); quotient rule.

Answer

\(f_{xx}=\dfrac{2y^2-2x^2}{(x^2+y^2)^2}\)

Full solution

\(f_x=\frac{2x}{x^2+y^2}\). \(f_{xx}=\frac{2(x^2+y^2)-2x\cdot2x}{(x^2+y^2)^2}=\frac{2y^2-2x^2}{(x^2+y^2)^2}\).

Question 2 (1.5 pts)

predicted

Find \(\frac{dy}{dx}\) for \(x^2+xy+y^2=7\) (in terms of \(x\) and \(y\)).

Hint

\(-\frac{F_x}{F_y}\).

Answer

\(-\dfrac{2x+y}{x+2y}\)

Full solution

\(F_x=2x+y\), \(F_y=x+2y\): \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\).

Predicted quiz · Gradient and tangent planes

14.5–14.6

Variant 1

Question 1 (1.5 pts)

predicted

For \(f(x,y)=x^2y-3xy^2\), find \(\nabla f(2,1)\) and the derivative at \((2,1)\) in the direction of \(3\mathbf i-4\mathbf j\).

Hint

\(\nabla f=\langle2xy-3y^2,\ x^2-6xy\rangle\); \(\mathbf u=\langle\frac35,-\frac45\rangle\).

Answer

\(\langle1,-8\rangle\); \(D_{\mathbf u}f=7\).

Full solution

\(\nabla f(2,1)=\langle4-3,\ 4-12\rangle=\langle1,-8\rangle\).
\(D_{\mathbf u}f=\frac{1\cdot3+(-8)(-4)}{5}=\frac{35}5=7\).

Question 2 (1.5 pts)

predicted

Find the tangent plane to the surface \(xyz=6\) at \((1,2,3)\).

Hint

\(\nabla F=\langle yz,xz,xy\rangle\).

Answer

\(6x+3y+2z=18\)

Full solution

\(\nabla F(1,2,3)=\langle6,3,2\rangle\): \(6(x-1)+3(y-2)+2(z-3)=0\iff6x+3y+2z=18\).

Variant 2

Question 1 (1.5 pts)

predicted

What is the maximum rate of change of \(f(x,y)=xe^{xy}\) at \((1,0)\)?

Hint

\(\nabla f=\langle e^{xy}+xye^{xy},\ x^2e^{xy}\rangle\).

Answer

\(\sqrt2\)

Full solution

\(\nabla f(1,0)=\langle1,1\rangle\), \(|\nabla f|=\sqrt2\).

Question 2 (1.5 pts)

predicted

Find the linearization of \(f(x,y)=e^x\cos y\) at \((0,0)\).

Hint

\(f(0,0)=1\), \(f_x=e^x\cos y\), \(f_y=-e^x\sin y\).

Answer

\(L=1+x\)

Full solution

\(f=1\), \(f_x=1\), \(f_y=0\) at the origin: \(L=1+x\).

Predicted quiz · Extreme values

14.7

Variant 1

Question 1 (1.5 pts)

predicted

Find the local extremum of \(f(x,y)=x^2+y^2-2x+4y+5\) and its value.

Hint

Complete the squares, or solve \(f_x=f_y=0\).

Answer

Local (and absolute) minimum \(0\) at \((1,-2)\).

Full solution

\(f_x=2x-2=0,\ f_y=2y+4=0\Rightarrow(1,-2)\). \(D=4>0\), \(f_{xx}=2>0\): minimum, \(f=(x-1)^2+(y+2)^2\ge0\).

Question 2 (1.5 pts)

predicted

Find and classify the critical points of \(f(x,y)=x^3+y^3-3xy\).

Hint

\(y=x^2\) and \(x=y^2\).

Answer

Saddle at \((0,0)\); local minimum at \((1,1)\).

Full solution

\(3x^2=3y,\ 3y^2=3x\Rightarrow x=x^4\Rightarrow x=0,1\): points \((0,0),(1,1)\).
\(D=36xy-9\): \(-9\) at \((0,0)\) (saddle); \(27\) at \((1,1)\) with \(f_{xx}=6>0\) (minimum).

Variant 2

Question 1 (1.5 pts)

predicted

Find and classify the critical points of \(f(x,y)=2x^2-4xy+y^4+2\).

Hint

\(f_x=4x-4y=0\Rightarrow x=y\); then \(y^3=y\).

Answer

Minima at \((\pm1,\pm1)\) (same signs); saddle at \((0,0)\).

Full solution

\(x=y\) and \(-4x+4y^3=0\Rightarrow y^3=y\Rightarrow y=0,\pm1\).
\(D=4\cdot12y^2-16\): \(-16\) at the origin (saddle), \(32\) at \((\pm1,\pm1)\) with \(f_{xx}=4>0\) (minima).

Question 2 (1.5 pts)

predicted

Find the absolute maximum and minimum of \(f(x,y)=xy\) on the rectangle \(0\le x\le2,\ -1\le y\le1\).

Hint

No interior critical points; check the edges and corners.

Answer

Max \(2\) at \((2,1)\); min \(-2\) at \((2,-1)\).

Full solution

The only critical point \((0,0)\) is on the boundary. On each edge \(f\) is linear, so the extremes are at corners: \(f(0,\pm1)=0\), \(f(2,1)=2\), \(f(2,-1)=-2\).

Predicted quiz · Double integrals and area

15.1–15.3

Variant 1

Question 1 (1.5 pts)

predicted

Evaluate \(\displaystyle\int_0^1\int_0^2(x^2+y)\,dx\,dy\).

Hint

Inner integral in \(x\): \(\frac83+2y\).

Answer

\(\frac{11}{3}\)

Full solution

\(\int_0^2(x^2+y)\,dx=\frac83+2y\); \(\int_0^1\left(\frac83+2y\right)dy=\frac83+1=\frac{11}3\).

Question 2 (1.5 pts)

predicted

Use a double integral to find the area between \(y=x^2\) and \(y=2x\).

Hint

They meet at \(x=0\) and \(x=2\).

Answer

\(\frac43\)

Full solution

\(\int_0^2\int_{x^2}^{2x}dy\,dx=\int_0^2(2x-x^2)\,dx=4-\frac83=\frac43\).

Variant 2

Question 1 (1.5 pts)

predicted

Evaluate \(\displaystyle\int_0^1\int_y^1\sin(x^2)\,dx\,dy\) by reversing the order of integration.

Hint

The region is \(0\le y\le x\le1\).

Answer

\(\frac{1-\cos1}{2}\)

Full solution

Reversed: \(\int_0^1\int_0^x\sin(x^2)\,dy\,dx=\int_0^1x\sin(x^2)\,dx\).
\(=\left[-\frac12\cos(x^2)\right]_0^1=\frac{1-\cos1}{2}\).

Question 2 (1.5 pts)

predicted

Evaluate \(\displaystyle\iint_R(x+2y)\,dA\), where \(R\) is bounded by \(y=x\), \(y=0\) and \(x=2\).

Hint

\(0\le x\le2,\ 0\le y\le x\).

Answer

\(\frac{16}{3}\)

Full solution

Inner: \(\left[xy+y^2\right]_0^x=2x^2\). Outer: \(\int_0^22x^2\,dx=\frac{16}{3}\).

Predicted quiz · Line integrals, work and flux

16.1–16.2

Variant 1

Question 1 (1.5 pts)

predicted

Evaluate \(\int_Cxy^2\,ds\), where \(C\) is the right half of the unit circle.

Hint

\(\mathbf r=\langle\cos t,\sin t\rangle\), \(-\frac\pi2\le t\le\frac\pi2\), \(ds=dt\).

Answer

\(\frac23\)

Full solution

\(\int_{-\pi/2}^{\pi/2}\cos t\sin^2t\,dt=\left[\frac{\sin^3t}{3}\right]_{-\pi/2}^{\pi/2}=\frac23\).

Question 2 (1.5 pts)

predicted

Find the work done by \(\mathbf F=\langle x^2,\ xy\rangle\) along \(\mathbf r(t)=\langle t,t^2\rangle\), \(0\le t\le1\).

Hint

\(\mathbf F(\mathbf r)\cdot\mathbf r'=t^2+t^3\cdot2t\).

Answer

\(\frac{11}{15}\)

Full solution

\(\int_0^1(t^2+2t^4)\,dt=\frac13+\frac25=\frac{11}{15}\).

Variant 2

Question 1 (1.5 pts)

predicted

Find the circulation of \(\mathbf F=\langle-y,x\rangle\) around the circle of radius \(2\) about the origin (counterclockwise).

Hint

\(\mathbf F\cdot\mathbf r'=4\sin^2t+4\cos^2t\).

Answer

\(8\pi\)

Full solution

\(\int_0^{2\pi}4\,dt=8\pi\).

Question 2 (1.5 pts)

predicted

Find the outward flux of \(\mathbf F=\langle x-y,\ y\rangle\) across the unit circle.

Hint

\(M\,dy-N\,dx=(\cos t-\sin t)\cos t\,dt+\sin^2t\,dt\).

Answer

\(2\pi\)

Full solution

Integrand \(1-\sin t\cos t\); \(\int_0^{2\pi}=2\pi\).

Predicted quiz · Conservative fields and potentials

16.3

Variant 1

Question 1 (1.5 pts)

predicted

Show that \(\mathbf F=\langle y^2+2x,\ 2xy+3\rangle\) is conservative and find a potential function.

Hint

\(M_y=2y=N_x\). Integrate \(M\) in \(x\).

Answer

\(f=xy^2+x^2+3y\) (+ constant)

Full solution

\(M_y=2y=N_x\): conservative.
\(f=\int(y^2+2x)\,dx=xy^2+x^2+g(y)\); \(f_y=2xy+g'(y)=2xy+3\Rightarrow g=3y\).

Question 2 (1.5 pts)

predicted

Find the work done by the same field along any path from \((0,0)\) to \((2,1)\).

Hint

\(f(2,1)-f(0,0)\).

Answer

\(9\)

Full solution

\(f(2,1)=2+4+3=9\), \(f(0,0)=0\).

Variant 2

Question 1 (1.5 pts)

predicted

Is \(\mathbf F=\langle2xz,\ z^2,\ x^2+2yz\rangle\) conservative? If so, find a potential.

Hint

Check \(P_y=N_z\), \(M_z=P_x\), \(N_x=M_y\).

Answer

Yes; \(f=x^2z+yz^2\).

Full solution

\(P_y=2z=N_z\), \(M_z=2x=P_x\), \(N_x=0=M_y\): conservative.
\(f=\int2xz\,dx=x^2z+g(y,z)\); \(g_y=z^2\Rightarrow g=yz^2+h(z)\); \(f_z=x^2+2yz+h'=x^2+2yz\Rightarrow h=0\).

Question 2 (1.5 pts)

predicted

Evaluate \(\int_C\mathbf F\cdot d\mathbf r\) for the same field along any path from \((0,0,0)\) to \((1,1,1)\).

Hint

\(f(1,1,1)-f(0,0,0)\).

Answer

\(2\)

Full solution

\(f(1,1,1)=1+1=2\), \(f(0,0,0)=0\).

Midterm practice · Chapters 10–13

\(\frac32\)

Full solution

\(S_N=1+\frac12-\frac1{N+1}-\frac1{N+2}\to\frac32\).

Final practice · emphasis on Chapters 13–16

Does not exist.

Full solution

Along \(x=my^2\): \(\frac{my^4}{m^2y^4+y^4}=\frac{m}{m^2+1}\), which depends on \(m\).

Going further

Hard problems, from easy to hard

Forty problems in five levels. Levels 1–3 stretch the quiz types; level 4 is proofs (ε–N limits, divergence proofs, error bounds, derivations); level 5 shows a wrong solution and asks you to find the broken line. Every problem has a model solution.

Level 1 · Warm-up

8 problems

A sum inside a root

level 1

Find \(\displaystyle\lim_{n\to\infty}\left(2^n+3^n\right)^{1/n}\).

Hint

Factor out \(3^n\).

Answer

\(3\)

Full solution

\(\left(2^n+3^n\right)^{1/n}=3\left(1+\left(\frac23\right)^n\right)^{1/n}\).
\(1\le1+\left(\frac23\right)^n\le2\), and \(2^{1/n}\to1\): by the sandwich theorem the bracket \(\to1\). Limit \(=3\).

Two geometric series

level 1

Find \(\displaystyle\sum_{n=1}^\infty\frac{3^n+(-2)^n}{5^n}\).

Hint

Split: \(r=\frac35\) and \(r=-\frac25\).

Answer

\(\frac{17}{14}\)

Full solution

\(\sum\left(\frac35\right)^n=\frac{3/5}{2/5}=\frac32\); \(\sum\left(-\frac25\right)^n=\frac{-2/5}{7/5}=-\frac27\).
Both converge, so the sum is \(\frac32-\frac27=\frac{17}{14}\).

For which p?

level 1

For which values of \(p\) does \(\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^p}\) converge?

Hint

Integral test with \(u=\ln x\) turns it into a \(p\)-integral.

Answer

\(p>1\)

Full solution

\(\int_2^\infty\frac{dx}{x(\ln x)^p}=\int_{\ln2}^\infty\frac{du}{u^p}\), which is finite iff \(p>1\).
For \(p\le0\) the terms are at least \(\frac{1}{n}\cdot(\ln 2)^{-p}\), so it diverges as well. Answer: \(p>1\).

Recognize a series

level 1

Find \(\displaystyle\sum_{n=0}^\infty\frac{(-1)^n\pi^{2n}}{(2n)!}\).

Hint

It is the Maclaurin series of a famous function at \(x=\pi\).

Answer

\(\cos\pi=-1\)

Full solution

\(\cos x=\sum\frac{(-1)^nx^{2n}}{(2n)!}\) for all \(x\); at \(x=\pi\) the sum is \(\cos\pi=-1\).

Distance between parallel planes

level 1

Find the distance between the planes \(x+2y-2z=3\) and \(2x+4y-4z=7\).

Hint

Write the second plane as \(x+2y-2z=\frac72\), then take any point of the first plane.

Answer

\(\frac16\)

Full solution

The point \((3,0,0)\) lies on the first plane.
Distance to \(x+2y-2z=\frac72\): \(\frac{|3-\frac72|}{3}=\frac16\).

Parallelogram in space

level 1

Find the area of the parallelogram with adjacent sides \(\overrightarrow{AB}=\langle1,2,3\rangle\) and \(\overrightarrow{AD}=\langle2,-1,1\rangle\).

Hint

\(|\overrightarrow{AB}\times\overrightarrow{AD}|\).

Answer

\(5\sqrt3\)

Full solution

\(\langle1,2,3\rangle\times\langle2,-1,1\rangle=\langle5,5,-5\rangle\), length \(5\sqrt3\).

A mixed partial

level 1

For \(f(x,y)=e^{x^2y}\), find \(f_{xy}\).

Hint

\(f_x=2xy\,e^{x^2y}\); product rule in \(y\).

Answer

\(2xe^{x^2y}\left(1+x^2y\right)\)

Full solution

\(f_x=2xye^{x^2y}\).
\(f_{xy}=2xe^{x^2y}+2xy\cdot x^2e^{x^2y}=2xe^{x^2y}(1+x^2y)\).

Exponential over a rectangle

level 1

Evaluate \(\displaystyle\iint_Re^{x+y}\,dA\), \(R=[0,\ln2]\times[0,\ln3]\).

Hint

\(e^{x+y}=e^xe^y\) splits.

Answer

\(2\)

Full solution

\(\left(\int_0^{\ln2}e^xdx\right)\left(\int_0^{\ln3}e^ydy\right)=(2-1)(3-1)=2\).

Level 2 · Standard+

8 problems

A 0 · ∞ limit

level 2

Find \(\displaystyle\lim_{n\to\infty}n^2\left(1-\cos\frac1n\right)\).

Hint

\(1-\cos u=\frac{u^2}{2}-\frac{u^4}{24}+\cdots\) with \(u=\frac1n\).

Answer

\(\frac12\)

Full solution

\(1-\cos\frac1n=\frac{1}{2n^2}-\frac{1}{24n^4}+\cdots\)
Multiply by \(n^2\): \(\frac12-\frac{1}{24n^2}+\cdots\to\frac12\).

Sum of n²/2ⁿ

level 2

Find \(\displaystyle\sum_{n=1}^\infty\frac{n^2}{2^n}\).

Hint

Differentiate \(\sum nx^n=\frac{x}{(1-x)^2}\) and multiply by \(x\): \(\sum n^2x^n=\frac{x(1+x)}{(1-x)^3}\).

Answer

\(6\)

Full solution

\(\sum n^2x^n=x\frac{d}{dx}\frac{x}{(1-x)^2}=\frac{x(1+x)}{(1-x)^3}\) for \(|x|<1\).
At \(x=\frac12\): \(\frac{\frac12\cdot\frac32}{\frac18}=6\).

Even powers only

level 2

Find the interval of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{x^{2n}}{n\,9^n}\).

Hint

Ratio test gives \(\frac{x^2}{9}\).

Answer

\((-3,3)\)

Full solution

\(\left|\frac{a_{n+1}}{a_n}\right|=\frac{x^2}{9}\cdot\frac{n}{n+1}\to\frac{x^2}{9}<1\iff|x|<3\).
At \(x=\pm3\): \(\sum\frac{9^n}{n9^n}=\sum\frac1n\), divergent at both ends.

Recognize ln 2

level 2

Find \(\displaystyle\sum_{n=1}^\infty\frac{1}{n\,2^n}\).

Hint

\(-\ln(1-x)=\sum\frac{x^n}{n}\) for \(-1\le x<1\).

Answer

\(\ln2\)

Full solution

At \(x=\frac12\): \(\sum\frac{1}{n2^n}=-\ln\left(1-\frac12\right)=\ln2\).

Closest point on a line

level 2

Find the point on the line \(\mathbf r(t)=\langle1,0,2\rangle+t\langle1,1,-1\rangle\) closest to the origin.

Hint

Minimize \(|\mathbf r(t)|^2\), or require \(\mathbf r(t)\perp\langle1,1,-1\rangle\).

Answer

\(\left(\frac43,\frac13,\frac53\right)\)

Full solution

\(\mathbf r(t)\cdot\langle1,1,-1\rangle=(1+t)+t-(2-t)=3t-1=0\Rightarrow t=\frac13\).
Point \(\left(\frac43,\frac13,\frac53\right)\).

Tangent plane to a quadric

level 2

Find the tangent plane to \(x^2-2y^2+z^2+yz=2\) at \((2,1,-1)\).

Hint

\(\nabla F=\langle2x,\ -4y+z,\ 2z+y\rangle\).

Answer

\(4x-5y-z=4\)

Full solution

\(\nabla F(2,1,-1)=\langle4,-5,-1\rangle\).
\(4(x-2)-5(y-1)-(z+1)=0\iff4x-5y-z=4\).

Maximum on a circle

level 2

Find the maximum of \(f(x,y)=x+2y\) on the circle \(x^2+y^2=5\).

Hint

Parametrize \(x=\sqrt5\cos t,\ y=\sqrt5\sin t\), or use \(\mathbf u\cdot\mathbf v\le|\mathbf u||\mathbf v|\).

Answer

\(5\), at \((1,2)\)

Full solution

\(f=\langle1,2\rangle\cdot\langle x,y\rangle\le|\langle1,2\rangle|\,|\langle x,y\rangle|=\sqrt5\cdot\sqrt5=5\).
Equality when \(\langle x,y\rangle\) points along \(\langle1,2\rangle\): \((1,2)\).

Reverse to integrate e^(x²)

level 2

Evaluate \(\displaystyle\int_0^2\int_{y/2}^1e^{x^2}\,dx\,dy\).

Hint

The region is \(0\le y\le2x\), \(0\le x\le1\).

Answer

\(e-1\)

Full solution

\(\int_0^1\int_0^{2x}e^{x^2}\,dy\,dx=\int_0^12xe^{x^2}dx=e-1\).

Level 3 · Challenge

8 problems

Almost the e-limit

level 3

Find \(\displaystyle\lim_{n\to\infty}\left(1+\frac1n+\frac1{n^2}\right)^n\).

Hint

Take logs: \(n\ln\left(1+\frac1n+\frac1{n^2}\right)\).

Answer

\(e\)

Full solution

\(n\ln\left(1+u\right)\) with \(u=\frac1n+\frac1{n^2}\): \(\ln(1+u)=u-\frac{u^2}{2}+\cdots\)
\(n\left(\frac1n+O\left(\frac1{n^2}\right)\right)\to1\), so the limit is \(e^1=e\).

sin(1/n)

level 3

Does \(\displaystyle\sum_{n=1}^\infty\sin\frac1n\) converge?

Hint

\(\sin u\approx u\) for small \(u\).

Answer

Diverges.

Full solution

Limit comparison with \(\frac1n\): \(\lim\frac{\sin(1/n)}{1/n}=1\).
\(\sum\frac1n\) diverges, so the series diverges.

1 − cos(1/n)

level 3

Does \(\displaystyle\sum_{n=1}^\infty\left(1-\cos\frac1n\right)\) converge?

Hint

\(1-\cos u\approx\frac{u^2}{2}\).

Answer

Converges.

Full solution

\(\lim\frac{1-\cos(1/n)}{1/n^2}=\frac12\) (Taylor series).
\(\sum\frac1{n^2}\) converges, so does the series (limit comparison).

Telescoping with n² − 1

level 3

Find \(\displaystyle\sum_{n=2}^\infty\frac{1}{n^2-1}\).

Hint

\(\frac{1}{n^2-1}=\frac12\left(\frac1{n-1}-\frac1{n+1}\right)\).

Answer

\(\frac34\)

Full solution

\(S_N=\frac12\left(1+\frac12-\frac1N-\frac1{N+1}\right)\to\frac12\cdot\frac32=\frac34\).

A limit by series

level 3

Find \(\displaystyle\lim_{x\to0}\frac{\sin x-x+\frac{x^3}{6}}{x^5}\).

Hint

Write out the Maclaurin series of \(\sin x\) to the \(x^5\) term.

Answer

\(\frac{1}{120}\)

Full solution

\(\sin x-x+\frac{x^3}{6}=\frac{x^5}{120}-\frac{x^7}{5040}+\cdots\)
Divide by \(x^5\): \(\frac1{120}+O(x^2)\to\frac1{120}\).

Volume of a tetrahedron

level 3

Find the volume of the tetrahedron with vertices \((0,0,0)\), \((1,2,0)\), \((2,0,1)\), \((0,1,3)\).

Hint

A tetrahedron is \(\frac16\) of the parallelepiped on the same three edges.

Answer

\(\frac{13}{6}\)

Full solution

\(\langle2,0,1\rangle\times\langle0,1,3\rangle=\langle-1,-6,2\rangle\); dot with \(\langle1,2,0\rangle\): \(-13\).
Volume \(=\frac{|-13|}{6}=\frac{13}{6}\).

A double-well function

level 3

Find and classify the critical points of \(f(x,y)=x^4-2x^2+y^2\).

Hint

\(f_x=4x^3-4x\), \(f_y=2y\).

Answer

Minima at \((\pm1,0)\), saddle at \((0,0)\).

Full solution

\(x(x^2-1)=0\), \(y=0\): points \((0,0),(\pm1,0)\).
\(D=(12x^2-4)\cdot2\): \(-8\) at the origin (saddle), \(16\) at \((\pm1,0)\) with \(f_{xx}=8>0\) (minima).

The vortex field

level 3

\(\mathbf F=\left\langle\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right\rangle\) satisfies \(N_x=M_y\) everywhere except the origin. Compute its circulation around the unit circle.

Hint

On the unit circle \(x^2+y^2=1\).

Answer

\(2\pi\ne0\): the field is not conservative on the punctured plane.

Full solution

On \(\mathbf r=\langle\cos t,\sin t\rangle\): \(\mathbf F=\langle-\sin t,\cos t\rangle\), \(\mathbf F\cdot\mathbf r'=1\), so the circulation is \(2\pi\).
The component test needs a simply connected domain; the plane minus the origin is not, so \(N_x=M_y\) does not imply conservative here.

Level 4 · Proofs

8 problems

An ε–N proof

level 4

Prove that \(\displaystyle\lim_{n\to\infty}\frac{3n+1}{n+2}=3\) from the definition. As a check: for \(\varepsilon=0.01\), what is the smallest \(N\) such that \(\left|a_n-3\right|<\varepsilon\) for every \(n\ge N\)?

Put the proof in order.

Let \(\varepsilon>0\) be given.
Compute \(\left|\frac{3n+1}{n+2}-3\right|=\frac{|3n+1-3n-6|}{n+2}=\frac{5}{n+2}\).
\(\frac{5}{n+2}<\varepsilon\iff n>\frac5\varepsilon-2\), so choose an integer \(N>\frac5\varepsilon-2\).
Then for every \(n>N\): \(|a_n-3|=\frac5{n+2}<\frac{5}{N+2}<\varepsilon\).
By the definition of the limit, \(a_n\to3\). ∎

Hint

\(|a_n-3|=\frac{5}{n+2}<0.01\iff n+2>500\).

Answer

\(N=499\) (at \(n=498\): \(\frac{5}{500}=0.01\) is not \(<0.01\)).

Full solution

The proof is the ordered list above. Its structure (given \(\varepsilon\), compute \(|a_n-L|\), solve for \(n\), choose \(N\), verify) works for every ε–N proof.

ε–N for 1/√n

level 4

Prove \(\frac{1}{\sqrt n}\to0\). For \(\varepsilon=0.05\), what is the smallest \(N\) with \(\frac1{\sqrt n}<\varepsilon\) for all \(n\ge N\)?

Hint

\(\frac{1}{\sqrt n}<\varepsilon\iff n>\frac1{\varepsilon^2}\).

Answer

\(N=401\)

Full solution

Given \(\varepsilon>0\), choose \(N>\frac{1}{\varepsilon^2}\). For \(n>N\): \(\frac{1}{\sqrt n}<\frac{1}{\sqrt N}<\varepsilon\).
For \(\varepsilon=0.05\): \(n>400\), so \(N=401\).

The harmonic series diverges (Oresme)

level 4

The classical proof shows \(S_{2^k}\ge1+\frac k2\). Using this bound, what is the smallest \(k\) that guarantees \(S_{2^k}>10\)?

Put the proof in order.

Group the terms: \(1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\cdots+\frac18\right)+\cdots\)
Each group \(\frac{1}{2^{j-1}+1}+\cdots+\frac{1}{2^j}\) has \(2^{j-1}\) terms, each at least \(\frac{1}{2^j}\).
So each group is at least \(2^{j-1}\cdot\frac{1}{2^j}=\frac12\).
Hence \(S_{2^k}\ge1+\frac k2\), which is unbounded as \(k\to\infty\).
The partial sums are unbounded, so \(\sum\frac1n\) diverges. ∎

Hint

\(1+\frac k2>10\iff k>18\).

Answer

\(k=19\) (so about \(2^{19}\approx524{,}288\) terms; the series diverges very slowly).

Full solution

See the ordered proof. Note how slowly the partial sums grow: \(S_N\approx\ln N\).

Taylor remainder bound

level 4

Using \(|R_n(1)|\le\frac{3}{(n+1)!}\) for \(e^x\) at \(x=1\) (since \(e<3\)), what is the smallest \(n\) for which \(P_n(1)\) approximates \(e\) with error less than \(10^{-6}\)?

Hint

Find the first \(n\) with \((n+1)!>3\,000\,000\).

Answer

\(n=9\) (\(10!=3{,}628{,}800\)).

Full solution

Remainder Estimation Theorem: \(|R_n(1)|\le M\frac{1^{n+1}}{(n+1)!}\) with \(M=\max_{[0,1]}e^x=e<3\).
\(9!=362{,}880<3\cdot10^6<10!\), so \(n+1=10\), \(n=9\).

Alternating series error

level 4

How many terms of \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}\) guarantee \(|S-S_N|<0.001\)?

Hint

\(|S-S_N|\le\frac{1}{(N+1)^2}\).

Answer

\(N=31\) (\(32^2=1024\)).

Full solution

Need \((N+1)^2>1000\). \(31^2=961\) fails, \(32^2=1024\) works: \(N+1=32\).

Derive and use Σ n xⁿ⁻¹

level 4

Differentiate \(\sum_{n=0}^\infty x^n=\frac1{1-x}\) to derive \(\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\), then compute \(\displaystyle\sum_{n=1}^\infty\frac{n}{3^{n-1}}\).

Hint

Plug in \(x=\frac13\).

Answer

\(\frac94\)

Full solution

Term-by-term differentiation is valid on \((-1,1)\): \(\sum nx^{n-1}=\frac{d}{dx}\frac1{1-x}=\frac1{(1-x)^2}\).
At \(x=\frac13\): \(\frac{1}{(2/3)^2}=\frac94\).

Derive the arctan series

level 4

Derive the Maclaurin series of \(\arctan x\) and give the coefficient of \(x^7\).

Put the derivation in order.

Start from the geometric series \(\frac{1}{1-u}=\sum u^n\) for \(|u|<1\).
Substitute \(u=-t^2\): \(\frac{1}{1+t^2}=\sum(-1)^nt^{2n}\) for \(|t|<1\).
Integrate term by term from \(0\) to \(x\): \(\arctan x=\sum\frac{(-1)^nx^{2n+1}}{2n+1}\).
So \(\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots\) (valid on \([-1,1]\)).

Hint

\(\frac{d}{dx}\arctan x=\frac1{1+x^2}\).

Answer

\(-\frac17\)

Full solution

At \(x=1\) this gives Leibniz's formula \(\frac\pi4=1-\frac13+\frac15-\cdots\)

Gradient fields have zero circulation

level 4

Prove that \(\oint_C\nabla f\cdot d\mathbf r=0\) for every closed curve \(C\). Check: compute \(\oint_C\nabla(x^2y^3)\cdot d\mathbf r\) around the unit circle.

Put the proof in order.

Parametrize \(C\) by \(\mathbf r(t)\), \(a\le t\le b\), with \(\mathbf r(a)=\mathbf r(b)\) because \(C\) is closed.
By the chain rule, \(\frac{d}{dt}f(\mathbf r(t))=\nabla f(\mathbf r(t))\cdot\mathbf r'(t)\).
So \(\oint_C\nabla f\cdot d\mathbf r=\int_a^b\frac{d}{dt}f(\mathbf r(t))\,dt=f(\mathbf r(b))-f(\mathbf r(a))\).
Since \(\mathbf r(b)=\mathbf r(a)\), the result is \(0\). ∎

Hint

The chain rule plus the Fundamental Theorem of Calculus.

Answer

\(0\)

Full solution

The ordered proof above is the Fundamental Theorem of line integrals in one line.

Level 5 · Spot the error

8 problems

Each solution below contains exactly one wrong line. Click it, then give the correct answer.

Comparison in the wrong direction

level 5

Claimed solution: “Does \(\sum_{n=2}^\infty\frac{1}{n\ln n}\) converge?”

Click the line that contains the mistake.

For \(n\ge3\), \(\ln n>1\).
So \(\frac{1}{n\ln n}<\frac1n\).
\(\sum\frac1n\) diverges.
Therefore \(\sum\frac{1}{n\ln n}\) diverges by the comparison test.

Hint

Which direction does the comparison test need for divergence?

Answer

Line 4 is wrong; the series does diverge (integral test).

Full solution

For divergence you need \(a_n\ge b_n\ge0\) with \(\sum b_n\) divergent.

Reading ρ = 1

level 5

Claimed solution: “Does \(\sum\frac{1}{n^2}\) converge?”

Click the line that contains the mistake.

\(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n^2}{(n+1)^2}\).
\(\lim\frac{n^2}{(n+1)^2}=1\).
Since \(\rho=1\), the ratio test shows that the series converges.

Hint

What does the ratio test say when \(\rho=1\)?

Answer

Line 3; the series converges (\(p\)-series).

Full solution

Always say “inconclusive” for \(\rho=1\) and switch tests.

Terms going to zero

level 5

Claimed solution: “Does \(\sum\frac{1}{\sqrt n}\) converge?”

Click the line that contains the mistake.

\(\lim_{n\to\infty}\frac{1}{\sqrt n}=0\).
Since the terms tend to \(0\), the series converges.

Hint

The \(n\)th-term test only ever proves divergence.

Answer

Line 2; the series diverges.

Full solution

Use the \(p\)-test (or the integral test).

Forgotten endpoints

level 5

Claimed solution: “Find the interval of convergence of \(\sum_{n\ge1}\frac{x^n}{n}\).”

Click the line that contains the mistake.

\(\left|\frac{a_{n+1}}{a_n}\right|=|x|\frac{n}{n+1}\to|x|\).
The series converges absolutely for \(|x|<1\).
Therefore the interval of convergence is \((-1,1)\).

Hint

Test \(x=1\) and \(x=-1\) separately.

Answer

Line 3; the interval is \([-1,1)\).

Full solution

\(x=1\): harmonic (diverges); \(x=-1\): alternating harmonic (converges).

The missing factorial

level 5

Claimed solution: “Find \(P_2(x)\) for \(e^{2x}\) at \(0\).”

Click the line that contains the mistake.

\(f=e^{2x}\), \(f'=2e^{2x}\), \(f''=4e^{2x}\).
\(f(0)=1\), \(f'(0)=2\), \(f''(0)=4\).
\(P_2(x)=1+2x+4x^2\).

Hint

\(c_k=\frac{f^{(k)}(0)}{k!}\).

Answer

Line 3; \(P_2=1+2x+2x^2\).

Full solution

The answer box would say “Did you forget to divide by k!?”: it detects exactly this mistake.

The j-component sign

level 5

Claimed solution: “Compute \(\langle1,2,3\rangle\times\langle4,5,6\rangle\).”

Click the line that contains the mistake.

\(\mathbf i\): \(2\cdot6-3\cdot5=-3\).
\(\mathbf j\): \(1\cdot6-3\cdot4=-6\).
\(\mathbf k\): \(1\cdot5-2\cdot4=-3\). So the cross product is \(\langle-3,-6,-3\rangle\).

Hint

Expand the determinant along the first row: \(+\mathbf i\), \(-\mathbf j\), \(+\mathbf k\).

Answer

Line 2; \(\langle-3,6,-3\rangle\).

Full solution

Always dot your result with both factors as a check.

Not a unit vector

level 5

Claimed solution: “Derivative of \(f=x^2+y^2\) at \((1,1)\) in the direction of \(\mathbf v=\langle3,4\rangle\).”

Click the line that contains the mistake.

\(\nabla f=\langle2x,2y\rangle=\langle2,2\rangle\) at \((1,1)\).
\(D_{\mathbf v}f=\langle2,2\rangle\cdot\langle3,4\rangle=14\).

Hint

\(|\mathbf v|=5\).

Answer

Line 2; \(\frac{14}5\).

Full solution

A directional derivative is never larger than \(|\nabla f|=2\sqrt2\approx2.83\); \(14\) is impossible.

Upside-down limits

level 5

Claimed solution: “Area between \(y=x^2\) and \(y=x\).”

Click the line that contains the mistake.

The curves meet at \(x=0\) and \(x=1\).
\(A=\int_0^1\int_x^{x^2}dy\,dx=\int_0^1(x^2-x)\,dx=-\frac16\).

Hint

Which curve is on top between \(0\) and \(1\)?

Answer

Line 2; area \(\frac16\).

Full solution

Sketch first, and put the lower curve as the lower limit.

Errata

Mistakes found while verifying

Every answer on this page was recomputed independently with SymPy and tested in the browser. All 103 final answers on the 13 quiz solution sheets are correct. The sheets contain the following mistakes in their working; the problems on this page explain each one where it occurs.

Quiz 1, CRN 10890, Variant B, Question 2 (see the problem). The question paper asks about \(\sum\frac{1}{5^n-n}\), but the solution sheet solves \(\sum\frac{1}{5^{n-1}}=\sum\left(\frac15\right)^{n-1}\) as a geometric series, which is a different series. Correction: \(5^n-n\ge\frac{5^n}{2}\), so \(0<\frac{1}{5^n-n}\le2\left(\frac15\right)^n\) and the series converges by comparison (or by limit comparison with \(\left(\frac15\right)^n\)). The verdict, converges, is unchanged.
Quiz 1, CRN 10890, Variant A, Question 1(a) (see the problem). The sheet bounds \(0<\frac{3^n}{n!}\le\frac9n\) “for \(n\ge3\)”. This is false for \(n=3,4,5\) (at \(n=3\): \(4.5>3\)) and only holds from \(n=6\). Correction: \(\frac{3^n}{n!}\le\frac{27}{2n}\) for \(n\ge3\). The limit \(0\) is unchanged.
Quiz 2, CRN 10893 (see the problem). The series is printed as \(\sum_{n=0}^\infty\frac{(-1)^nx^n}{n2^n}\), but the \(n=0\) term divides by zero. Correction: start at \(n=1\), as the solution itself does when it substitutes the endpoints. The answer \(R=2\), \((-2,2]\) is unchanged.
Quiz 2 (handwritten, CRN not stated), Variant 2, Exercise 1 (see the problem). After substituting \(u=x^2+4\), the solution keeps the old limits \(1\) and \(t\) and writes the result as \(\frac12\ln|t|\). Correction: \(\int_1^t\frac{x\,dx}{x^2+4}=\frac12\int_5^{t^2+4}\frac{du}{u}=\frac12\ln\frac{t^2+4}{5}\to\infty\). The conclusion, divergence, is unchanged.
Quiz 2 (handwritten), Variant 2, Question 2 (see the problem). The limit comparison is written as \(\frac{\lim\sqrt{1/n^3}}{\lim 1/n^{3/2}}\), a quotient of two limits that are both \(0\) (an undefined \(\frac00\)). Correction: take the limit of the ratio, \(\lim n^{3/2}\sqrt{\frac{n+4}{n^4+4}}=1\).
Syllabus, weeks 9–11. The course table lists “14.3–14.4”, “14.4–14.5” and “14.6–14.7”. In Thomas 13e the topics are §14.4 (chain rule), §14.5 (directional derivatives and gradient), §14.6 (tangent planes and differentials) and §14.7 (extreme values). This page uses the 13e numbering (see the study path).

A note on the verification itself

The checks caught one error in the checking software: SymPy 1.14's convergence test reports \(\sum\left(1-\cos\frac1n\right)\) as divergent. That is wrong: \(1-\cos\frac1n\sim\frac{1}{2n^2}\) and the partial sums settle at \(0.77876\ldots\) That answer key (hard problem 3.3) is verified by an explicit limit comparison instead. Every other key was confirmed by SymPy and by the page's own grader, and the past-quiz keys were also matched against the final answers printed on the sheets.

Sources

Sources and credits

Course: MATH 1222 Calculus II, ADA University, School of IT and Engineering, Fall 2026 (syllabus: topics, weekly schedule, grading). The syllabus itself is not published here.
Textbook: G. B. Thomas Jr., M. D. Weir, J. Hass, Thomas' Calculus: Early Transcendentals, 13th edition, Pearson. Theorem and section labels follow this book.
Quiz solution sheets (CRNs 10890–10897): linked from each paper in Past quizzes; copies are in the pdfs/ folder.
Software: math typesetting by KaTeX 0.16.11; fonts Schibsted Grotesk, Source Serif 4 and JetBrains Mono (Google Fonts); answer keys verified with SymPy. The grading engine, plots and tools are written from scratch for this page.