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AP Calculus BC – Series, Parametric and Polar

Convergence tests, Taylor and Maclaurin series, parametric and polar calculus.

12 questions  ·  73 marks  ·  answers included  ·  AP Calculus BC
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This worksheet covers:

  • Convergence tests
  • Power series and their radius of convergence
  • Taylor and Maclaurin series with error bounds
  • Calculus with parametric and polar curves

Section A is non-calculator; Sections B and C allow a graphing calculator. Allow about 75 minutes.

Questions

Section A – Series Skills

Non-calculator.
1

Determine whether each series converges or diverges. Name the test used.(a) $\sum_{n=1}^{\infty} \frac{n}{3n+1}$(b) $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}$(c) $\sum_{n=1}^{\infty} \frac{2^n}{n!}$

Show answer
(a) Diverges: $n$th-term test, $\lim \frac{n}{3n+1} = \frac{1}{3} e 0$(b) Converges: $p$-series with $p = \frac{3}{2} > 1$(c) Converges: ratio test, $\lim \frac{2}{n+1} = 0 < 1$
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2

Use the limit comparison test to determine whether $\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 + 2}$ converges.

Show answer
Compare with $\sum \frac{1}{n^2}$: $\lim \frac{(3n+1)/(n^3+2)}{1/n^2} = 3$ (finite, positive), and $\sum \frac{1}{n^2}$ converges, so the series converges.
[3]
3

(a) Find the sum of the geometric series $\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n$.(b) Determine whether $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges absolutely, converges conditionally, or diverges.

Show answer
(a) $\frac{2/3}{1 – 2/3} = 2$(b) Converges conditionally: alternating series test applies (terms decrease to 0) but $\sum \frac{1}{n}$ diverges (harmonic).
[5]
4

Find the interval of convergence of the power series $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}$. Check the endpoints.

Show answer
Ratio test: $\frac{|x-2|}{3} < 1 \Rightarrow -1 < x < 5$. At $x = 5$: harmonic series, diverges. At $x = -1$: alternating harmonic, converges. Interval $[-1, 5)$, radius $R = 3$.
[6]
5

(a) Write the first four nonzero terms of the Maclaurin series for $e^x$.(b) Use your answer to write the first four nonzero terms of the Maclaurin series for $x e^{x^2}$.(c) Hence find $\frac{d^5}{dx^5}\left(x e^{x^2}\right)$ at $x = 0$.

Show answer
(a) $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$(b) $x + x^3 + \frac{x^5}{2} + \frac{x^7}{6}$(c) Coefficient of $x^5$ is $\frac{f^{(5)}(0)}{5!} = \frac{1}{2} \Rightarrow f^{(5)}(0) = 60$
[6]

Section B – Exam-style

Graphing calculator allowed.
6

Let $f(x) = \sin x$.(a) Write the third-degree Taylor polynomial $P_3(x)$ for $f$ about $x = 0$.(b) Use the Lagrange error bound to show that $|f(0.5) – P_3(0.5)| < 0.0003$.(c) Use $P_3$ to approximate $\int_0^{0.5} \frac{\sin x}{x}\,dx$.

Show answer
(a) $P_3(x) = x – \frac{x^3}{6}$(b) Since the $x^4$ term of the series is zero, $P_3 = P_4$, so the error is bounded by the degree-5 Lagrange term: $|R| \le \frac{\max|f^{(5)}|}{5!}(0.5)^5 \le \frac{0.5^5}{120} = 0.00026 < 0.0003$(c) $\int_0^{0.5}\left(1 – \frac{x^2}{6}\right)dx = 0.5 – \frac{0.125}{18} = 0.4931$
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7

The Taylor series for a function $g$ about $x = 1$ is $\sum_{n=0}^{\infty} \frac{(-1)^n (x-1)^n}{2^n (n+1)}$.(a) Find $g(1)$ and $g'(1)$.(b) Find the radius of convergence.(c) The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ is approximated by its first four terms. Use the alternating series error bound to find the maximum possible error.

Show answer
(a) $g(1) = 1$; $g'(1) = $ coefficient of $(x-1)$ $= -\frac{1}{4}$(b) Ratio: $\frac{|x-1|}{2} < 1 \Rightarrow R = 2$(c) Error $\le \frac{1}{5^2} = 0.04$ (sum of four terms $= 0.7986$)
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8

A particle moves in the plane with position $x(t) = t^2 – 3t$, $y(t) = 2t^3$ for $t \ge 0$.(a) Find the velocity vector at $t = 1$ and the speed of the particle at $t = 1$.(b) Find $\frac{dy}{dx}$ at $t = 1$.(c) Find the time when the particle is moving vertically.

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(a) $\langle -1, 6 \rangle$; speed $= \sqrt{37} = 6.08$(b) $\frac{6t^2}{2t-3} = -6$ at $t = 1$(c) $\frac{dx}{dt} = 0 \Rightarrow t = 1.5$ (and $\frac{dy}{dt} e 0$ there)
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9

The curve $C$ is given by $x = t^2$, $y = t^3$ for $0 \le t \le 2$.(a) Find $\frac{d^2y}{dx^2}$ in terms of $t$.(b) Find the length of $C$.

Show answer
(a) $\frac{dy}{dx} = \frac{3t}{2}$, so $\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$(b) $\int_0^2 t\sqrt{4 + 9t^2}\,dt = \frac{1}{27}\left[(4 + 9t^2)^{3/2}\right]_0^2 = \frac{40^{3/2} – 8}{27} = 9.07$
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10

The polar curve $r = 2 + 2\cos\theta$ (a cardioid) is drawn for $0 \le \theta \le 2\pi$.(a) Find the area enclosed by the curve.(b) Find $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$.

Show answer
(a) $\frac{1}{2}\int_0^{2\pi}(2 + 2\cos\theta)^2 d\theta = \frac{1}{2}(8\pi + 4\pi) = 6\pi = 18.8$(b) $x = r\cos\theta$, $y = r\sin\theta$; at $\theta = \frac{\pi}{2}$: $\frac{dx}{d\theta} = -2$, $\frac{dy}{d\theta} = -2$, so $\frac{dy}{dx} = 1$
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Section C – Challenge

Graphing calculator allowed. FRQ style.
11

The polar curves $r = 3$ and $r = 2 + 2\cos\theta$ are drawn.(a) Find the values of $\theta$ in $[0, 2\pi)$ where the curves intersect.(b) Set up and evaluate an integral for the area inside $r = 2 + 2\cos\theta$ and outside $r = 3$.(c) For the cardioid, find the rate of change of $r$ with respect to $\theta$ at $\theta = \frac{\pi}{3}$ and interpret its sign.

Show answer
(a) $2 + 2\cos\theta = 3 \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}, \frac{5\pi}{3}$(b) $A = 2 \cdot \frac{1}{2}\int_0^{\pi/3}\left[(2 + 2\cos\theta)^2 – 9\right]d\theta = \int_0^{\pi/3}(8\cos\theta + 4\cos^2\theta – 5)\,d\theta = 4\sqrt{3} + \frac{\sqrt{3}}{2} – \pi = 4.65$(c) $\frac{dr}{d\theta} = -2\sin\theta = -\sqrt{3} = -1.73$; negative, so the distance from the origin is decreasing as $\theta$ increases.
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12

Let $h(x) = \frac{1}{1 + x^2}$.(a) Write the Maclaurin series for $h(x)$ and state its interval of convergence.(b) By integrating your series, write the first four nonzero terms of the Maclaurin series for $\arctan x$.(c) Use the first three nonzero terms to approximate $\arctan(0.2)$ and use the alternating series bound to estimate the error.

Show answer
(a) $\sum_{n=0}^{\infty}(-1)^n x^{2n} = 1 – x^2 + x^4 – x^6 + \cdots$, $-1 < x < 1$(b) $x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7}$(c) $0.2 – \frac{0.008}{3} + \frac{0.00032}{5} = 0.197397$; error $\le \frac{0.2^7}{7} = 1.8 \times 10^{-6}$
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