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ap19-frq-calculus-bc

Question 1

(a) Topic-6.9- Integrating Using Substitution

(b) Topic-8.1 Finding the Average Value of a Function on an Interval

(c) Topic-8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

(d) Topic-8.2- Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Fish enter a lake at a rate modeled by the function E given by \(E(t)=20 + 15 sin\left ( \frac{\pi t}{6} \right ).\) Fish leave the lake at a rate modeled by the function L given by \(L(t) = 4 + 2^{0.1t^2}\). Both E (t) and L (t) are measured in fish per hour, and t is measured in hours since midnight (t = 0).
(a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ? Give your answer to the nearest whole number.
(b) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ?
(c) At what time t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake? Justify your answer.
(d) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. (t = 5) ? Explain your reasoning. 

▶️Answer/Explanation

1(a)

\(\int_{0}^{5} E(t) \, dt = 153.457690 \)

\(\text{To the nearest whole number, 153 fish enter the lake from midnight to 5 A.M.} \)

1(b)

\(\frac{1}{5 – 0} \int_{0}^{5} L(t) \, dt = 6.059038 \)

\(\text{The average number of fish that leave the lake per hour from midnight to 5 A.M. is 6.059 fish per hour.} \)

1(c)

\(\text{The rate of change in the number of fish in the lake at time } t \text{ is given by } E(t) – L(t). \)

\(E(t) – L(t) = 0 \implies t = 6.20356 \)

\(E(t) – L(t) > 0 \text{ for } 0 \leq t < 6.20356, \text{ and } E(t) – L(t) < 0 \text{ for } 6.20356 < t \leq 8\).
\(\text{ Therefore, the greatest number of fish in the lake is at time } t = 6.204 \text{ (or 6.203).} \)

1(d)

\(E'(5) – L'(5) = -10.7228 < 0 \)

\(\text{Because } E'(5) – L'(5) < 0, \text{ the rate of change in the number of fish is decreasing at time } t = 5. \)

Question 2

(a) Topic-9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

(b) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

(c) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

(d) Topic-9.1 Defining and Differentiating Parametric Equations

Let $S$ be the region bounded by the graph of the polar curve $r(\theta) = 3\sqrt{\theta}\sin(\theta^2)$ for $0 \leq \theta \leq \sqrt{\pi}$, as shown in the figure above.

(a) Find the area of $S$.

(b) What is the average distance from the origin to a point on the polar curve $r(\theta) = 3\sqrt{\theta}\sin(\theta^2)$ for $0 \leq \theta \leq \sqrt{\pi}$?

(c) There is a line through the origin with positive slope $m$ that divides the region $S$ into two regions with equal areas. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of $m$.

(d) For $k > 0$, let $A(k)$ be the area of the portion of region $S$ that is also inside the circle $r = k \cos \theta$. Find
\(\lim_{k \to \infty} A(k).\)

▶️Answer/Explanation

\(\textbf{2(a)}\)
\(
r(\theta) = 3\sqrt{\theta} \sin(\theta^2) \quad \text{for } 0 \leq \theta \leq \sqrt{\pi}.
\)
\(
\text{Area} = \frac{1}{2} \int_{0}^{\sqrt{\pi}} \left(r(\theta)\right)^2 \, d\theta \approx 3.534291735.
\)
\(\textbf{2(b)}\)
The distance from the origin to a point on the curve is $r$.
\(
r_{\text{avg}} = \frac{1}{\sqrt{\pi} – 0} \int_{0}^{\sqrt{\pi}} r(\theta) \, d\theta \approx 1.57993277.
\)
\(\textbf{2(c)}\)
Since the slope of the line is $m$, then $m = \tan \theta$. So $\theta = \tan^{-1} m$ is where the line intersects $r(\theta)$.
So, the area from $0$ to $\tan^{-1} m$ is equal to the area from $\tan^{-1} m$ to $\sqrt{\pi}$:
\(
\frac{1}{2} \int_{0}^{\tan^{-1} m} \left(r(\theta)\right)^2 \, d\theta = \frac{1}{2} \int_{\tan^{-1} m}^{\sqrt{\pi}} \left(r(\theta)\right)^2 \, d\theta.
\)
\(\textbf{2(d)}\)
As $k \to \infty$, $r = k \cos \theta$ is a circle that gets bigger, and $\theta \to \frac{\pi}{2}$.
\(
\lim_{k \to \infty} A(k) = \frac{1}{2} \int_{0}^{\pi/2} \left(r(\theta)\right)^2 \, d\theta \approx 3.324470722.
\)

Question 3

(a) Topic-6.6 Applying Properties of Definite Integrals

(b) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals

(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(d) Topic-1.4 Estimating Limit Values from Tables

The continuous function f is defined on the closed interval −6 ≤ x ≤ 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5, 3). It is known that the point \((3,3-\sqrt{5})\) is on the graph of f. 

(a) If \(\int_{-6}^{5}f(x)dx=7,\) find the value of \(\int_{-6}^{-2}f(x)dx.\) Show the work that leads to your answer
(b) Evaluate \(\int_{3}^{5}(2f'(x)+4)dx.\)
(c) The function g is given by \(g(x)=\int_{-2}^{x}f(t)dt.\) Find the absolute maximum value of g on the interval − 2 ≤ x ≤ 5. Justify your answer.
(d) Find \(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}.\)

▶️Answer/Explanation

3(a)

\(\int_{-6}^{5} f(x) \, dx = \int_{-6}^{-2} f(x) \, dx + \int_{-2}^{5} f(x) \, dx \)

\(\implies 7 = \int_{-6}^{-2} f(x) \, dx + 2 + \left(9 – \frac{9\pi}{4}\right) \)

\(\implies \int_{-6}^{-2} f(x) \, dx = 7 – \left(11 – \frac{9\pi}{4}\right) = \frac{9\pi}{4} – 4 \)

3(b)

\(\int_{3}^{5} \big(2f'(x) + 4\big) \, dx = 2\int_{3}^{5} f'(x) \, dx + \int_{3}^{5} 4 \, dx\)

\(= 2\big(f(5) – f(3)\big) + 4(5 – 3)\)

\(= 2\big(0 – (3 – \sqrt{5})\big) + 8\)

\(= 2(-3 + \sqrt{5}) + 8 = 2 + 2\sqrt{5}\)

3(c)

\(g'(x) = f(x) = 0 \implies x = -1, \, x = \frac{1}{2}, \, x = 5\)

\(\text{On the interval } -2 \leq x \leq 5, \text{ the absolute maximum value of } g \text{ is } g(5) = 11 – \frac{9\pi}{4}.\)

3(d)

\(\lim_{x \to 1} \frac{10^x – 3f'(x)}{f(x) – \arctan x} = \frac{10^1 – 3f'(1)}{f(1) – \arctan 1}\)

\(\quad = \frac{10 – 3 \cdot 2}{1 – \arctan 1} = \frac{4}{1 – \frac{\pi}{4}}\)

Question 4

(a) Topic-4.5 Solving Related Rates Problems

(b) Topic-4.5 Solving Related Rates Problems

(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(d) Topic-7.6 Finding General Solutions Using Separation of Variables

A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height h of the water in the barrel with respect to time t is modeled by \(\frac{dh}{dt}=-\frac{1}{10}\sqrt{h},\) where h is measured in feet and t is measured in seconds. (The volume V of a cylinder with radius r and height h is V = πr2h.)
(a) Find the rate of change of the volume of water in the barrel with respect to time when the height of the water is 4 feet. Indicate units of measure.
(b) When the height of the water is 3 feet, is the rate of change of the height of the water with respect to time increasing or decreasing? Explain your reasoning.
(c) At time t = 0 seconds, the height of the water is 5 feet. Use separation of variables to find an expression for h in terms of t. 

▶️Answer/Explanation

\(\textbf{4(a)} \quad V = \pi r^2 h = \pi (1)^2 h = \pi h \)

\(\frac{dV}{dt}\bigg|_{h=4} = \pi \frac{dh}{dt}\bigg|_{h=4} = \pi \left(-\frac{1}{10} \sqrt{4}\right) = -\frac{\pi}{5} \text{ cubic feet per second.} \)

\(\textbf{4(b)} \quad \frac{d^2h}{dt^2} = -\frac{1}{20\sqrt{h}} \cdot \frac{dh}{dt} = -\frac{1}{20\sqrt{h}} \cdot \left(-\frac{1}{10} \sqrt{h}\right) = \frac{1}{200} \)

\(\text{Because } \frac{d^2h}{dt^2} = \frac{1}{200} > 0 \text{ for } h > 0, \text{ the rate of change of the height is increasing when the height of the water is 3 feet.} \)

\(\textbf{4(c)} \quad \frac{dh}{\sqrt{h}} = -\frac{1}{10} \, dt \)

\(\int \frac{dh}{\sqrt{h}} = \int -\frac{1}{10} \, dt \)

\(2\sqrt{h} = -\frac{1}{10} t + C \)

\(2\sqrt{5} = -\frac{1}{10} \cdot 0 + C \implies C = 2\sqrt{5} \)

\(2\sqrt{h} = -\frac{1}{10} t + 2\sqrt{5} \)

\(h(t) = \left(-\frac{1}{20} t + \sqrt{5}\right)^2 \)

Question 5

(a) Topic-4.5 Solving Related Rates Problems

(b) Topic-4.5 Solving Related Rates Problems

(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(d) Topic-7.6 Finding General Solutions Using Separation of Variables

Consider the family of functions $f(x) = \frac{1}{x^2 – 2x + k}$, where $k$ is a constant.

(a) Find the value of $k$, for $k > 0$, such that the slope of the line tangent to the graph of $f$ at $x = 0$ equals 6.

(b) For $k = -8$, find the value of $\int_{0}^{1} f(x) \, dx$.

(c) For $k = 1$, find the value of $\int_{0}^{2} f(x) \, dx$ or show that it diverges.

▶️Answer/Explanation

\(\textbf{5(a)}\)
\(
f(x) = \frac{1}{x^2 – 2x + k} = (x^2 – 2x + k)^{-1}
\)
\(
f'(x) = -(x^2 – 2x + k)^{-2}(2x – 2)
\)
\(
f'(0) = \frac{-2}{k^2} = \frac{2}{k^2} \implies 6 = \frac{2}{k^2} \implies k = \sqrt{\frac{1}{3}}.
\)
\(\text{Note: only the positive square root since $k > 0$.}\)
\(\textbf{5(b)}\)
\(
\int_{0}^{1} \frac{1}{x^2 – 2x – 8} \, dx = \int_{0}^{1} \frac{1}{(x-4)(x+2)} \, dx.
\)
Partial fractions:
\(
\frac{1}{(x-4)(x+2)} = \frac{A}{x-4} + \frac{B}{x+2}.
\)
\(
1 = A(x+2) + B(x-4).
\)
For $x = -2$:
\(
B = \frac{-1}{6}.
\)
For $x = 4$:
\(
A = \frac{1}{6}.
\)
\(
\int_{0}^{1} \frac{1}{x^2 – 2x – 8} \, dx = \int_{0}^{1} \frac{1}{(x-4)(x+2)} \, dx = \frac{1}{6} \int_{0}^{1} \left(\frac{1}{x-4} – \frac{1}{x+2}\right) \, dx.
\)
\(
= \frac{1}{6} \left[ \ln |x-4| – \ln |x+2| \right]_0^1.
\)
\(
= \frac{1}{6} \left[ \ln |1-4| – \ln |1+2| \right].
\)
\(
= \frac{1}{6} (\ln 1 – \ln 2) = -\frac{1}{6} \ln 2.
\)
\(\textbf{(c)}\)
\(
\int_{0}^{2} \frac{1}{x^2 – 2x + 1} \, dx = \int_{0}^{2} \frac{1}{(x-1)^2} \, dx \quad \text{(Improper integral)}.
\)
\(
= \lim_{a \to 1^-} \int_{0}^{a} (x-1)^{-2} \, dx + \lim_{b \to 1^+} \int_{b}^{2} (x-1)^{-2} \, dx.
\)
\(
= \lim_{a \to 1^-} \left[-(x-1)^{-1}\right]_{0}^{a} + \lim_{b \to 1^+} \left[-(x-1)^{-1}\right]_{b}^{2}.
\)
\(
= -\lim_{a \to 1^-} \left[\frac{1}{a-1} – \frac{1}{0-1}\right] – \lim_{b \to 1^+} \left[\frac{1}{2-1} – \frac{1}{b-1}\right].
\)
\(
= \lim_{a \to 1^-} \frac{1}{a-1} \quad \text{diverges, so the integral diverges.}
\)

Question 6

(a) Topic-10.11 Finding Taylor Polynomial Approximations of Functions

(b) Topic-10.14 Finding Taylor or Maclaurin Series for a Function

(c) Topic-10.1 Defining Convergent and Divergent Infinite Series

(d) Topic-10.10 Alternating Series Error Bound

A function $f$ has derivatives of all orders for all real numbers $x$. A portion of the graph of $f$ is shown above, along with the line tangent to the graph of $f$ at $x = 0$. Selected derivatives of $f$ at $x = 0$ are given in the table above.

(a) Write the third-degree Taylor polynomial for $f$ about $x = 0$.

(b) Write the first three nonzero terms of the Maclaurin series for $e^x$. Write the second-degree Taylor polynomial for $e^x f(x)$ about $x = 0$.

(c) Let $h$ be the function defined by $h(x) = \int_{0}^{x} f(t) \, dt$. Use the Taylor polynomial found in part (a) to find an approximation for $h(1)$.

(d) It is known that the Maclaurin series for $h$ converges to $h(x)$ for all real numbers $x$. It is also known that the individual terms of the series for $h(1)$ alternate in sign and decrease in absolute value to $0$. Use the alternating series error bound to show that the approximation found in part (c) differs from $h(1)$ by at most $0.45$.

▶️Answer/Explanation

$\textbf{6(a)}$
The graph shown is the graph of $f$ and a tangent line.
$f(0) = 3 \quad \text{from the graph.}$
$f'(0) = \frac{-3}{3/2} = -2 \quad \text{from the tangent line.}$
$f^{”}(0) = 3 \quad \text{from the chart.}$
$f^{(3)}(0) = \frac{-23}{2} \quad \text{from the chart.}$
The third-degree Taylor polynomial is:
$T_3(x) = 3 – 2x + \frac{3}{2!}x^2 – \frac{23/2}{3!}x^3.$
$\textbf{6(b)}$
The Maclaurin series for $e^x$ is:
$e^x = 1 + x + \frac{x^2}{2!} + \cdots$
The product $e^x f(x)$ is:
$e^x f(x) \approx \left(1 + x + \frac{x^2}{2!} + \cdots\right)\left(3 – 2x + \frac{3}{2!}x^2 – \frac{23/2}{3!}x^3 + \cdots\right).$
$\approx \left(3 – 2x + \frac{3}{2!}x^2 + \cdots\right) + \left(3x – 2x^2 + \cdots\right) + \left(\frac{3x^2}{2!} + \cdots\right).$
Simplifying:
$e^x f(x) \approx 3 + x + \left(\frac{3}{2!} – 2 + \frac{3}{2!}\right)x^2.$
$\textbf{6(c)}$
$h(1) = \int_{0}^{1} f(t) \, dt \approx \int_{0}^{1} \left(3 – 2x + \frac{3}{2!}x^2 – \frac{23/2}{3!}x^3\right) \, dx$
$= \left[3x – x^2 + \frac{1}{2}x^3 – \frac{23}{48}x^4\right]_{0}^{1}.$
$= 3 – 1 + \frac{1}{2} – \frac{23}{48} = \frac{97}{48}.$
$\textbf{6(d)}$
The alternating series error bound is the first unused term of $h(x)$ found in part (c).
The first unused term of $f(x)$ from part (a) is $\frac{54}{4!}x^4$, so the first unused term of the approximation in part (c) $h(x)$ is:
$\int_{0}^{1} \frac{54}{4!}x^4 \, dx = \frac{54}{4!} \int_{0}^{1} x^4 \, dx.$
$= \frac{54}{4!} \cdot \frac{x^5}{5}\Big|_{0}^{1} = \frac{54}{4! \cdot 5}.$
$= \frac{54}{120} = \frac{9}{20} = 0.45.$
This is the alternating series error bound, so the approximation found in part (c) differs from $h(1)$ by at most $0.45$.

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