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Question 1

(a) Topic-8.4-Finding the Area Between Curves Expressed as Functions of x

(b) Topic-8.9-Volume with Disc Method Revolving Around the x or y Axis

(c) Topic-8.7-Volumes with Cross Sections Squares and Rectangles

Let $R$ be the region in the first and second quadrants bounded above by the graph of
\(
y = \frac{20}{1 + x^2}
\)
and below by the horizontal line $y = 2$.
(a) Find the area of $R$.
(b) Find the volume of the solid generated when $R$ is rotated about the $x$-axis.
(c) The region $R$ is the base of a solid. For this solid, the cross sections perpendicular to the $x$-axis are semicircles. Find the volume of this solid.

▶️Answer/Explanation

\(
\frac{20}{1 + x^2} = 2 \text{ when } x = \pm 3
\)

\(\textbf{1(a)}\)
\(
\text{Area} = \int_{-3}^{3} \left(\frac{20}{1 + x^2} – 2\right) dx = 37.961 \text{ or } 37.962
\)

\(\textbf{1(b)}\)
\(
\text{Volume} = \pi \int_{-3}^{3} \left(\left(\frac{20}{1 + x^2}\right)^2 – 2^2\right) dx = 1871.190
\)

\(\textbf{1(c)}\)
\(
\text{Volume} = \frac{\pi}{2} \int_{-3}^{3} \left(\frac{1}{2}\left(\frac{20}{1 + x^2} – 2\right)\right)^2 dx
\)
\(
= \frac{\pi}{8} \int_{-3}^{3} \left(\frac{20}{1 + x^2} – 2\right)^2 dx = 174.268
\)

Question 2

(a) Topic-6.3-Riemann Sums Summation Notation and Definite Integral Notation

(b) Topic-5.3-Determining Intervals on Which a Function Is Increasing or Decreasing

(c) Topic-6.5-Interpreting the Behavior of Accumulation Functions Involving Area


The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval $0 \leq t \leq 7$, where $t$ is measured in hours. In this model, rates are given as follows:

(i) The rate at which water enters the tank is $f(t) = 100t^2 \sin(\sqrt{t})$ gallons per hour for $0 \leq t \leq 7$.
(ii) The rate at which water leaves the tank is
\(
g(t) =
\begin{cases}
250 & \text{for } 0 \leq t < 3, \\
2000 & \text{for } 3 \leq t \leq 7.
\end{cases}
\)
The graphs of $f$ and $g$, which intersect at $t = 1.617$ and $t = 5.076$, are shown in the figure above. At time $t = 0$, the amount of water in the tank is 5000 gallons.
(a) How many gallons of water enter the tank during the time interval $0 \leq t \leq 7$? Round your answer to the nearest gallon.
(b) For $0 \leq t \leq 7$, find the time intervals during which the amount of water in the tank is decreasing. Give a reason for each answer.
(c) For $0 \leq t \leq 7$, at what time $t$ is the amount of water in the tank greatest? To the nearest gallon, compute the amount of water at this time. Justify your answer.

▶️Answer/Explanation

\(\textbf{2(a)}\)
$\int_0^7 f(t) \, dt \approx 8264 \, \text{gallons}$

\(\textbf{2(b)}\)
The amount of water in the tank is decreasing on the intervals $0 \leq t \leq 1.617$ and $3 \leq t \leq 5.076$
because $f(t) < g(t)$ for $0 \leq t < 1.617$ and $3 < t < 5.076$.

\(\textbf{2(c)}\)
Since $f(t) – g(t)$ changes sign from positive to negative only at $t = 3$, the candidates for the absolute maximum are at $t = 0$, $3$, and $7$.
\( \begin{array}{|c|c|} \hline t \, (\text{hours}) & \text{gallons of water} \\ \hline 0 & 5000 \\ 3 & 5000 + \int_0^3 f(t) \, dt – 250(3) = 5126.591 \\ 7 & 5126.591 + \int_3^7 f(t) \, dt – 2000(4) = 4513.807 \\ \hline \end{array} \)
The amount of water in the tank is greatest at $t = 3$ hours. At that time, the amount of water in the tank, rounded to the nearest gallon, is $5127$ gallons.

Question 3

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

(b) Topic-9.7-Defining Polar Coordinates and Differentiating in Polar Form

(c) Topic-9.7-Defining Polar Coordinates and Differentiating in Polar Form


The graphs of the polar curves $r = 2$ and $r = 3 + 2\cos\theta$ are shown in the figure above.
The curves intersect when $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$.
(a) Let $R$ be the region that is inside the graph of $r = 2$ and also inside the graph of $r = 3 + 2\cos\theta$, as shaded in the figure above. Find the area of $R$.
(b) A particle moving with nonzero velocity along the polar curve given by $r = 3 + 2\cos\theta$ has position $(x(t), y(t))$ at time $t$, with $\theta = 0$ when $t = 0$.
This particle moves along the curve so that $\frac{dr}{dt} = \frac{dr}{d\theta}$.
Find the value of $\frac{dr}{dt}$ at $\theta = \frac{\pi}{3}$ and interpret your answer in terms of the motion of the particle.
(c) For the particle described in part (b), $\frac{dy}{dt} = \frac{dy}{d\theta}$.
Find the value of $\frac{dy}{dt}$ at $\theta = \frac{\pi}{3}$ and interpret your answer in terms of the motion of the particle.

▶️Answer/Explanation

\(\textbf{3(a)}\)
\(
\text{Area} = \frac{2}{3}\pi(2)^2 + \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (3 + 2\cos\theta)^2 \, d\theta
\)
\(
= 10.370
\)

\(\textbf{3(b)}\)
\(
\left.\frac{dr}{dt}\right|_{\theta = \frac{\pi}{3}} = \left.\frac{dr}{d\theta}\right|_{\theta = \frac{\pi}{3}} = -1.732
\)
The particle is moving closer to the origin, since $\frac{dr}{dt} < 0$ and $r > 0$ when $\theta = \frac{\pi}{3}$.

\(\textbf{3(c)}\)
\(
y = r\sin\theta = (3 + 2\cos\theta)\sin\theta
\)
\(
\left.\frac{dy}{dt}\right|_{\theta = \frac{\pi}{3}} = \left.\frac{dy}{d\theta}\right|_{\theta = \frac{\pi}{3}} = 0.5
\)
The particle is moving away from the $x$-axis, since $\frac{dy}{dt} > 0$ and $y > 0$ when $\theta = \frac{\pi}{3}$.

Question 4

(a) Topic-5.8-Sketching Graphs of Functions and Their Derivatives

(b) Topic-5.6-Determining Concavity of Functions over Their Domains

(c) Topic-6.8-Finding Antiderivatives and Indefinite Integrals Basic Rules and Notation

Let $f$ be the function defined for $x > 0$, with $f(e) = 2$ and $f’$, the first derivative of $f$, given by
\(
f'(x) = x^2 \ln x.
\)
(a) Write an equation for the line tangent to the graph of $f$ at the point $(e, 2)$.
(b) Is the graph of $f$ concave up or concave down on the interval $1 < x < 3$? Give a reason for your answer.
(c) Use antidifferentiation to find $f(x)$.

▶️Answer/Explanation

\(\textbf{4(a)}\)
\(
f'(e) = e^2
\)
An equation for the line tangent to the graph of $f$ at the point $(e, 2)$ is
\(
y – 2 = e^2(x – e).
\)

\(\textbf{4(b)}\)
\(
f”(x) = x + 2\ln x.
\)
For $1 < x < 3$, $x > 0$ and $\ln x > 0$, so $f”(x) > 0$. Thus, the graph of $f$ is concave up on $(1, 3)$.

\(\textbf{4(c)}\)
Since $f(x) = \int (x^2 \ln x) \, dx$, we consider integration by parts.
\(
u = \ln x \quad dv = x^2 dx
\)
\(
du = \frac{1}{x} dx \quad v = \int x^2 dx = \frac{1}{3}x^3
\)
Therefore,
\(
f(x) = \int (x^2 \ln x) \, dx = \frac{1}{3}x^3 \ln x – \int \left(\frac{1}{3}x^3 \cdot \frac{1}{x}\right) dx
\)
\(
= \frac{1}{3}x^3 \ln x – \frac{1}{9}x^3 + C.
\)
Since $f(e) = 2$,
\(
2 = \frac{e^3}{3} \ln e – \frac{e^3}{9} + C = \frac{e^3}{3} – \frac{e^3}{9} + C.
\)
\(
C = 2 – \frac{2}{9}e^3.
\)
Thus,
\(
f(x) = \frac{x^3}{3} \ln x – \frac{1}{9}x^3 + 2 – \frac{2}{9}e^3.
\)

Question 5

(a) Topic-4.6-Approximating Values of a Function Using Local Linearity and Linearization

(b) Topic-6.6-Applying Properties of Definite Integrals

(c) Topic-6.2-Approximating Areas with Riemann Sums

(d) Topic-6.2-Approximating Areas with Riemann Sums


The volume of a spherical hot air balloon expands as the air inside the balloon is heated.
The radius of the balloon, in feet, is modeled by a twice-differentiable function $r$ of time $t$, where $t$ is measured in minutes.
For $0 < t < 12$, the graph of $r$ is concave down.
The table above gives selected values of the rate of change, $r'(t)$, of the radius of the balloon over the time interval $0 \leq t \leq 12$. The radius of the balloon is 30 feet when $t = 5$.

(Note: The volume of a sphere of radius $r$ is given by $V = \frac{4}{3} \pi r^3$.)
(a) Estimate the radius of the balloon when $t = 5.4$ using the tangent line approximation at $t = 5$. Is your estimate greater than or less than the true value? Give a reason for your answer.
(b) Find the rate of change of the volume of the balloon with respect to time when $t = 5$. Indicate units of measure.
(c) Use a right Riemann sum with the five subintervals indicated by the data in the table to approximate $\int_0^{12} r'(t) \, dt$. Using correct units, explain the meaning of $\int_0^{12} r'(t) \, dt$ in terms of the radius of the balloon.
(d) Is your approximation in part (c) greater than or less than $\int_0^{12} r'(t) \, dt$? Give a reason for your answer.

▶️Answer/Explanation

\(\textbf{5(a)}\)
\(
r(5.4) \approx r(5) + r'(5)\Delta t = 30 + 2(0.4) = 30.8 \, \text{ft}
\)
Since the graph of $r’$ is concave down on the interval $5 < t < 5.4$, this estimate is greater than $r(5.4)$.

\(\textbf{5(b)}\)
\(
\frac{dV}{dt} = 3 \left(\frac{4}{3}\pi r^2 \frac{dr}{dt}\right)
\)
\(
\left.\frac{dV}{dt}\right|_{t=5} = 4\pi(30)^2(2) = 7200\pi \, \text{ft}^3/\text{min}
\)

\(\textbf{5(c)}\)
\(
\int_0^{12} r'(t) \, dt \approx 2(4.0) + 3(2.0) + 2(1.2) + 4(0.6) + 1(0.5) = 19.3 \, \text{ft}
\)
\(
\int_0^{12} r'(t) \, dt \text{ is the change in the radius, in feet, from } t = 0 \text{ to } t = 12 \text{ minutes.}
\)

\(\textbf{5(d)}\)
Since $r$ is concave down, $r’$ is decreasing on $0 < t < 12$. Therefore, this approximation, $19.3 \, \text{ft}$, is less than $\int_0^{12} r'(t) \, dt$.
\( \text{Units of ft}^3/\text{min in part (b) and ft in part (c)}. \)

Question 6

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

(b) Topic-10.9-Determining Absolute or Conditional Convergence

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

(d) Topic-10.12-Lagrange Error Bound

Let $f$ be the function given by $f(x) = e^{-x^2}$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 0$.
(b) Use your answer to part (a) to find
\(
\lim_{x \to 0} \frac{1 – x^2 – f(x)}{x^4}.
\)
(c) Write the first four nonzero terms of the Taylor series for $\int_0^x e^{-t^2} \, dt$ about $x = 0$. Use the first two terms of your answer to estimate
\(
\int_0^{1/2} e^{-t^2} \, dt.
\)
(d) Explain why the estimate found in part (c) differs from the actual value of $\int_0^{1/2} e^{-t^2} \, dt$ by less than $\frac{1}{200}$.

▶️Answer/Explanation

\(\textbf{6(a)}\)
\(
e^{-x^2} = 1 + \frac{(-x^2)}{1!} + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \dots + \frac{(-x^2)^n}{n!} + \dots
\)
\(
= 1 – x^2 + \frac{x^4}{2} – \frac{x^6}{6} + \dots + \frac{(-1)^n x^{2n}}{n!}.
\)

\(\textbf{6(b)}\)
\(
\frac{1 – x^2 – f(x)}{x^4} = \frac{-\frac{1}{2}x^2 + \frac{x^4}{6} + \sum_{n=4}^\infty \frac{(-1)^{n+1} x^{2n-4}}{n!}}{x^4}.
\)
Thus,
\(
\lim_{x \to 0} \frac{1 – x^2 – f(x)}{x^4} = -\frac{1}{2}.
\)

\(\textbf{6(c)}\)
\(
\int_0^x e^{-t^2} \, dt = \int_0^x \left( 1 – t^2 + \frac{t^4}{2} – \frac{t^6}{6} + \dots + \frac{(-1)^n t^{2n}}{n!} \right) \, dt
\)
\(
= x – \frac{x^3}{3} + \frac{x^5}{10} – \frac{x^7}{42} + \dots.
\)
Using the first two terms of this series, we estimate that
\(
\int_0^{1/2} e^{-t^2} \, dt \approx \frac{1}{2} – \frac{\left(\frac{1}{2}\right)^3}{3} = \frac{11}{24}.
\)

\(\textbf{6(d)}\)
\(
\int_0^{1/2} e^{-t^2} \, dt – \frac{11}{24} < \left(\frac{1}{2}\right)^5 \frac{1}{10} = \frac{1}{320} < \frac{1}{200},
\)
since
\(
\int_0^{1/2} e^{-t^2} \, dt = \sum_{n=0}^\infty \frac{(-1)^n \left(\frac{1}{2}\right)^{2n+1}}{n!(2n+1)},
\)
which is an alternating series with individual terms that decrease in absolute value to $0$.

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