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Question 1

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

(b) Topic-8.10 Volume with Disc Method Revolving Around Other Axes

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


Let $R$ be the shaded region bounded by the graphs of $y = \sqrt{x}$ and $y = e^{-3x}$ and the vertical line $x = 1$, as shown in the figure above.
(a) Find the area of $R$.
(b) Find the volume of the solid generated when $R$ is revolved about the horizontal line $y = 1$.
(c) The region $R$ is the base of a solid. For this solid, each cross section perpendicular to the $x$-axis is a rectangle whose height is 5 times the length of its base in region $R$. Find the volume of this solid.

▶️Answer/Explanation

\(\textbf{1(a)}\)
Point of intersection
$e^{-3x} = \sqrt{x}$ at $(T, S) = (0.238734, 0.488604)$
$\text{Area} = \int_T^1 (\sqrt{x} – e^{-3x}) , dx = 0.442 \text{ or } 0.443$

\(\textbf{1(b)}\)
$\text{Volume} = \pi \int_T^1 \left((1 – e^{-3x})^2 – (1 – \sqrt{x})^2\right) dx = 0.453\pi \text{ or } 1.423 \text{ or } 1.424$

\(\textbf{1(c)}\)
Length $= \sqrt{x} – e^{-3x}$
Height $= 5(\sqrt{x} – e^{-3x})$
$\text{Volume} = \int_T^1 5(\sqrt{x} – e^{-3x})^2 dx = 1.554$

Question 2

(a) Topic-9.4 Analyzing the Signs of Derivatives to Determine Behavior

(b) Topic-9.3 Critical Points and Undefined Slopes

(c) Topic-9.6 Tangent Line Approximation and Velocity Calculations

(d) Topic-7.5 Distance Between Points Using Particle Trajectory


A particle starts at point $A$ on the positive $x$-axis at time $t = 0$ and travels along the curve from $A$ to $B$ to $C$ to $D$, as shown above. The coordinates of the particle’s position $(x(t), y(t))$ are differentiable functions of $t$, where
$x'(t) = \frac{dx}{dt} = -9\cos\left(\frac{\pi}{6}\right) \sin\left(\frac{\pi \sqrt{t} + 1}{2}\right)$
and $y'(t) = \frac{dy}{dt}$ is not explicitly given. At time $t = 9$, the particle reaches its final position at point $D$ on the positive $x$-axis.
(a) At point $C$, is $\frac{dy}{dt}$ positive? At point $C$, is $\frac{dx}{dt}$ positive? Give a reason for each answer.
(b) The slope of the curve is undefined at point $B$. At what time $t$ is the particle at point $B$?
(c) The line tangent to the curve at the point $(x(8), y(8))$ has equation $y = \frac{5}{9}x – 2$. Find the velocity vector and the speed of the particle at this point.
(d) How far apart are points $A$ and $D$, the initial and final positions, respectively, of the particle?

▶️Answer/Explanation

\(\textbf{2(a)}\)
At point $C$, $\frac{dy}{dt}$ is not positive because $y(t)$ is decreasing along the arc $BD$ as $t$ increases.
At point $C$, $\frac{dx}{dt}$ is not positive because $x(t)$ is decreasing along the arc $BD$ as $t$ increases.

\(\textbf{2(b)}\)
\(
\frac{dx}{dt} = 0; \cos\left(\frac{\pi}{6}\right) = 0 \ \text{or} \ \sin\left(\frac{\pi\sqrt{t} + 1}{2}\right) = 0
\)
\(
\frac{\pi}{6} = \frac{\pi}{2} \ \text{or} \ \frac{\pi\sqrt{t} + 1}{2} = \pi; \ t = 3 \ \text{for both.}
\)
Particle is at point $B$ at $t = 3$.

\(\textbf{2(c)}\)
\(
x'(8) = -9\cos\left(\frac{4\pi}{3}\right)\sin\left(\frac{3\pi}{2}\right) = -\frac{9}{2}
\)
\(
\frac{y'(8)}{x'(8)} = \frac{dy}{dx} = \frac{5}{9}, \ \therefore y'(8) = \frac{5}{9}x'(8) = -\frac{5}{2}.
\)
The velocity vector is $\langle -4.5, -2.5 \rangle$.
Speed $= \sqrt{4.5^2 + 2.5^2} = 5.147 \ \text{or} \ 5.148$.

\(\textbf{2(d)}\)
\(
x(9) – x(0) = \int_0^9 x'(t)dt = -39.255
\)
The initial and final positions are $39.255$ apart.

Question 3

(a) Topic-6.1 Derivatives and Slopes of Curves

(b) Topic-8.4 Finding the Area Between Curves Expressed as Functions of yy

(c) Topic-11.3 Converting Cartesian Equations to Polar Equations

(d) Topic-11.4 Setting Up Integrals in Polar Coordinates for Area


The figure above shows the graphs of the line $x = \frac{5}{3}y$ and the curve $C$ given by $x = \sqrt{1 + y^2}$. Let $S$ be the shaded region bounded by the two graphs and the $x$-axis. The line and the curve intersect at point $P$.
(a) Find the coordinates of point $P$ and the value of $\frac{dx}{dy}$ for curve $C$ at point $P$.
(b) Set up and evaluate an integral expression with respect to $y$ that gives the area of $S$.
(c) Curve $C$ is a part of the curve $x^2 – y^2 = 1$. Show that $x^2 – y^2 = 1$ can be written as the polar equation
\(
r^2 = \frac{1}{\cos^2\theta – \sin^2\theta}.
\)
(d) Use the polar equation given in part (c) to set up an integral expression with respect to the polar angle $\theta$ that represents the area of $S$.

▶️Answer/Explanation

\(\textbf{3(a)}\)
At $P$, $\frac{5}{3}y = \sqrt{1 + y^2}$, so $y = \frac{3}{4}$.
Since $x = \frac{5}{3}y$, $x = \frac{5}{4}$.
\(
\frac{dx}{dy} = \frac{y}{\sqrt{1 + y^2}} = \frac{y}{x}. \quad \text{At } P, \, \frac{dx}{dy} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}.
\)

\(\textbf{3(b)}\)
Area $= \int_{0}^{3/4} \left( \sqrt{1 + y^2} – \frac{5}{3}y \right) \, dy = 0.346 \, \text{or} \, 0.347$.

\(\textbf{3(c)}\)
$x = r \cos\theta; \, y = r \sin\theta$
$x^2 – y^2 = 1 \implies r^2 \cos^2\theta – r^2 \sin^2\theta = 1$
\(
r^2 = \frac{1}{\cos^2\theta – \sin^2\theta}.
\)

\(\textbf{3(d)}\)
Let $\beta$ be the angle that segment $OP$ makes with the $x$-axis. Then $\tan \beta = \frac{y}{x} = \frac{3/4}{5/4} = \frac{3}{5}$.
\(
\text{Area} = \int_{0}^{\tan^{-1}(3/5)} \frac{1}{2} r^2 \, d\theta
\)
\(
= \frac{1}{2} \int_{0}^{\tan^{-1}(3/5)} \frac{1}{\cos^2\theta – \sin^2\theta} \, d\theta.
\)

Question 4

(a) Topic-5.4 Intervals of Increase and Decrease Using Derivatives

(b) Topic-5.6 Identifying Points of Inflection Using the Second Derivative

(c) Topic-6.1 Derivatives and Tangent Lines

(d) Topic-4.1 Computing Definite Integrals and Using the Fundamental Theorem of Calculus


Let $f$ be a function defined on the closed interval $-3 \leq x \leq 4$ with $f(0) = 3$. The graph of $f’$, the derivative of $f$, consists of one line segment and a semicircle, as shown above.
(a) On what intervals, if any, is $f$ increasing? Justify your answer.
(b) Find the $x$-coordinate of each point of inflection of the graph of $f$ on the open interval $-3 < x < 4$. Justify your answer.
(c) Find an equation for the line tangent to the graph of $f$ at the point $(0, 3)$.
(d) Find $f(-3)$ and $f(4)$. Show the work that leads to your answers.

▶️Answer/Explanation

\(\textbf{4(a)}\)
The function $f$ is increasing on $[-3, -2]$ since $f’ > 0$ for $-3 \leq x < -2$.

\(\textbf{4(b)}\)
$x = 0$ and $x = 2$.
$f’$ changes from decreasing to increasing at $x = 0$ and from increasing to decreasing at $x = 2$.

\(\textbf{4(c)}\)
$f'(0) = -2$
Tangent line is $y = -2x + 3$

\(\textbf{4(d)}\)
$f(0) – f(-3) = \int_{-3}^0 f'(t) \, dt$
\(
= \frac{1}{2}(1)(1) – \frac{1}{2}(2)(2) = -\frac{3}{2}
\)
\(
f(-3) = f(0) + \frac{3}{2} = \frac{9}{2}
\)
$f(4) – f(0) = \int_0^4 f'(t) \, dt$
\(
= \left(-8 – \frac{1}{2}(2)^2 \pi \right) = -8 + 2\pi
\)
\(
f(4) = f(0) – 8 + 2\pi = -5 + 2\pi
\)

Question 5

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

(b) Topic-7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

(c) Topic-7.8 Exponential Models with Differential Equations

A coffeepot has the shape of a cylinder with radius $5$ inches, as shown in the figure above. Let $h$ be the depth of the coffee in the pot, measured in inches, where $h$ is a function of time $t$, measured in seconds. The volume $V$ of coffee in the pot is changing at the rate of $-5\pi \sqrt{h}$ cubic inches per second. (The volume $V$ of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$.)
(a) Show that $\frac{dh}{dt} = -\frac{\sqrt{h}}{5}$.
(b) Given that $h = 17$ at time $t = 0$, solve the differential equation $\frac{dh}{dt} = -\frac{\sqrt{h}}{5}$ for $h$ as a function of $t$.
(c) At what time $t$ is the coffeepot empty?

▶️Answer/Explanation

\(\textbf{5(a)}\)
The volume of the coffee is given by $V = 25\pi h$.
Differentiating with respect to $t$,
$\frac{dV}{dt} = 25\pi \frac{dh}{dt} = -5\pi \sqrt{h}$.
Dividing through by $25\pi$, we get
$\frac{dh}{dt} = -\frac{\sqrt{h}}{5}$.

\(\textbf{5(b)}\)
The differential equation $\frac{dh}{dt} = -\frac{\sqrt{h}}{5}$ can be solved by separating variables: $\frac{1}{\sqrt{h}} , dh = -\frac{1}{5} , dt$.
Integrating both sides, $2\sqrt{h} = -\frac{1}{5}t + C$. At $t = 0$, $h = 17$, so $2\sqrt{17} = C$.
Substituting $C$ back, we have $2\sqrt{h} = -\frac{1}{5}t + 2\sqrt{17}$.
Squaring both sides, $h = \left(-\frac{1}{10}t + \sqrt{17}\right)^2$.

\(\textbf{5(c)}\)
To find when the coffeepot is empty, set $h = 0$.
Solving $\left(-\frac{1}{10}t + \sqrt{17}\right)^2 = 0$, we get
$-\frac{1}{10}t + \sqrt{17} = 0$.
Solving for $t$, we find
$t = 10\sqrt{17}$.

Question 6

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

(b) Topic-10.4 Integral Test for Convergence

(c) Topic-7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

The function $f$ is defined by the power series
\(
f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n+1)!} = 1 – \frac{x^2}{3!} + \frac{x^4}{5!} – \frac{x^6}{7!} + \cdots
\)
for all real numbers $x$.
(a) Find $f'(0)$ and $f”(0)$. Determine whether $f$ has a local maximum, a local minimum, or neither at $x = 0$. Give a reason for your answer.
(b) Show that $1 – \frac{1}{3!}$ approximates $f(1)$ with error less than $\frac{1}{100}$.
(c) Show that $y = f(x)$ is a solution to the differential equation $xy’ + y = \cos x$.

▶️Answer/Explanation

\(\textbf{6(a)}\)
\(
f'(0) = \text{coefficient of } x \text{ term} = 0
\)
\(
f”(0) = 2 \cdot (\text{coefficient of } x^2 \text{ term}) = 2 \cdot \left(-\frac{1}{3!}\right) = -\frac{1}{3}
\)
\(
f \text{ has a local maximum at } x = 0 \text{ because } f'(0) = 0 \text{ and } f”(0) < 0.
\)

\(\textbf{6(b)}\)
\(
f(1) = 1 – \frac{1}{3!} + \frac{1}{5!} – \frac{1}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}.
\)
This is an alternating series whose terms decrease in absolute value with limit 0. Thus, the error is less than the first omitted term, so
\(
\left| f(1) – \left(1 – \frac{1}{3!}\right) \right| \leq \frac{1}{5!} = \frac{1}{120} < \frac{1}{100}.
\)

\(\textbf{6(c)}\)

\(
y’ = -\frac{2x}{3!} + \frac{4x^3}{5!} – \frac{6x^5}{7!} + \cdots = \sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n-1}}{(2n+1)!}.
\)
\(
xy’ = -\frac{2x^2}{3!} + \frac{4x^4}{5!} – \frac{6x^6}{7!} + \cdots = \sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n}}{(2n+1)!}.
\)
\(
xy’ + y = 1 – \left(\frac{2}{3!}\right)x^2 + \left(\frac{4}{5!}\right)x^4 – \left(\frac{6}{7!}\right)x^6 + \cdots
\)
\(
+ \sum_{n=1}^\infty \frac{(-1)^n \left(2n + 1\right)x^{2n}}{(2n+1)!}.
\)
\(
= 1 – \frac{1}{2!}x^2 + \frac{1}{4!}x^4 – \frac{1}{6!}x^6 + \cdots = \cos x.
\)
OR
\(
xy = x f(x) = x – \frac{x^3}{3!} + \cdots + (-1)^n \frac{x}{(2n+1)!}.
\)
\(
= \sin x.
\)
\(
xy’ + y = (xy)’ = (\sin x)’ = \cos x.
\)

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