Question 1
(a) Topic-8.4-Finding the Area Between Curves Expressed as Functions of x
(b) Topic-8.4-Finding the Area Between Curves Expressed as Functions of x
(c) Topic-8.10-Volume with Disc Method Revolving Around Other Axes
Let $f$ and $g$ be the functions given by $f(x) = \frac{1}{4} + \sin(\pi x)$ and $g(x) = 4^{-x}$. Let $R$ be the shaded region in the first quadrant enclosed by the $y$-axis and the graphs of $f$ and $g$, and let $S$ be the shaded region in the first quadrant enclosed by the graphs of $f$ and $g$, as shown in the figure above.
(a) Find the area of $R$.
(b) Find the area of $S$.
(c) Find the volume of the solid generated when $S$ is revolved about the horizontal line $y = -1$.
▶️Answer/Explanation
$f(x) = g(x)$ when $\frac{1}{4} + \sin(\pi x) = 4^{-x}$.
$f$ and $g$ intersect when $x = 0.178218$ and when $x = 1$. Let $a = 0.178218$.
\(\textbf{1(a)}\)
$\int_0^a \left(g(x) – f(x)\right) dx = 0.064 \text{ or } 0.065.$
\(\textbf{1(b)}\)
$\int_a^1 \left(f(x) – g(x)\right) dx = 0.410.$
\(\textbf{1(c)}\)
$\pi \int_a^1 \left((f(x) + 1)^2 – (g(x) + 1)^2\right) dx = 4.558 \text{ or } 4.559.$
Question 2
(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
(d) Topic-9.9-Finding the Area of the Region Bounded by Two Polar Curves
The curve above is drawn in the $xy$-plane and is described by the equation in polar coordinates $r = \theta + \sin(2\theta)$ for $0 \leq \theta \leq \pi$, where $r$ is measured in meters and $\theta$ is measured in radians. The derivative of $r$ with respect to $\theta$ is given by
\(
\frac{dr}{d\theta} = 1 + 2\cos(2\theta).
\)
(a) Find the area bounded by the curve and the $x$-axis.
(b) Find the angle $\theta$ that corresponds to the point on the curve with $x$-coordinate $-2$.
(c) For $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $\frac{dr}{d\theta}$ is negative. What does this fact say about $r$? What does this fact say about the curve?
(d) Find the value of $\theta$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$ that corresponds to the point on the curve in the first quadrant with greatest distance from the origin. Justify your answer.
▶️Answer/Explanation
\(\textbf{2(a)}\)
$\text{Area} = \frac{1}{2} \int_0^\pi r^2 \, d\theta$
$= \frac{1}{2} \int_0^\pi \left(\theta + \sin(2\theta)\right)^2 \, d\theta = 4.382$
\(\textbf{2(b)}\)
$-2 = r\cos(\theta) = \left(\theta + \sin(2\theta)\right)\cos(\theta)$
$\theta = 2.786$
\(\textbf{2(c)}\)
Since $\frac{dr}{d\theta} < 0$ for $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $r$ is decreasing on this interval. This means the curve is getting closer to the origin.
\(\textbf{2(d)}\)
The only value in $\left[0, \frac{\pi}{2}\right]$ where $\frac{dr}{d\theta} = 0$ is $\theta = \frac{\pi}{3}$.
\( \begin{array}{|c|c|} \hline \theta & r \\ \hline 0 & 0 \\ \frac{\pi}{3} & 1.913 \\ \frac{\pi}{2} & 1.571 \\ \hline \end{array} \)
Question 3
(a) Topic-2.3-Estimating Derivatives of a Function at a Point
(b) Topic-8.1-Finding the Average Value of a Function on an Interval
(c) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-5.6-Determining Concavity of Functions over Their Domains
\(
\begin{array}{|c|c|}
\hline
\text{Distance } x \text{ (cm)} & \text{Temperature } T(x) \text{ (°C)} \\
\hline
0 & 100 \\
1 & 93 \\
5 & 70 \\
6 & 62 \\
8 & 55 \\
\hline
\end{array}
\)
A metal wire of length 8 centimeters (cm) is heated at one end. The table above gives selected values of the temperature $T(x)$, in degrees Celsius (°C), of the wire $x$ cm from the heated end. The function $T$ is decreasing and twice differentiable.
(a) Estimate $T'(7)$. Show the work that leads to your answer. Indicate units of measure.
(b) Write an integral expression in terms of $T(x)$ for the average temperature of the wire. Estimate the average temperature of the wire using a trapezoidal sum with the four subintervals indicated by the data in the table. Indicate units of measure.
(c) Find $\int_0^8 T'(x) \, dx$, and indicate units of measure. Explain the meaning of $\int_0^8 T'(x) \, dx$ in terms of the temperature of the wire.
(d) Are the data in the table consistent with the assertion that $T”(x) > 0$ for every $x$ in the interval $0 < x < 8$? Explain your answer.
▶️Answer/Explanation
\(\textbf{3(a)}\)
$T'(7) \approx \frac{T(8) – T(6)}{8 – 6} = \frac{55 – 62}{2} = -\frac{7}{2} \, \text{°C/cm}$
\(\textbf{3(b)}\)
$\frac{1}{8} \int_0^8 T(x) \, dx$
$Trapezoidal approximation for $\int_0^8 T(x) \, dx$:$A = \frac{100 + 93}{2} \cdot 1 + \frac{93 + 70}{2} \cdot 4 + \frac{70 + 62}{2} \cdot 1 + \frac{62 + 55}{2} \cdot 2$
$\text{Average temperature} \approx \frac{1}{8} A = 75.6875 \, \text{°C}$
\(\textbf{3(c)}\)
$\int_0^8 T'(x) \, dx = T(8) – T(0) = 55 – 100 = -45 \, \text{°C}$
The temperature drops $45 \, \text{°C}$ from the heated end of the wire to the other end of the wire.
\(\textbf{3(d)}\)
$\text{Average rate of change of temperature on } [1, 5] \text{ is } \frac{70 – 93}{5 – 1} = -5.75$.
$\text{Average rate of change of temperature on } [5, 6] \text{ is } \frac{62 – 70}{6 – 5} = -8.$
No. By the Mean Value Theorem (MVT), $T'(c_1) = -5.75$ for some $c_1$ in the interval $(1, 5)$ and $T'(c_2) = -8$ for some $c_2$ in the interval $(5, 6)$.
It follows that $T’$ must decrease somewhere in the interval $(c_1, c_2)$.
Therefore $T”$ is not positive for every $x$ in $[0, 8]$. Units of °C/cm in (a), and °C in (b) and (c).
Question 4
(a) Topic-7.3-Sketching Slope Fields
(b) Topic-7.7-Finding Particular Solutions Using Initial Conditions and Separation of Variables
(c) Topic-7.5-Approximating Solutions Using Euler’s Method BC Only
(d) Topic-7.2-Verifying Solutions for Differential Equations
Consider the differential equation
\(
\frac{dy}{dx} = 2x – y.
\)
(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated and sketch the solution curve that passes through the point $(0, 1)$.
(Note: Use the axes provided in the pink test booklet.)
(b) The solution curve that passes through the point $(0, 1)$ has a local minimum at $x = \ln\left(\frac{3}{2}\right)$. What is the $y$-coordinate of this local minimum?
(c) Let $y = f(x)$ be the particular solution to the given differential equation with the initial condition $f(0) = 1$. Use Euler’s method, starting at $x = 0$ with two steps of equal size, to approximate $f(-0.4)$. Show the work that leads to your answer.
(d) Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$. Determine whether the approximation found in part (c) is less than or greater than $f(-0.4)$. Explain your reasoning.
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(\textbf{4(b)}\)
\(
\frac{dy}{dx} = 0 \text{ when } 2x = y
\)
The $y$-coordinate is $2 \ln\left(\frac{3}{2}\right)$.
\(\textbf{4(c)}\)
\(
f(-0.2) \approx f(0) + f'(0)(-0.2)
\)
\(
= 1 + (-1)(-0.2) = 1.2
\)
\(
f(-0.4) \approx f(-0.2) + f'(-0.2)(-0.2)
\)
\(
\approx 1.2 + (-1.6)(-0.2) = 1.52
\)
\(\textbf{4(d)}\)
\(
\frac{d^2y}{dx^2} = 2 – \frac{dy}{dx} = 2 – 2x + y
\)
\(
\frac{d^2y}{dx^2} \text{ is positive in quadrant II because } x < 0 \text{ and } y > 0.
\)
\(
1.52 < f(-0.4) \text{ since all solution curves in quadrant II are concave up.}
\)
Question 5
(a) Topic-8.2-Connecting Position Velocity and Acceleration of Functions Using Integrals
(b) Topic-5.6-Determining Concavity of Functions over Their Domains
(c) Topic-5.9-Connecting a Function, Its First Derivative, and Its Second Derivative
(d) Topic-5.1-Using the Mean Value Theorem
A car is traveling on a straight road. For $0 \leq t \leq 24$ seconds, the car’s velocity $v(t)$, in meters per second, is modeled by the piecewise-linear function defined by the graph above.
(a) Find $\int_0^{24} v(t) \, dt$. Using correct units, explain the meaning of $\int_0^{24} v(t) \, dt$.
(b) For each of $v'(4)$ and $v'(20)$, find the value or explain why it does not exist. Indicate units of measure.
(c) Let $a(t)$ be the car’s acceleration at time $t$, in meters per second per second. For $0 < t < 24$, write a piecewise-defined function for $a(t)$.
(d) Find the average rate of change of $v$ over the interval $8 \leq t \leq 20$. Does the Mean Value Theorem guarantee a value of $c$, for $8 < c < 20$, such that $v'(c)$ is equal to this average rate of change? Why or why not?
▶️Answer/Explanation
\(\textbf{5(a)}\)
\( \int_0^{24} v(t) \, dt = \frac{1}{2}(4)(20) + (12)(20) + \frac{1}{2}(8)(20) = 360 \)
The car travels 360 meters in these 24 seconds.
\(\textbf{5(b)}\)
\(
v'(4) \text{ does not exist because }
\)
\(
\lim_{t \to 4^+} \frac{v(t) – v(4)}{t – 4} = 5 \neq 0 = \lim_{t \to 4^-} \frac{v(t) – v(4)}{t – 4}.
\)
\(
v'(20) = \frac{20 – 0}{16 – 24} = -\frac{5}{2} \, \text{m/sec}^2
\)
\(\textbf{5(c)}\)
\(
a(t) =
\begin{cases}
5 & \text{if } 0 < t < 4, \\
0 & \text{if } 4 < t < 16, \\
-\frac{5}{2} & \text{if } 16 < t < 24.
\end{cases}
\)
\(
a(t) \text{ does not exist at } t = 4 \text{ and } t = 16.
\)
\(\textbf{5(d)}\)
\(
\text{The average rate of change of } v \text{ on } [8, 20] \text{ is }
\frac{v(20) – v(8)}{20 – 8} = \frac{-5}{6} \, \text{m/sec}^2.
\)
No, the Mean Value Theorem does not apply to $v$ on $[8, 20]$ because $v$ is not differentiable at $t = 16$.
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-10.13-Radius and Interval of Convergence of Power Series
Let $f$ be a function with derivatives of all orders and for which $f(2) = 7$. When $n$ is odd, the $n$th derivative of $f$ at $x = 2$ is $0$. When $n$ is even and $n \geq 2$, the $n$th derivative of $f$ at $x = 2$ is given by
\(
f^{(n)}(2) = \frac{(n-1)!}{3^n}.
\)
(a) Write the sixth-degree Taylor polynomial for $f$ about $x = 2$.
(b) In the Taylor series for $f$ about $x = 2$, what is the coefficient of $(x-2)^{2n}$ for $n \geq 1$?
(c) Find the interval of convergence of the Taylor series for $f$ about $x = 2$. Show the work that leads to your answer.
▶️Answer/Explanation
\(\textbf{6(a)}\)
\( P_6(x) = 7 + \frac{1!}{3^2} \cdot \frac{1}{2!} (x – 2)^2 + \frac{3!}{3^4} \cdot \frac{1}{4!} (x – 2)^4 + \frac{5!}{3^6} \cdot \frac{1}{6!} (x – 2)^6 \)
\(\textbf{6(b)}\)
$\frac{(2n-1)!}{3^{2n}} \cdot \frac{1}{(2n)!} = \frac{1}{3^{2n} \cdot (2n)}$
\(\textbf{6(c)}\)
The Taylor series for $f$ about $x = 2$ is
\(
f(x) = 7 + \sum_{n=1}^\infty \frac{1}{2n \cdot 3^{2n}} (x – 2)^{2n}.
\)
\(
L = \lim_{n \to \infty} \frac{\frac{1}{2(n+1) \cdot 3^{2(n+1)}} (x – 2)^{2(n+1)}}{\frac{1}{2n \cdot 3^{2n}} (x – 2)^{2n}}
\)
\(
= \lim_{n \to \infty} \frac{2n \cdot 3^{2n} \cdot (x – 2)^2}{2(n+1) \cdot 3^{2n} \cdot 3^2} = \frac{(x – 2)^2}{9}.
\)
$L < 1$ when $|x – 2| < 3$.
Thus, the series converges when $-1 < x < 5$.
When $x = 5$, the series is
\(
7 + \sum_{n=1}^\infty \frac{1}{2n \cdot 3^{2n}} = 7 + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n},
\)
which diverges, because $\sum_{n=1}^\infty \frac{1}{n}$, the harmonic series, diverges.
When $x = -1$, the series is
\(
7 + \sum_{n=1}^\infty \frac{(-3)^{2n}}{2n \cdot 3^{2n}} = 7 + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n},
\)
which diverges, because $\sum_{n=1}^\infty \frac{1}{n}$, the harmonic series, diverges.
The interval of convergence is $(-1, 5)$.