Question 1

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

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

Let $R$ be the shaded region bounded by the graph of $y = \ln x$ and the line $y = x – 2$, as shown above.
(a) Find the area of $R$.
(b) Find the volume of the solid generated when $R$ is rotated about the horizontal line $y = -3$.
(c) Write, but do not evaluate, an integral expression that can be used to find the volume of the solid generated when $R$ is rotated about the $y$-axis.

▶️Answer/Explanation

$\textbf{1(a)}$
$
2x + 8y y’ = 3y + 3xy’
$
$
(8y – 3x)y’ = 3y – 2x
$
$
y’ = \frac{3y – 2x}{8y – 3x}
$

$\textbf{1(b)}$
$
\frac{3y – 2x}{8y – 3x} = 0; \quad 3y – 2x = 0
$
When $x = 3$, $3y = 6$, so $y = 2$.
$
3^2 + 4 \cdot 2^2 = 25 \quad \text{and} \quad 7 + 3 \cdot 3 \cdot 2 = 25
$
Therefore, $P = (3, 2)$ is on the curve and the slope is $0$ at this point.

$\textbf{1(c)}$
$
\frac{d^2y}{dx^2} = \frac{(8y – 3x)(3y’ – 2) – (3y – 2x)(8y’ – 3)}{(8y – 3x)^2}
$
At $P = (3, 2)$,
$
\frac{d^2y}{dx^2} = \frac{(16 – 9)(-2) – (16 – 9)(-3)}{(16 – 9)^2} = \frac{-14}{49} = -\frac{2}{7}.
$
Since $y’ = 0$ and $y” < 0$ at $P$, the curve has a local maximum at $P$.

Question 2

(a) Topic-6.1-Exploring Accumulations of Change

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

At an intersection in Thomasville, Oregon, cars turn left at the rate $L(t) = 60\sqrt{t} \sin^2\left(\frac{t}{3}\right)$ cars per hour over the time interval $0 \leq t \leq 18$ hours. The graph of $y = L(t)$ is shown above.
(a) To the nearest whole number, find the total number of cars turning left at the intersection over the time interval $0 \leq t \leq 18$ hours.
(b) Traffic engineers will consider turn restrictions when $L(t) \geq 150$ cars per hour. Find all values of $t$ for which $L(t) \geq 150$ and compute the average value of $L$ over this time interval. Indicate units of measure.
(c) Traffic engineers will install a signal if there is any two-hour time interval during which the product of the total number of cars turning left and the total number of oncoming cars traveling straight through the intersection is greater than $200,000$. In every two-hour time interval, $500$ oncoming cars travel straight through the intersection. Does this intersection require a traffic signal? Explain the reasoning that leads to your conclusion.

▶️Answer/Explanation

$\textbf{2(a)}$
$\int_0^{18} L(t) \, dt \approx 1658 \, \text{cars}$

$\textbf{2(b)}$
$
L(t) = 150 \text{ when } t = 12.42831, \, 16.12166.
$
Let $R = 12.42831$ and $S = 16.12166$.
$L(t) \geq 150$ for $t$ in the interval $[R, S]$.
$
\frac{1}{S – R} \int_R^S L(t) \, dt = 199.426 \, \text{cars per hour.}
$

$\textbf{2(c)}$
For the product to exceed 200,000, the number of cars turning left in a two-hour interval must be greater than 400.
$ \int_{13}^{15} L(t) \, dt = 431.931 > 400 $
OR
The number of cars turning left will be greater than 400 on a two-hour interval if $L(t) \geq 200$ on that interval.
$L(t) \geq 200$ on any two-hour subinterval of $[13.25304, 15.32386]$.
Yes, a traffic signal is required.

Question 3

(a) Topic-4.2-Straight-Line Motion Connecting Position Velocity and Acceleration

(b) Topic-4.3-Rates of Change in Applied Contexts Other Than Motion

(d) Topic-6.13-Evaluating Improper Integrals BC Only

An object moving along a curve in the $xy$-plane is at position $(x(t), y(t))$ at time $t$, where
$
\frac{dx}{dt} = \sin^{-1}(1 – 2e^{-t}) \quad \text{and} \quad \frac{dy}{dt} = \frac{4t}{1 + t^3}
$
for $t \geq 0$. At time $t = 2$, the object is at the point $(6, -3)$. (Note: $\sin^{-1} x = \arcsin x$)
(a) Find the acceleration vector and the speed of the object at time $t = 2$.
(b) The curve has a vertical tangent line at one point. At what time $t$ is the object at this point?
(c) Let $m(t)$ denote the slope of the line tangent to the curve at the point $(x(t), y(t))$. Write an expression for $m(t)$ in terms of $t$ and use it to evaluate $\lim_{t \to \infty} m(t)$.
(d) The graph of the curve has a horizontal asymptote $y = c$. Write, but do not evaluate, an expression involving an improper integral that represents this value $c$.

▶️Answer/Explanation

$\textbf{3(a)}$
$
a(2) = \langle 0.395 \text{ or } 0.396, -0.741 \text{ or } -0.740 \rangle
$
$
\text{Speed} = \sqrt{x'(2)^2 + y'(2)^2} = 1.207 \text{ or } 1.208
$

$\textbf{3(b)}$
$
\sin^{-1}(1 – 2e^{-t}) = 0
$
$
1 – 2e^{-t} = 0
$
$
t = \ln 2 = 0.693 \quad \text{and } \frac{dy}{dt} \neq 0 \text{ when } t = \ln 2
$

$\textbf{3(c)}$
$
m(t) = \frac{\frac{4t}{1 + t^3}}{\sin^{-1}(1 – 2e^{-t})}
$
$
\lim_{t \to \infty} m(t) = \lim_{t \to \infty} \frac{\frac{4t}{1 + t^3}}{\sin^{-1}(1 – 2e^{-t})}
$
$
= \frac{0}{\sin^{-1}(1)} = 0
$

$\textbf{3(d)}$
Since $\lim_{t \to \infty} x(t) = \infty$,
$ c = \lim_{t \to \infty} y(t) = -3 + \int_2^\infty \frac{4t}{1 + t^3} \, dt $

Question 4

(a) Topic-4.2-Straight-Line Motion Connecting Position Velocity and Acceleration

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

$
\begin{array}{|c|c|}
\hline
t \text{ (seconds)} & v(t) \text{ (feet per second)} \\
\hline
0 & 5 \\
10 & 14 \\
20 & 22 \\
30 & 29 \\
40 & 35 \\
50 & 40 \\
60 & 44 \\
70 & 47 \\
80 & 49 \\
\hline
\end{array}
$
Rocket A has positive velocity $v(t)$ after being launched upward from an initial height of 0 feet at time $t = 0$ seconds. The velocity of the rocket is recorded for selected values of $t$ over the interval $0 \leq t \leq 80$ seconds, as shown in the table above.
(a) Find the average acceleration of rocket A over the time interval $0 \leq t \leq 80$ seconds. Indicate units of measure.
(b) Using correct units, explain the meaning of $\int_{10}^{70} v(t) \, dt$ in terms of the rocket’s flight. Use a midpoint Riemann sum with 3 subintervals of equal length to approximate $\int_{10}^{70} v(t) \, dt$.
(c) Rocket B is launched upward with an acceleration of $a(t) = \frac{3}{\sqrt{t + 1}}$ feet per second per second. At time $t = 0$ seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at time $t = 80$ seconds? Explain your answer.

▶️Answer/Explanation

$\textbf{4(a)}$
Average acceleration of rocket A is
$\frac{v(80) – v(0)}{80 – 0} = \frac{49 – 5}{80} = \frac{11}{20} \, \text{ft/sec}^2 $

$\textbf{4(b)}$
Since the velocity is positive, $\int_{10}^{70} v(t) \, dt$ represents the distance, in feet, traveled by rocket A from $t = 10$ seconds to $t = 70$ seconds.
A midpoint Riemann sum is
$
20[v(20) + v(40) + v(60)]
$
$
= 20[22 + 35 + 44] = 2020 \, \text{ft}.
$

$\textbf{4(c)}$
Let $v_B(t)$ be the velocity of rocket B at time $t$.
$
v_B(t) = \int \frac{3}{\sqrt{t + 1}} \, dt = 6\sqrt{t + 1} + C
$
$
2 = v_B(0) = 6\sqrt{0 + 1} + C
$
$
C = -4
$
$
v_B(t) = 6\sqrt{t + 1} – 4
$
$
v_B(80) = 50 > v(80) = 49
$
Rocket B is traveling faster at time $t = 80$ seconds.
Units of ft/sec$^2$ in (a) and ft in (b).

Question 5

(a) Topic-7.2-Verifying Solutions for Differential Equations

(b) Topic-5.7-Using the Second Derivative Test to Determine Extrema

(d) Topic-7.5-Approximating Solutions Using Euler’s Method BC Only

Consider the differential equation
$
\frac{dy}{dx} = 5x^2 – \frac{6}{y – 2} \quad \text{for } y \neq 2.
$
Let $y = f(x)$ be the particular solution to this differential equation with the initial condition $f(-1) = -4$.
(a) Evaluate $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $(-1, -4)$.
(b) Is it possible for the $x$-axis to be tangent to the graph of $f$ at some point? Explain why or why not.
(c) Find the second-degree Taylor polynomial for $f$ about $x = -1$.
(d) Use Euler’s method, starting at $x = -1$ with two steps of equal size, to approximate $f(0)$. Show the work that leads to your answer.

▶️Answer/Explanation

$\textbf{5(a)}$
$
\left. \frac{dy}{dx} \right|_{(-1, -4)} = 6
$
$
\frac{d^2y}{dx^2} = 10x + 6(y – 2)^{-2} \frac{dy}{dx}
$
$
\left. \frac{d^2y}{dx^2} \right|_{(-1, -4)} = -10 + 6\left(\frac{1}{(-6)^2}\right)(6) = -9
$

$\textbf{5(b)}$
The $x$-axis will be tangent to the graph of $f$ if $\left. \frac{dy}{dx} \right|_{(k, 0)} = 0$.
The $x$-axis will never be tangent to the graph of $f$ because
$
\frac{dy}{dx} = 5k^2 + 3 > 0 \quad \text{for all } k.
$

$\textbf{5(c)}$
$P(x) = -4 + 6(x + 1) – \frac{9}{2}(x + 1)^2$

$\textbf{5(d)}$
$
f(-1) = -4
$
$
f\left(-\frac{1}{2}\right) \approx -4 + \frac{1}{2}(6) = -1
$
$
f(0) \approx -1 + \frac{1}{2}\left(\frac{5}{4} + 2\right) = \frac{5}{8}
$

Question 6

(a) Topic-10.13-Radius and Interval of Convergence of Power Series

(b) Topic-5.7-Using the Second Derivative Test to Determine Extrema

The function $f$ is defined by the power series
$
f(x) = -\frac{x}{2} + \frac{2x^2}{3} – \frac{3x^3}{4} + \dots + \frac{(-1)^n n x^n}{n + 1} + \dots
$
for all real numbers $x$ for which the series converges. The function $g$ is defined by the power series
$
g(x) = 1 – \frac{x}{2!} + \frac{x^2}{4!} – \frac{x^3}{6!} + \dots + \frac{(-1)^n x^n}{(2n)!} + \dots
$
for all real numbers $x$ for which the series converges.
(a) Find the interval of convergence of the power series for $f$. Justify your answer.
(b) The graph of $y = f(x) – g(x)$ passes through the point $(0, -1)$. Find $y'(0)$ and $y”(0)$. Determine whether $y$ has a relative minimum, a relative maximum, or neither at $x = 0$. Give a reason for your answer.

▶️Answer/Explanation

$\textbf{6(a)}$
$
\left| \frac{(-1)^{n+1} (n + 1)x^{n+1}}{n + 2} \cdot \frac{n + 1}{(-1)^n n x^n} \right| = \frac{(n + 1)^2}{(n + 2)(n)} \cdot |x|
$
$
\lim_{n \to \infty} \frac{(n + 1)^2}{(n + 2)(n)} |x| = |x|
$
The series converges when $-1 < x < 1$.
When $x = 1$, the series is
$
-\frac{1}{2} + \frac{2}{3} – \frac{3}{4} + \dots
$
This series does not converge, because the limit of the individual terms is not zero.
When $x = -1$, the series is
$
\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \dots
$
This series does not converge, because the limit of the individual terms is not zero.
Thus, the interval of convergence is $-1 < x < 1$.

$\textbf{6(b)}$
$
f'(x) = -\frac{1}{2} + \frac{4}{3}x – \frac{9}{4}x^2 + \dots \quad \text{and } f'(0) = -\frac{1}{2}.
$
$
g'(x) = -\frac{1}{2!} + \frac{2}{4!}x – \frac{3}{6!}x^2 + \dots \quad \text{and } g'(0) = -\frac{1}{2}.
$
$
y'(0) = f'(0) – g'(0) = 0
$
$
f”(0) = \frac{4}{3} \quad \text{and } g”(0) = \frac{2}{4!} = \frac{1}{12}.
$
$
y”(0) = \frac{4}{3} – \frac{1}{12} = \frac{15}{12} = \frac{5}{4}.
$
Thus, $y”(0) > 0$.
Since $y'(0) = 0$ and $y”(0) > 0$, $y$ has a relative minimum at $x = 0$.

Question 2

Question 3

Question 5

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