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
(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
Let $f$ and $g$ be the functions given by $f(x) = e^x$ and $g(x) = \ln x$.
(a) Find the area of the region enclosed by the graphs of $f$ and $g$ between $x = \frac{1}{2}$ and $x = 1$.
(b) Find the volume of the solid generated when the region enclosed by the graphs of $f$ and $g$ between $x = \frac{1}{2}$ and $x = 1$ is revolved about the line $y = 4$.
(c) Let $h$ be the function given by $h(x) = f(x) – g(x)$.
Find the absolute minimum value of $h(x)$ on the closed interval $\frac{1}{2} \leq x \leq 1$, and find the absolute maximum value of $h(x)$ on the closed interval $\frac{1}{2} \leq x \leq 1$.
Show the analysis that leads to your answers.
▶️Answer/Explanation
\(\textbf{1(a)}\)
Area = $\int_{\frac{1}{2}}^{1} \left(e^x – \ln x\right) dx = 1.222$ or $1.223$
\(\textbf{1(b)}\)
Volume = $\pi \int_{\frac{1}{2}}^{1} \left( (4 – \ln x)^2 – (4 – e^x)^2 \right) dx = 7.515\pi$ or $23.609$
\(\textbf{1(c)}\)
$h'(x) = f'(x) – g'(x) = e^x – \frac{1}{x} = 0$
$x = 0.567143$
Absolute minimum value and absolute maximum value occur at the critical point or at the endpoints.
$h(0.567143) = 2.330$
$h(0.5) = 2.3418$
$h(1) = 2.718$
The absolute minimum is $2.330$.
The absolute maximum is $2.718$.
Question 2
(a) Topic-6.1 Exploring Accumulations of Change
(b) Topic-8.1 Finding the Average Value of a Function on an Interval
(c) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
(d) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
The rate at which people enter an amusement park on a given day is modeled by the function $E$ defined by:
$ E(t) = \frac{15600}{t^2 – 24t + 160}. $
The rate at which people leave the same amusement park on the same day is modeled by the function $L$ defined by:
$ L(t) = \frac{9890}{t^2 – 38t + 370}. $
Both $E(t)$ and $L(t)$ are measured in people per hour and time $t$ is measured in hours after midnight. These functions are valid for $9 \leq t \leq 23$, the hours during which the park is open. At time $t = 9$, there are no people in the park.
(a) How many people have entered the park by 5:00 P.M. ($t = 17$)? Round your answer to the nearest whole number.
(b) The price of admission to the park is $15$ until 5:00 P.M. ($t = 17$). After 5:00 P.M., the price of admission to the park is $11$. How many dollars are collected from admissions to the park on the given day? Round your answer to the nearest whole number.
(c) Let $H(t) = \int_9^t (E(x) – L(x)) \, dx$ for $9 \leq t \leq 23$. The value of $H(17)$ to the nearest whole number is $3725$. Find the value of $H'(17)$, and explain the meaning of $H(17)$ and $H'(17)$ in the context of the amusement park.
(d) At what time $t$, for $9 \leq t \leq 23$, does the model predict that the number of people in the park is a maximum?
▶️Answer/Explanation
\(\textbf{2(a)}\)
\(
\int_9^{17} E(t) \, dt = 6004.270
\)
6004 people entered the park by 5 pm.
\(\textbf{2(b)}\)
\(
15 \int_9^{17} E(t) \, dt + 11 \int_{17}^{23} E(t) \, dt = 104048.165
\)
The amount collected was $104,048$.
Alternatively:
\(
\int_{23}^{17} E(t) \, dt = 1271.283
\)
1271 people entered the park between 5 pm and 11 pm, so the amount collected was:
\(
15 \cdot (6004) + 11 \cdot (1271) = 104,041.
\)
\(\textbf{2(c)}\)
\(
H'(17) = E(17) – L(17) = -380.281
\)
There were 3725 people in the park at \(t = 17\). The number of people in the park was decreasing at the rate of approximately 380 people/hr at time \(t = 17\).
\(\textbf{2(d)}\)
\(
H'(t) = E(t) – L(t) = 0
\)
\(
t = 15.794 \text{ or } 15.795
\)
Question 3
(a) Topic-9.1 Defining and Differentiating Parametric Equations
(b) Topic-9.4 Defining and Differentiating Vector-Valued Functions
(c) Topic-9.3 Finding Arc Lengths of Curves Given by Parametric Equations
(d) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
The figure above shows the path traveled by a roller coaster car over the time interval \(0 \leq t \leq 18\) seconds. The position of the car at time \(t\) seconds can be modeled parametrically by
\(
x(t) = 10t + 4 \sin t
\)
\(
y(t) = (20 – t)(1 – \cos t),
\)
where \(x\) and \(y\) are measured in meters. The derivatives of these functions are given by
\(
x'(t) = 10 + 4 \cos t
\)
\(
y'(t) = (20 – t) \sin t + \cos t – 1.
\)
(a) Find the slope of the path at time \(t = 2\). Show the computations that lead to your answer.
(b) Find the acceleration vector of the car at the time when the car’s horizontal position is \(x = 140\).
(c) Find the time \(t\) at which the car is at its maximum height, and find the speed, in m/sec, of the car at this time.
(d) For \(0 < t < 18\), there are two times at which the car is at ground level (\(y = 0\)). Find these two times and write an expression that gives the average speed, in m/sec, of the car between these two times. Do not evaluate the expression.
▶️Answer/Explanation
\(\textbf{3(a)}\)
Slope = $\frac{dy}{dx} \Big|_{t=2} = \frac{y'(2)}{x'(2)} = \frac{18 \sin 2 + \cos 2 – 1}{10 + 4 \cos 2} = 1.793 \text{ or } 1.794$
\(\textbf{3(b)}\)
$x(t) = 10t + 4 \sin t = 140; \ t_0 = 13.647083$
$x”(t_0) = -3.529, \ y”(t_0) = 1.225 \text{ or } 1.226$
Acceleration vector is $\langle -3.529, 1.225 \rangle$
or $\langle -3.529, 1.226 \rangle$
\(\textbf{3(c)}\)
$y'(t) = (20 – t) \sin t + \cos t – 1 = 0$
$t_1 = 3.023 \text{ or } 3.024 \text{ at maximum height}$
Speed = $\sqrt{\big(x'(t_1)\big)^2 + \big(y'(t_1)\big)^2} = |x'(t_1)| = 6.027 \text{ or } 6.028$
\(\textbf{3(d)}\)
$y(t) = 0 \text{ when } t = 2\pi \text{ and } t = 4\pi$
Average speed = $\frac{1}{2\pi} \int_{2\pi}^{4\pi} \sqrt{\big(x'(t)\big)^2 + \big(y'(t)\big)^2} , dt$
= $\frac{1}{2\pi} \int_{2\pi}^{4\pi} \sqrt{(10 + 4 \cos t)^2 + ((20 – t) \sin t + \cos t – 1)^2} , dt$
Question 4
(a) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
(b) Topic-5.3 Determining Intervals on Which a Function Is Increasing or Decreasing
(c) Topic-5.6 Determining Concavity of Functions Over Their Domains
(d) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
The graph of the function $f$ shown above consists of two line segments. Let $g$ be the function given by
\( g(x) = \int_0^x f(t) \, dt. \)
(a) Find $g(-1)$, $g'(-1)$, and $g”(-1)$.
(b) For what values of $x$ in the open interval $(-2, 2)$ is $g$ increasing? Explain your reasoning.
(c) For what values of $x$ in the open interval $(-2, 2)$ is the graph of $g$ concave down? Explain your reasoning.
(d) On the axes provided, sketch the graph of $g$ on the closed interval $[-2, 2]$.
▶️Answer/Explanation
\(\textbf{4(a)}\)
$g(-1) = \int_0^{-1} f(t) , dt = -\int_{-1}^0 f(t) , dt = -\frac{3}{2}$
$g'(-1) = f(-1) = 0$
$g”(-1) = f'(-1) = 3$
\(\textbf{4(b)}\)
$g$ is increasing on $-1 < x < 1$ because $g'(x) = f(x) > 0$ on this interval.
\(\textbf{4(c)}\)
The graph of $g$ is concave down on $0 < x < 2$ because $g”(x) = f'(x) < 0$ on this interval.
or
because $g'(x) = f(x)$ is decreasing on this interval.
\(\textbf{4(d)}\)
Question 5
(a) Topic-7.3 Sketching Slope Fields
(b) Topic-7.5 Approximating Solutions Using Euler’s Method
(c) Topic-7.6 Finding General Solutions Using Separation of Variables
(d) Topic-5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
Consider the differential equation $\frac{dy}{dx} = 2y – 4x$.
(a) The slope field for the given differential equation is provided. Sketch the solution curve that passes through the point $(0, 1)$ and sketch the solution curve that passes through the point $(0, -1)$.
(b) Let $f$ be the function that satisfies the given differential equation with the initial condition $f(0) = 1$.
Use Euler’s method, starting at $x = 0$ with a step size of $0.1$, to approximate $f(0.2)$. Show the work that leads to your answer.
(c) Find the value of $b$ for which $y = 2x + b$ is a solution to the given differential equation. Justify your answer.
(d) Let $g$ be the function that satisfies the given differential equation with the initial condition $g(0) = 0$.
Does the graph of $g$ have a local extremum at the point $(0, 0)$? If so, is the point a local maximum or a local minimum? Justify your answer.
▶️Answer/Explanation
\(\textbf{5(a)}\)
\(\textbf{5(b)}\)
$f(0.1) \approx f(0) + f'(0)(0.1)$
$= 1 + (2 – 0)(0.1) = 1.2$
$f(0.2) \approx f(0.1) + f'(0.1)(0.1)$
$\approx 1.2 + (2.4 – 0.4)(0.1) = 1.4$
\(\textbf{5(c)}\)
Substitute $y = 2x + b$ in the DE:
$2 = 2(2x + b) – 4x = 2b$, so $b = 1$
OR
Guess $b = 1$, $y = 2x + 1$
Verify: $2y – 4x = (4x + 2) – 4x = 2 = \frac{dy}{dx}$.
\(\textbf{5(d)}\)
$g$ has local maximum at $(0, 0)$.
$g'(0) = \frac{dy}{dx}\big|_{(0, 0)} = 2(0) – 4(0) = 0$, and
$g”(x) = \frac{d^2y}{dx^2} = 2\frac{dy}{dx} – 4$, so
$g”(0) = 2g'(0) – 4 = -4 < 0$.
Question 6
(a) Topic-10.13 Radius and Interval of Convergence of Power Series
(b) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
(c) Topic-10.12 Lagrange Error Bound
The Maclaurin series for the function $f$ is given by
\(
f(x) = \sum_{n=0}^\infty \frac{(2x)^{n+1}}{n+1} = 2x + \frac{4x^2}{2} + \frac{8x^3}{3} + \frac{16x^4}{4} + \cdots + \frac{(2x)^{n+1}}{n+1} + \cdots
\)
on its interval of convergence.
(a) Find the interval of convergence of the Maclaurin series for $f$. Justify your answer.
(b) Find the first four terms and the general term for the Maclaurin series for $f'(x)$.
(c) Use the Maclaurin series you found in part (b) to find the value of $f’\left(-\frac{1}{3}\right)$.
▶️Answer/Explanation
\(\textbf{6(a)}\)
\(
\lim_{n \to \infty} \frac{(2x)^{n+2}}{n+2} \cdot \frac{n+1}{(2x)^{n+1}} = \lim_{n \to \infty} \left| \frac{(n+1)}{(n+2)} 2x \right| = |2x|
\)
\(
|2x| < 1 \text{ for } -\frac{1}{2} < x < \frac{1}{2}
\)
At $x = \frac{1}{2}$, the series is $\sum_{n=0}^\infty \frac{1}{n+1}$, which diverges since this is the harmonic series.
At $x = -\frac{1}{2}$, the series is $\sum_{n=0}^\infty (-1)^{n+1} \frac{1}{n+1}$, which converges by the Alternating Series Test.
Hence, the interval of convergence is $-\frac{1}{2} \leq x < \frac{1}{2}$.
\(\textbf{6(b)}\)
\(
f'(x) = 2 + 4x + 8x^2 + 16x^3 + \dots + 2(2x)^n + \dots
\)
\(\textbf{6(c)}\)
The series in (b) is a geometric series.
\(
f’\left(-\frac{1}{3}\right) = 2 + 4\left(-\frac{1}{3}\right) + 8\left(-\frac{1}{3}\right)^2 + \dots + 2\left(2\left(-\frac{1}{3}\right)\right)^n + \dots
\)
\(
= 2 – \frac{4}{3} + \frac{8}{9} – \frac{16}{27} + \dots + 2\left(-\frac{2}{3}\right)^n + \dots
\)
\(
= \frac{2}{1 + \frac{2}{3}} = \frac{6}{5}
\)
OR
\(
f'(x) = \frac{2}{1-2x} \text{ for } -\frac{1}{2} < x < \frac{1}{2}.
\)
Therefore,
\(
f’\left(-\frac{1}{3}\right) = \frac{2}{1 + \frac{2}{3}} = \frac{6}{5}.
\)