Question 1
(a) Topic-2.1- Defining Average and Instantaneous Rates of Change at a Point
(b) Topic-2.3- Estimating Derivatives of a Function at a Point
(c) Topic-8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
(d) Topic-4.6 Approximating Values of a Function Using Local Linearity and Linearization
Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A (t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A (t) over the interval 0 ≤ t ≤ 30 . Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30.
(d) For t > 30, L(t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.
▶️Answer/Explanation
\(\textbf{1(a) } \frac{A(30) – A(0)}{30 – 0} = -0.197 \ (\text{or } -0.196) \ \text{lbs/day}\)
\(\textbf{1(b) } A'(15) = -0.164 \ (\text{or } -0.163)\)
\(\text{The amount of grass clippings in the bin is decreasing at a rate of } 0.164 \ (\text{or } 0.163) \ \text{lbs/day at time } t = 15 \ \text{days.}\)
\(\textbf{1(c) } A(t) = \frac{1}{30} \int_0^{30} A(t) \, dt \implies t = 12.415 \ (\text{or } 12.414)\)
\(\textbf{1(d) } L(t) = A(30) + A'(30) \cdot (t – 30)\)
\(A'(30) = -0.055976 \quad A(30) = 0.782928\)
\(L(t) = 0.5 \implies t = 35.054\)
Question 2
(a) Topic-9.9-Finding the Area of the Region Bounded by Two Polar Curves
(b) Topic-9.1-Defining and Differentiating Parametric Equations
(c) Topic-9.6-Solving Motion Problems Using Parametric and Vector-Valued Functions
(d) Topic-9.1-Defining and Differentiating Parametric Equations
The graphs of the polar curves $r = 3$ and $r = 3 – 2\sin(2\theta)$ are shown in the figure above for $0 \leq \theta \leq \pi$.
(a) Let $R$ be the shaded region that is inside the graph of $r = 3$ and inside the graph of $r = 3 – 2\sin(2\theta)$. Find the area of $R$.
(b) For the curve $r = 3 – 2\sin(2\theta)$, find the value of $\frac{dx}{d\theta}$ at $\theta = \frac{\pi}{6}$.
(c) The distance between the two curves changes for $0 < \theta < \frac{\pi}{2}$.
Find the rate at which the distance between the two curves is changing with respect to $\theta$ when $\theta = \frac{\pi}{3}$.
(d) A particle is moving along the curve $r = 3 – 2\sin(2\theta)$ so that $\frac{d\theta}{dt} = 3$ for all times $t \geq 0$. Find the value of $\frac{dr}{dt}$ at $\theta = \frac{\pi}{6}$.
▶️Answer/Explanation
The polar curves are circle $r_1 = 3$ and $r_2 = 3 – 2\sin(2\theta)$ for $0 \leq \theta \leq \pi$.
Note: $r_1 = r_2$ when $\theta = 0$, $\frac{\pi}{2}$, and $\pi$.
\(\textbf{2(a)}\)
The area, $R$, inside of $r_1$ and inside of $r_2$ on this interval is:
\(
\frac{1}{2} \int_{0}^{\pi/2} r_2^2 \, d\theta + \frac{1}{2} \int_{\pi/2}^{\pi} r_1^2 \, d\theta \approx 9.707963268 \approx 9.708.
\)
OR
\(
\frac{1}{2} \int_{0}^{\pi/2} r_2^2 \, d\theta + \frac{1}{4}\pi(3)^2 = 2.639379797 + \frac{9\pi}{4} \approx 9.708.
\)
\(\textbf{2(b)}\)
For $x = r_2 \cos\theta = (3 – 2\sin(2\theta))\cos\theta$:
Using your calculator:
\(
\frac{dx}{d\theta}\Big|_{\theta = \pi/6} \approx -2.366025009 \approx -2.366.
\)
\(\textbf{2(c)}\)
The distance between the two curves on $\left(0, \frac{\pi}{2}\right)$ is $d = r_1 – r_2$.
We want:
\(
\frac{d}{d\theta}(d)\Big|_{\theta = \pi/3} = -1.999998667 \approx -2.
\)
\(\textbf{2(d)}\)
\(
r = 3 – 2\sin(2\theta) \quad \text{and} \quad \frac{d\theta}{dt} = 3.
\)
\(
\frac{dr}{dt} = -2\cos(2\theta) \cdot 2 \cdot \frac{d\theta}{dt}.
\)
\(
\frac{dr}{dt}\Big|_{\theta = \pi/6} = -2\cos\left(2 \cdot \frac{\pi}{6}\right) \cdot 2(3) = -12\cos\left(\frac{\pi}{3}\right) = -12\left(\frac{1}{2}\right) = -6.
\)
Question 3
(a) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(b) Topic-5.3 Determining Intervals on Which a Function is Increasing or Decreasing
(c) Topic-2.9 The Quotient Rule
(d) Topic-2.8 The Product Rule
The function f is defined on the closed interval [-5, 4] . The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by \(g(x)=\int_{-3}^{x}f(t)dt.\)
(a) Find g(3) .
(b) On what open intervals contained in -5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer.
(c) The function h is defined by \(h(x)= \frac{g(x)}{5x}\). Find h'(3).
(d) The function p is defined by p(x) = f(x2 – x). Find the slope of the line tangent to the graph of p at the point where x = -1.
▶️Answer/Explanation
\(
\textbf{3(a) } g(3) = \int_{-3}^{3} f(t) \, dt = 6 + 4 – 1 = 9
\)
\(
\textbf{3(b) } g'(x) = f(x)
\)
\(
\text{The graph of } g \text{ is increasing and concave down on the intervals } -5 < x < -3\)
\(\text{ and } 0 < x < 2 \text{ because } g’ = f \text{ is positive and decreasing on these intervals.}\)
\(
\textbf{3(c) } h'(x) = \frac{5x g'(x) – g(x) 5}{(5x)^2} = \frac{5x g'(x) – 5 g(x)}{25x^2}
\)
\(
h'(3) = \frac{(5)(3)g'(3) – 5g(3)}{25 \cdot 3^2}
\)
\(
= \frac{15(-2) – 5(9)}{225} = \frac{-75}{225} = -\frac{1}{3}
\)
\(
\textbf{3(d) } p'(x) = f'(x^2 – x)(2x – 1)
\)
\(
p'(-1) = f'(2)(-3) = (-2)(-3) = 6
\)
Question 4
(a) Topic-2.1 Defining Average and Instantaneous Rates of Change at a Point
(b) Topic-1.16 Working with the Intermediate Value Theorem (IVT)
(c) Topic-6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
(d) Topic-3.2 Implicit Differentiation
t (minutes) | 0 | 2 | 5 | 8 | 12 |
vA (t) (meters/ minute) | 0 | 100 | 40 | -120 | -150 |
Train A runs back and forth on an east-west section of railroad track. Train A’s velocity, measured in meters per minute, is given by a differentiable function vA (t) where time t is measured in minutes. Selected values for vA (t) are given in the table above.
(a) Find the average acceleration of train A over the interval 2 ≤ t ≤ 8.
(b) Do the data in the table support the conclusion that train A’s velocity is -100 meters per minute at some time t with 5 < t < 8 ? Give a reason for your answer.
(c) At time t = 2, train A’s position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train A, in meters from the Origin Station, at time t = 12. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time t = 12.
(d) A second train, train B, travels north from the Origin Station. At time t the velocity of train B is given by vB (t) = -5t2 + 60t + 25, and at time t = 2 the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train A and train B is changing at time t = 2.
▶️Answer/Explanation
\(
\textbf{4(a)} \text{average accel} = \frac{v_A(8) – v_A(2)}{8 – 2} = \frac{-120 – 100}{6} = \frac{-110}{3} \, \text{m/min}^2
\)
\(
\textbf{4(b) } v_A \text{ is differentiable } \implies v_A \text{ is continuous}
\)
\(
v_A(8) = -120 < -100 < 40 = v_A(5)
\)
\(
\text{Therefore, by the Intermediate Value Theorem, there is a time } t, \, 5 < t < 8, \text{ such that } v_A(t) = -100.
\)
\(
\textbf{4(c) } s_A(12) = s_A(2) + \int_{2}^{12} v_A(t) \, dt = 300 + \int_{2}^{12} v_A(t) \, dt
\)
\(
\int_{2}^{12} v_A(t) \, dt \approx 3 \cdot \frac{100 + 40}{2} + 3 \cdot \frac{40 – 120}{2} + 4 \cdot \frac{-120 – 150}{2}
\)
\(
= -450
\)
\(
s_A(12) \approx 300 – 450 = -150
\)
\(
\text{The position of Train A at time } t = 12 \text{ minutes is approximately 150 meters west of Origin Station.}
\)
\(
\textbf{4(d)} \text{Let } x \text{ be train A’s position, } y \text{ train B’s position, and } z \text{ the distance between train A and train B.}
\)
\(
z^2 = x^2 + y^2 \implies 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\)
\(
x = 300, \, y = 400 \implies z = 500
\)
\(
v_B(2) = -20 + 120 + 25 = 125
\)
\(
500 \frac{dz}{dt} = (300)(100) + (400)(125)
\)
\(
\frac{dz}{dt} = \frac{80000}{500} = 160 \, \text{meters per minute}
\)
Question 5
(a) Topic-9.9-Finding the Area of the Region Bounded by Two Polar Curves
(b) Topic-8.9-Volume with Disc Method Revolving Around the x or y Axis
(c) Topic-8.13-The Arc Length of a Smooth Planar Curve and Distance Traveled
Let $R$ be the shaded region bounded by the graph of $y = xe^{x^2}$, the line $y = -2x$, and the vertical line $x = 1$, as shown in the figure above.
(a) Find the area of $R$.
(b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when $R$ is rotated about the horizontal line $y = -2$.
(c) Write, but do not evaluate, an expression involving one or more integrals that gives the perimeter of $R$.
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(
\text{Area} = \int_{0}^{1} (y_1 – y_2) \, dx = \int_{0}^{1} \left(xe^{x^2} – (-2x)\right) \, dx = \int_{0}^{1} xe^{x^2} + 2x \, dx.
\)
\(
= \left[\frac{1}{2} e^{x^2} + x^2\right]_0^1 = \left(\frac{1}{2}e + 1\right) – \left(\frac{1}{2} + 0\right) = \frac{1}{2}e + \frac{1}{2}.
\)
\(\textbf{4(b)}\)
When rotated about $y = -2$, the volume obtained is:
\(
\pi \int_{0}^{1} \left[(y_1 – (-2))^2 – (y_2 – (-2))^2\right] \, dx.
\)
\(\textbf{4(c)}\)
The perimeter of $R$ is the sum of the length of curves $y_1$ and $y_2$ and the vertical line from $y_1(1) = e$ to $y_2(1) = -2$:
\(
\int_{0}^{1} \sqrt{1 + \left(\frac{dy_1}{dx}\right)^2} \, dx + \int_{0}^{1} \sqrt{1 + \left(\frac{dy_2}{dx}\right)^2} \, dx + e – (-2).
\)
\(
\frac{dy_1}{dx} = 2x^2e^{x^2} + e^{x^2} \quad \text{and} \quad \frac{dy_2}{dx} = -2.
\)
\(
\text{Perimeter} = \int_{0}^{1} \sqrt{1 + (2x^2e^{x^2} + e^{x^2})^2} \, dx + \int_{0}^{1} \sqrt{1 + (-2)^2} \, dx + e + 2.
\)
Question 6
(a) Topic-10.8-Ratio Test for Convergence
(b) Topic-10.14-Finding Taylor or Maclaurin Series for a Function
(c) Topic-10.2-Working with Geometric Series
The Taylor series for a function $f$ about $x = 1$ is given by
\(
\sum_{n=1}^\infty (-1)^{n+1} \frac{2^n}{n} (x-1)^n
\)
and converges to $f(x)$ for $|x-1| < R$, where $R$ is the radius of convergence of the Taylor series.
(a) Find the value of $R$.
(b) Find the first three nonzero terms and the general term of the Taylor series for $f’$, the derivative of $f$, about $x = 1$.
(c) The Taylor series for $f’$ about $x = 1$, found in part (b), is a geometric series.
Find the function $f’$ to which the series converges for $|x-1| < R$. Use this function to determine $f$ for $|x-1| < R$.
▶️Answer/Explanation
\(
f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{2^n}{n} (x-1)^n \quad \text{for } |x-1| < R,
\)
where $R$ is the radius of convergence.
\(
f(x) = 2(x-1) – \frac{2^2}{2} (x-1)^2 + \frac{2^3}{3} (x-1)^3 – \dots
\)
\(\textbf{6(a)}\)
Using the Ratio Test:
\(
\lim_{n \to \infty} \left| \frac{2^{n+1} (x-1)^{n+1}}{n+1} \cdot \frac{n}{2^n (x-1)^n} \right| =
\lim_{n \to \infty} \left| \frac{n}{n+1} \cdot 2(x-1) \right| = 2|x-1| < 1.
\)
\(
|x-1| < \frac{1}{2} \implies R = \frac{1}{2}.
\)
\(\textbf{6(b)}\)
Taking the derivative of $f(x)$:
\(
f'(x) = 2 – 2^2(x-1) + 2^3(x-1)^2 – \dots
\)
General term:
\(
f'(x) = \sum_{n=1}^\infty (-1)^{n+1} 2^n (x-1)^{n-1}.
\)
\(\textbf{6(c)}\)
Since $f’$ is geometric, then $f’ = \frac{a_1}{1 – r}$, where the first term $a_1 = 2$ and the common ratio $r = -2(x-1)$. So:
\(
f'(x) = \frac{2}{1 – (-2(x-1))} = \frac{2}{2x-1}.
\)
Now,
\(
f(x) = \int \frac{2}{2x-1} \, dx = \ln|2x-1| + C.
\)
From the original series at $f(1) = 0$:
\(
0 = \ln|2(1)-1| + C \implies C = 0.
\)
Since $R = \frac{1}{2}$ and $2x-1 > 0$ on this interval:
\(
f(x) = \ln(2x-1).
\)