Question 1
(a) Topic-4.3-Rates of Change in Applied Contexts Other Than Motion
(b) Topic-6.2-Approximating Areas with Riemann Sums
(c) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(d) Topic-8.2-Connecting Position Velocity and Acceleration of Functions Using Integrals
Caren rides her bicycle along a straight road from home to school, starting at home at time \( t = 0 \) minutes and arriving at school at time \( t = 12 \) minutes.
During the time interval \( 0 \leq t \leq 12 \) minutes, her velocity \( v(t) \), in miles per minute, is modeled by the piecewise-linear function whose graph is shown below.
(a) Find the acceleration of Caren’s bicycle at time \( t = 7.5 \) minutes. Indicate units of measure.
(b) Using correct units, explain the meaning of \( \int_{0}^{12} |v(t)| \, dt \) in terms of Caren’s trip.
Find the value of \( \int_{0}^{12} |v(t)| \, dt. \)
(c) Shortly after leaving home, Caren realizes she left her calculus homework at home, and she returns to get it.
At what time does she turn around to go back home? Give a reason for your answer.
(d) Larry also rides his bicycle along a straight road from home to school in 12 minutes.
His velocity is modeled by the function \( w \) given by \( w(t) = \frac{\pi}{15} \sin\left(\frac{\pi}{12} t\right), \) where \( w(t) \) is in miles per minute for \( 0 \leq t \leq 12 \).
Who lives closer to school: Caren or Larry? Show the work that leads to your answer.
▶️Answer/Explanation
\(\textbf{1(a)}\)
The acceleration of Caren’s bicycle at \( t = 7.5 \) minutes is given by:
\( a(7.5) = v'(7.5) = \frac{v(8) – v(7)}{8 – 7} = \frac{0.1 – 0.2}{1} = -0.1 \, \text{miles/minute}^2. \)
\(\textbf{1(b)}\)
The integral \( \int_{0}^{12} |v(t)| \, dt \) represents the total distance, in miles, that Caren rode during the 12 minutes from \( t = 0 \) to \( t = 12 \).
We compute: \( \int_{0}^{12} |v(t)| \, dt = \int_{0}^{2} v(t) \, dt – \int_{2}^{4} v(t) \, dt + \int_{4}^{12} v(t) \, dt. \)
Substituting the values: \( \int_{0}^{12} |v(t)| \, dt = 0.2 + 0.2 + 1.4 = 1.8 \, \text{miles}. \)
\(\textbf{1(c)}\)
Caren turns around to go back home at time \( t = 2 \) minutes. This is the time at which her velocity changes from positive to negative.
\(\textbf{1(d)}\)
Larry’s total distance from home to school is given by: \( \int_{0}^{12} w(t) \, dt = 1.6 \, \text{miles}. \)
Caren’s total distance from home to school is given by: \( \int_{0}^{12} |v(t)| \, dt = 1.4 \, \text{miles}. \)
Therefore, Caren lives closer to school.
Question 2
(a) Topic-6.1-Exploring Accumulations of Change
(b) Topic-5.3-Determining Intervals on Which a Function Is Increasing or Decreasing
(c) Topic-6.5-Interpreting the Behavior of Accumulation Functions Involving Area
(d) Topic-6.5-Interpreting the Behavior of Accumulation Functions Involving Area
The rate at which people enter an auditorium for a rock concert is modeled by the function $R$ given by \( R(t) = 1380t^2 – 675t^3 \quad \text{for } 0 \leq t \leq 2 \text{ hours}, \)
where $R(t)$ is measured in people per hour. No one is in the auditorium at time $t = 0$, when the doors open. The doors close and the concert begins at time $t = 2$.
(a) How many people are in the auditorium when the concert begins?
(b) Find the time when the rate at which people enter the auditorium is a maximum. Justify your answer.
(c) The total wait time for all the people in the auditorium is found by adding the time each person waits,
starting at the time the person enters the auditorium and ending when the concert begins.
The function $w$ models the total wait time for all the people who enter the auditorium before time $t$.
The derivative of $w$ is given by \( w'(t) = (2 – t)R(t). \) Find $w(2) – w(1)$, the total wait time for those who enter the auditorium after time $t = 1$.
(d) On average, how long does a person wait in the auditorium for the concert to begin? Consider all people who enter the auditorium after the doors open, and use the model for total wait time from part (c).
▶️Answer/Explanation
\(\textbf{2(a)}\)
$\int_0^2 R(t) \, dt = 980 \text{ people.}$
\(\textbf{2(b)}\)
\(
R'(t) = 0 \text{ when } t = 0 \text{ and } t = 1.36296.
\)
The maximum rate may occur at $0$, $a = 1.36296$, or $2$.
\(
R(0) = 0, \quad R(a) \approx 854.527, \quad R(2) = 120.
\)
The maximum rate occurs when $t = 1.362 \text{ or } 1.363$.
\(\textbf{2(c)}\)
\( w(2) – w(1) = \int_1^2 w'(t) \, dt = \int_1^2 (2 – t)R(t) \, dt = 387.5. \)
The total wait time for those who enter the auditorium after time $t = 1$ is $387.5$ hours.
\(\textbf{2(d)}\)
\( \frac{1}{980} w(2) = \frac{1}{980} \int_0^2 (2 – t)R(t) \, dt \approx 0.77551. \)
On average, a person waits $0.775 \text{ or } 0.776$ hours.
Question 3
(a) Topic-5.10-Introduction to Optimization Problems
(b) Topic-6.1-Exploring Accumulations of Change
(c) Topic-6.3-Riemann Sums Summation Notation and Definite Integral Notation
(d) Topic-3.1-The Chain Rule
A diver leaps from the edge of a diving platform into a pool below. The figure below shows the initial position of the diver and her position at a later time.
At time $t$ seconds after she leaps, the horizontal distance from the front edge of the platform to the diver’s shoulders is given by $x(t)$,
and the vertical distance from the water surface to her shoulders is given by $y(t)$, where $x(t)$ and $y(t)$ are measured in meters.|
Suppose that the diver’s shoulders are $11.4$ meters above the water when she makes her leap and that \( \frac{dx}{dt} = 0.8 \quad \text{and} \quad \frac{dy}{dt} = 3.6 – 9.8t, \)
for $0 \leq t \leq A$, where $A$ is the time that the diver’s shoulders enter the water.
(a) Find the maximum vertical distance from the water surface to the diver’s shoulders.
(b) Find $A$, the time that the diver’s shoulders enter the water.
(c) Find the total distance traveled by the diver’s shoulders from the time she leaps from the platform until the time her shoulders enter the water.
(d) Find the angle $\theta$, $0 < \theta < \frac{\pi}{2}$, between the path of the diver and the water at the instant the diver’s shoulders enter the water.
▶️Answer/Explanation
\(\textbf{3(a)}\)
$\frac{dy}{dt} = 0$ only when $t = 0.36735$. Let $b = 0.36735$.
The maximum vertical distance from the water surface to the diver’s shoulders is \( y(b) = 11.4 + \int_{0}^{b} \frac{dy}{dt} \, dt = 12.061 \, \text{meters}. \)
Alternatively, \( y(t) = 11.4 + 3.6t – 4.9t^2, \quad \text{so } y(b) = 12.061 \, \text{meters}. \)
\(\textbf{3(b)}\)
$y(A) = 11.4 + \int_{0}^{A} \frac{dy}{dt} \, dt = 11.4 + 3.6A – 4.9A^2 = 0$\\ when $A = 1.936$ seconds.
\(\textbf{3(c)}\)
\( \int_{0}^{A} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = 12.946 \, \text{meters}. \)
\(\textbf{3(d)}\)
At time $A$, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \Big|_{t=A} = -19.21913. \)
The angle between the path of the diver and the water is \( \tan^{-1}(19.21913) = 1.518 \, \text{or } 1.519. \)
Question 4
(a) Topic-7.5-Approximating Solutions Using Euler’s Method BC Only
(b) Topic-10.11-Finding Taylor Polynomial Approximations of Functions
(c) Topic-7.7-Finding Particular Solutions Using Initial Conditions and Separation of Variables
Consider the differential equation $\frac{dy}{dx} = 6x^2 – x^2y$. Let $y = f(x)$ be a particular solution to this differential equation with the initial condition $f(-1) = 2$.
(a) Use Euler’s method with two steps of equal size, starting at $x = -1$, to approximate $f(0)$. Show the work that leads to your answer.
(b) At the point $(-1, 2)$, the value of $\frac{d^2y}{dx^2}$ is $-12$. Find the second-degree Taylor polynomial for $f$ about $x = -1$.
(c) Find the particular solution $y = f(x)$ to the given differential equation with the initial condition $f(-1) = 2$.
▶️Answer/Explanation
\(\textbf{4(a)}\)
Using Euler’s method:
\(
f\left(-\frac{1}{2}\right) \approx f(-1) + \left.\frac{dy}{dx}\right|_{(-1, 2)} \cdot \Delta x
\)
\(
= 2 + 4 \cdot \frac{1}{2} = 4
\)
\(
f(0) \approx f\left(-\frac{1}{2}\right) + \left.\frac{dy}{dx}\right|_{\left(-\frac{1}{2}, 4\right)} \cdot \Delta x
\)
\(
\approx 4 + \frac{1}{2} \cdot 1 = \frac{17}{4}
\)
\(\textbf{4(b)}\)
The second-degree Taylor polynomial is: \( P_2(x) = 2 + 4(x+1) – 6(x+1)^2 \)
\(\textbf{4(c)}\)
Solving the differential equation:
\(
\frac{dy}{dx} = x^2 (6 – y)
\)
Separating variables:
\(
\int \frac{1}{6 – y} \, dy = \int x^2 \, dx
\)
\(
-\ln|6 – y| = \frac{1}{3}x^3 + C
\)
Using the initial condition $f(-1) = 2$:
\(
-\ln 4 = \frac{1}{3}(-1)^3 + C
\)
\(
C = \frac{1}{3} \ln 4
\)
Substituting back:
\(
\ln|6 – y| = -\frac{1}{3}x^3 + \left(\frac{1}{3} \ln 4\right)
\)
Solving for $y$:
\(
|6 – y| = 4e^{-\frac{1}{3}(x^3 + 1)}
\)
\(
y = 6 – 4e^{-\frac{1}{3}(x^3 + 1)}
\)
Question 5
(a) Topic-2.3-Estimating Derivatives of a Function at a Point
(b) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(c) Topic-6.2-Approximating Areas with Riemann Sums
(d) Topic-5.9-Connecting a Function, Its First Derivative, and Its Second Derivative
Let $f$ be a function that is twice differentiable for all real numbers. The table below gives values of $f$ for selected points in the closed interval $2 \leq x \leq 13$.
(a) Estimate $f'(4)$. Show the work that leads to your answer.
(b) Evaluate $\int_2^{13} \left(3 – 5f'(x)\right) dx$. Show the work that leads to your answer.
(c) Use a left Riemann sum with subintervals indicated by the data in the table to approximate $\int_2^{13} f(x) \, dx$.
Show the work that leads to your answer.
(d) Suppose $f'(5) = 3$ and $f”(x) < 0$ for all $x$ in the closed interval $5 \leq x \leq 8$.
Use the line tangent to the graph of $f$ at $x = 5$ to show that $f(7) \leq 4$.
Use the secant line for the graph of $f$ on $5 \leq x \leq 8$ to show that $f(7) \geq \frac{4}{3}$.
▶️Answer/Explanation
\(\textbf{5(a)}\)
$f'(4) \approx \frac{f(5) – f(3)}{5 – 3} = \frac{-2 – 4}{5 – 3} = -3.$
\(\textbf{5(b)}\)
\(
\int_2^{13} \left(3 – 5f'(x)\right) dx = \int_2^{13} 3 \, dx – 5 \int_2^{13} f'(x) \, dx,
\)
\(
= 3(13 – 2) – 5\left(f(13) – f(2)\right) = 33 – 5(6 – 1) = 33 – 25 = 8.
\)
\(\textbf{5(c)}\)
\(
\int_2^{13} f(x) \, dx \approx f(2)(3 – 2) + f(3)(5 – 3) + f(5)(8 – 5) + f(8)(13 – 8),
\)
\(
= 1(1) + 4(2) + (-2)(3) + 3(5) = 1 + 8 – 6 + 15 = 18.
\)
\(\textbf{5(d)}\)
An equation for the tangent line is
\(
y = -2 + 3(x – 5).
\)
Since $f”(x) < 0$ for all $x$ in the interval $5 \leq x \leq 8$, the line tangent to the graph of $y = f(x)$ at $x = 5$ lies above the graph for all $x$ in the interval $5 \leq x \leq 8$. Therefore,
\(
f(7) \leq -2 + 3(2) = 4.
\)
An equation for the secant line is
\(
y = -2 + \frac{5}{3}(x – 5).
\)
Since $f”(x) < 0$ for all $x$ in the interval $5 \leq x \leq 8$, the secant line connecting $(5, f(5))$ and $(8, f(8))$ lies below the graph of $y = f(x)$ for all $x$ in the interval $5 \leq x \leq 8$. Therefore,
\(
f(7) \geq -2 + \frac{5}{3}(2) = -2 + \frac{10}{3} = \frac{4}{3}.
\)
Question 6
(a) Topic-10.11-Finding Taylor Polynomial Approximations of Functions
(b) Topic-10.11-Finding Taylor Polynomial Approximations of Functions
(c) Topic-10.13-Radius and Interval of Convergence of Power Series
(d) Topic-10.15-Representing Functions as Power Series
The Maclaurin series for $e^x$ is
\(
e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots + \frac{x^n}{n!} + \cdots.
\)
The continuous function $f$ is defined by
\(
f(x) = \frac{e^{(x-1)^2} – 1}{(x-1)^2} \quad \text{for } x \neq 1 \quad \text{and} \quad f(1) = 1.
\)
The function $f$ has derivatives of all orders at $x = 1$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $e^{(x-1)^2}$ about $x = 1$.
(b) Use the Taylor series found in part (a) to write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 1$.
(c) Use the ratio test to find the interval of convergence for the Taylor series found in part (b).
(d) Use the Taylor series for $f$ about $x = 1$ to determine whether the graph of $f$ has any points of inflection.
▶️Answer/Explanation
\(\textbf{6(a)}\)
The Taylor series for $e^{(x-1)^2}$ about $x = 1$ is:
\(
1 + \frac{(x-1)^2}{2} + \frac{(x-1)^4}{6} + \frac{(x-1)^6}{24} + \cdots + \frac{(x-1)^{2n}}{n!} + \cdots
\)
\(\textbf{6(b)}\)
Using the result from part (a), the Taylor series for $f(x)$ about $x = 1$ is:
\(
1 + \frac{(x-1)^2}{2} + \frac{(x-1)^4}{6} + \frac{(x-1)^6}{24} + \cdots + \frac{(x-1)^{2n}}{(n+1)!} + \cdots
\)
\(\textbf{6(c)}\)
Using the ratio test, the interval of convergence for the Taylor series found in part (b) is:
\(
\lim_{n \to \infty} \left| \frac{\frac{(x-1)^{2n+2}}{(n+2)!}}{\frac{(x-1)^{2n}}{(n+1)!}} \right| = \lim_{n \to \infty} \frac{(n+1)!}{(n+2)!} \cdot (x-1)^2 = \lim_{n \to \infty} \frac{(x-1)^2}{n+2} = 0.
\)
Therefore, the interval of convergence is $(-\infty, \infty)$.
\(\textbf{6(d)}\)
The second derivative of $f(x)$ is:
\( f”(x) = 1 + \frac{4 \cdot 3}{6}(x-1)^2 + \frac{6 \cdot 5}{24}(x-1)^4 + \cdots + \frac{2n(2n-1)}{(n+1)!}(x-1)^{2n-2} + \cdots \)
Since every term of this series is nonnegative, $f”(x) \geq 0$ for all $x$. Therefore, the graph of $f$ has no points of inflection.