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Question 1

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

(b) Topic-6.8-Finding Antiderivatives and Indefinite Integrals Basic Rules and Notation

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

(d) Topic-6.4-The Fundamental Theorem of Calculus and Accumulation Functions

There is no snow on Janet’s driveway when snow begins to fall at midnight. From midnight to 9 A.M., snow accumulates on the driveway at a rate
modeled by $f(t) = 7t e^{\cos t}$ cubic feet per hour, where $t$ is measured in hours since midnight. Janet starts removing snow at 6 A.M. ($t = 6$).
The rate $g(t)$, in cubic feet per hour, at which Janet removes snow from the driveway at time $t$ hours after midnight is modeled by
\( g(t) = \begin{cases} 0 & \text{for } 0 \leq t < 6, \\ 125 & \text{for } 6 \leq t < 7, \\ 108 & \text{for } 7 \leq t \leq 9. \end{cases} \)
(a) How many cubic feet of snow have accumulated on the driveway by 6 A.M.?
(b) Find the rate of change of the volume of snow on the driveway at 8 A.M.
(c) Let $h(t)$ represent the total amount of snow, in cubic feet, that Janet has removed from the driveway at time $t$ hours after midnight.
Express $h$ as a piecewise-defined function with domain $0 \leq t \leq 9$.
(d) How many cubic feet of snow are on the driveway at 9 A.M.?

▶️Answer/Explanation

\(\textbf{1(a)}\)
$\int_{0}^{6} f(t) \, dt = 142.274 \text{ or } 142.275 \text{ cubic feet.}$

\(\textbf{1(b)}\)
$\text{Rate of change is } f(8) – g(8) = -59.582 \text{ or } -59.583 \text{ cubic feet per hour.}$

\(\textbf{1(c)}\)
\(
h(0) = 0
\)
\(
\text{For } 0 < t \leq 6, \quad h(t) = h(0) + \int_{0}^{t} g(s) \, ds = 0 + \int_{0}^{t} 0 \, ds = 0.
\)
\(
\text{For } 6 < t \leq 7, \quad h(t) = h(6) + \int_{6}^{t} g(s) \, ds = 0 + \int_{6}^{t} 125 \, ds = 125(t – 6).
\)
\(
\text{For } 7 < t \leq 9, \quad h(t) = h(7) + \int_{7}^{t} g(s) \, ds = 125 + \int_{7}^{t} 108 \, ds = 125 + 108(t – 7).
\)
\(
\text{Thus, } h(t) =
\begin{cases}
0 & \text{for } 0 \leq t \leq 6, \\
125(t – 6) & \text{for } 6 < t \leq 7, \\
125 + 108(t – 7) & \text{for } 7 < t \leq 9.
\end{cases}
\)

\(\textbf{1(d)}\)
$ \text{Amount of snow is } \int_{0}^{9} f(t) \, dt – h(9) = 26.334 \text{ or } 26.335 \text{ cubic feet.}$

Question 2

(a) Topic-6.1-Exploring Accumulations of Change

(b) Topic-6.2-Approximating Areas with Riemann Sums

(c) Topic-6.8-Finding Antiderivatives and Indefinite Integrals Basic Rules and Notation

(d) Topic-5.6-Determining Concavity of Functions over Their Domains


A zoo sponsored a one-day contest to name a new baby elephant. Zoo visitors deposited entries in a special box between noon ($t = 0$) and 8 P.M. ($t = 8$).
The number of entries in the box $t$ hours after noon is modeled by a differentiable function $E$ for $0 \leq t \leq 8$.
Values of $E(t)$, in hundreds of entries, at various times $t$ are shown in the table below.
(a) Use the data in the table to approximate the rate, in hundreds of entries per hour, at which entries were being deposited at time $t = 6$.
Show the computations that lead to your answer.
(b) Use a trapezoidal sum with the four subintervals given by the table to approximate the value of $\frac{1}{8} \int_{0}^{8} E(t) \, dt$.
Using correct units, explain the meaning of $\frac{1}{8} \int_{0}^{8} E(t) \, dt$ in terms of the number of entries.
(c) At 8 P.M., volunteers began to process the entries.
They processed the entries at a rate modeled by the function $P$, where \(P(t) = t^3 – 30t^2 + 298t – 976 \) hundreds of entries per hour for $8 \leq t \leq 12$.
According to the model, how many entries had not yet been processed by midnight ($t = 12$)?

▶️Answer/Explanation

\(\textbf{2(a)}\)
$E'(6) \approx \frac{E(7) – E(5)}{7 – 5} = \frac{60 – 49}{7 – 5} = 4$ hundred entries per hour.

\(\textbf{2(b)}\)
\(
\frac{1}{8} \int_{0}^{8} E(t) \, dt \approx \frac{1}{8} \left( 2 \cdot \frac{E(0) + E(2)}{2} + 3 \cdot \frac{E(2) + E(5)}{2} + 2 \cdot \frac{E(5) + E(7)}{2} + 1 \cdot \frac{E(7) + E(8)}{2} \right)
\)
\(
= \frac{1}{8} \left( 2 \cdot \frac{0 + 7}{2} + 3 \cdot \frac{7 + 24}{2} + 2 \cdot \frac{24 + 49}{2} + 1 \cdot \frac{49 + 60}{2} \right)
\)
\(
= 10.687 \text{ or } 10.688
\)
$\frac{1}{8} \int_{0}^{8} E(t) \, dt$ is the average number of hundreds of entries in the box between noon and 8 P.M.

\(\textbf{2(c)}\)
$ 23 – \int_{8}^{12} P(t) \, dt = 23 – 16 = 7 \text{ hundred entries.}$

\(\textbf{2(d)}\)
$P'(t) = 0$ when $t = 9.183503$ and $t = 10.816497$. From the table:
\( \begin{array}{|c|c|} \hline t & P(t) \\ \hline 8 & 0 \\ 9.183503 & 5.088662 \\ 10.816497 & 2.911338 \\ 12 & 8 \\ \hline \end{array} \)
Entries are being processed most quickly at time $t = 12$.

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

(c) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema

(d.i) Topic-5.8-Sketching Graphs of Functions and Their Derivatives

(d.ii) Topic-5.9-Connecting a Function Its First Derivative and Its Second Derivative

(d.iii) Topic-4.1-Interpreting the Meaning of the Derivative in Context

A particle is moving along a curve so that its position at time $t$ is $(x(t), y(t))$, where $x(t) = t^2 – 4t + 8$ and $y(t)$ is not explicitly given.
Both $x$ and $y$ are measured in meters, and $t$ is measured in seconds. It is known that \( \frac{dy}{dt} = t e^{t^3 – 1}. \)
a) Find the speed of the particle at time $t = 3$ seconds.
b) Find the total distance traveled by the particle for $0 \leq t \leq 4$ seconds.
c) Find the time $t$, $0 \leq t \leq 4$, when the line tangent to the path of the particle is horizontal. Is the direction of motion of the particle toward
the left or toward the right at that time? Give a reason for your answer.
d) There is a point with $x$-coordinate 5 through which the particle passes twice. Find each of the following.
(i) The two values of $t$ when that occurs.
(ii) The slopes of the lines tangent to the particle’s path at that point.
(iii) The $y$-coordinate of that point, given $y(2) = 3 + \frac{1}{e}$.

▶️Answer/Explanation

\(\textbf{3(a)}\)
Speed = $\sqrt{(x'(3))^2 + (y'(3))^2} = 2.828$ meters per second.

\(\textbf{3(b)}\)
$x'(t) = 2t – 4$
Distance = $\int_0^4 \sqrt{(2t – 4)^2 + (te^{-t^3} – 1)^2} , dt = 11.587$ or $11.588$ meters.

\(\textbf{3(c)}\)
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 0$ when $te^{-t^3} – 1 = 0$ and $2t – 4 \neq 0$.
This occurs at $t = 2.20794$.
Since $x'(2.20794) > 0$, the particle is moving toward the right at time $t = 2.207$ or $2.208$.

\(\textbf{3(d)}\)
   (i) $x(t) = 5$ at $t = 1$ and $t = 3$.
At time $t = 1$, the slope is $\frac{dy}{dx}\big|{t=1} = \frac{dy/dt}{dx/dt}\big|{t=1} = 0.432$.
   (ii) At time $t = 3$, the slope is $\frac{dy}{dx}\big|{t=3} = \frac{dy/dt}{dx/dt}\big|{t=3} = 1$.
   (iii) $y(1) = y(3) = 3 + \frac{1}{e} + \int_2^3 \frac{dy}{dt} , dt = 4$.

Question 4

(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-8.7-Volumes with Cross Sections Squares and Rectangles



Let \( R \) be the region in the first quadrant bounded by the graph of \( y = 2\sqrt{x} \), the horizontal line \( y = 6 \), and the \( y \)-axis, as shown in the figure.
(a), the area of \( R \) is calculated as \( \int_0^9 (6 – 2\sqrt{x}) \, dx \).
(b), the volume of the solid generated when \( R \) is rotated about the horizontal line \( y = 7 \) is expressed as \( \pi \int_0^9 \left[ (7 – 2\sqrt{x})^2 – (7 – 6)^2 \right] \, dx \).
(c), the volume of the solid where the cross-sections perpendicular to the \( y \)-axis are rectangles with height three times the base is given by \( \int_0^6 3 \cdot \left( \frac{y^2}{4} \right)^2 \, dy \).

▶️Answer/Explanation

\(\begin{aligned}
\textbf{4(a)} Area &= \int_0^9 \left(6 – 2\sqrt{x}\right) \, dx = \left( 6x – \frac{4}{3}x^{3/2} \right) \Big|_{x=0}^{x=9} = 18.
\end{aligned}\)

\(\begin{aligned}
\textbf{4(b)} Volume &= \pi \int_0^9 \left[ \left(7 – 2\sqrt{x}\right)^2 – \left(7 – 6\right)^2 \right] \, dx.
\end{aligned}\)

$\textbf{4(c)}$ Solving  $y = 2\sqrt{x} \text{ for } x \text{ yields } x = \frac{y^2}{4}$.
Each rectangular cross section has area $\left( 3 \frac{y^2}{4} \right) \left( \frac{y^2}{4} \right) = \frac{3}{16}y^4.$
Volume = $\int_0^6 \frac{3}{16} y^4 \, dy.$

Question 5

(a) Topic-7.5-Approximating Solutions Using Euler’s Method

(b) Topic-7.8-Exponential Models with Differential Equations

(c) Topic-7.6-Finding General Solutions Using Separation of Variables

Consider the differential equation \( \frac{dy}{dx} = 1 – y \). Let \( y = f(x) \) be the particular solution to this differential equation with the initial condition \( f(1) = 0 \).
For this particular solution, \( f(x) < 1 \) for all values of \( x \).
(a) 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.
(b) Find \( \lim_{x \to 1} \frac{f(x)}{x^3 – 1} \). Show the work that leads to your answer.
(c) Find the particular solution \( y = f(x) \) to the differential equation \( \frac{dy}{dx} = 1 – y \) with the initial condition \( f(1) = 0 \).

▶️Answer/Explanation

\(\textbf{5(a)}\)
\( f\left(\frac{1}{2}\right) \approx f(1) + \left(\frac{dy}{dx}\Big|_{(1, 0)}\right) \cdot \Delta x = 0 + 1 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} \).
\( f(0) \approx f\left(\frac{1}{2}\right) + \left(\frac{dy}{dx}\Big|_{\left(\frac{1}{2}, -\frac{1}{2}\right)}\right) \cdot \Delta x \approx -\frac{1}{2} + \frac{3}{2} \cdot \left(-\frac{1}{2}\right) = -\frac{5}{4} \).

\(\textbf{5(b)}\)
Since \( f \) is differentiable at \( x = 1 \), \( f \) is continuous at \( x = 1 \). So,
\(
\lim_{x \to 1} f(x) = 0 = \lim_{x \to 1} (x^3 – 1),
\)
and we may apply L’Hospital’s Rule:
\(
\lim_{x \to 1} \frac{f(x)}{x^3 – 1} = \lim_{x \to 1} \frac{f'(x)}{3x^2} = \lim_{x \to 1} \frac{f'(x)}{3x^2} = \frac{1}{3}.
\)

\(\textbf{5(c)}\)
For \( \frac{dy}{dx} = 1 – y \), separate variables:
\(
\int \frac{1}{1 – y} \, dy = \int 1 \, dx.
\)
Solving,
\(
-\ln|1 – y| = x + C.
\)
Substitute \( -\ln 1 = 1 + C \), so \( C = -1 \). Then,
\(
\ln|1 – y| = 1 – x.
\)
Exponentiate:
\(
|1 – y| = e^{1-x}.
\)
Simplify:
\(
f(x) = 1 – e^{1-x}.
\)

Question 6

(a) Topic-10.14-Finding Taylor or Maclaurin Series for a Function

(b) Topic-10.15-Representing Functions as Power Series

(c) Topic-10.11-Finding Taylor Polynomial Approximations of Functions

(d) Topic-10.12-Lagrange Error Bound

\(
f(x) =
\begin{cases}
\frac{\cos x – 1}{x^2} & \text{for } x \neq 0, \\
-\frac{1}{2} & \text{for } x = 0.
\end{cases}
\)
The function \( f \), defined above, has derivatives of all orders. Let \( g \) be the function defined by
\(
g(x) = 1 + \int_0^x f(t) \, dt.
\)
(a) Write the first three nonzero terms and the general term of the Taylor series for \( \cos x \) about \( x = 0 \). Use this series to write the first three nonzero terms and the general term of the Taylor series for \( f \) about \( x = 0 \).
(b) Use the Taylor series for \( f \) about \( x = 0 \) found in part (a) to determine whether \( f \) has a relative maximum, relative minimum, or neither at \( x = 0 \). Give a reason for your answer.
(c) Write the fifth-degree Taylor polynomial for \( g \) about \( x = 0 \).
(d) The Taylor series for \( g \) about \( x = 0 \), evaluated at \( x = 1 \), is an alternating series with individual terms that decrease in absolute value to 0. Use the third-degree Taylor polynomial for \( g \) about \( x = 0 \) to estimate the value of \( g(1) \). Explain why this estimate differs from the actual value of \( g(1) \) by less than \( \frac{1}{6!} \).

▶️Answer/Explanation

$\textbf{6(a)}$
\( \cos(x) = 1 – \frac{x^2}{2} + \frac{x^4}{4!} – \cdots + (-1)^n \frac{x^{2n}}{(2n)!} + \cdots \)
\( f(x) = -\frac{1}{2} + \frac{x^2}{4!} – \frac{x^4}{6!} + \cdots + (-1)^{n+1} \frac{x^{2n}}{(2n+2)!} + \cdots \)

$\textbf{6(b)}$
\( f'(0) \) is the coefficient of \( x \) in the Taylor series for \( f \) about \( x = 0 \), so \( f'(0) = 0 \).
\( \frac{f”(0)}{2!} = \frac{1}{4!} \) is the coefficient of \( x^2 \) in the Taylor series for \( f \) about \( x = 0 \), so \( f”(0) = \frac{1}{12} \).
Therefore, by the Second Derivative Test, \( f \) has a relative minimum at \( x = 0 \).

$\textbf{6(c)}$
\( P_5(x) = 1 – \frac{x}{2} + \frac{x^3}{3 \cdot 4!} – \frac{x^5}{5 \cdot 6!} \)

$\textbf{6(d)}$
\( g(1) \approx 1 – \frac{1}{2} + \frac{1}{3 \cdot 4!} = \frac{37}{72} \).
Since the Taylor series for \( g \) about \( x = 0 \) evaluated at \( x = 1 \) is alternating and the terms decrease in absolute value to 0, we know
\(
\left| g(1) – \frac{37}{72} \right| < \frac{1}{5 \cdot 6!} < \frac{1}{6!}.
\)

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