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Question 1

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

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

(c) Topic-8.7-Volumes with Cross Sections Squares and Rectangles

(d) Topic-8.8-Volumes with Cross Sections Triangles and Semicircles


Let $R$ be the region bounded by the graphs of $y = \sin(\pi x)$ and $y = x^3 – 4x$, as shown in the figure above.
(a) Find the area of $R$.
(b) The horizontal line $y = -2$ splits the region $R$ into two parts. Write, but do not evaluate, an integral expression for the area of the part of $R$ that is below this horizontal line.
(c) The region $R$ is the base of a solid. For this solid, each cross section perpendicular to the $x$-axis is a square. Find the volume of this solid.
(d) The region $R$ models the surface of a small pond. At all points in $R$ at a distance $x$ from the $y$-axis, the depth of the water is given by $h(x) = 3 – x$.
Find the volume of water in the pond.

▶️Answer/Explanation

\(\textbf{1(a)}\)
$\sin(\pi x) = x^3 – 4x$ at $x = 0$ and $x = 2$.
\(
\text{Area} = \int_0^2 \left( \sin(\pi x) – \left(x^3 – 4x\right) \right) dx = 4.
\)

\(\textbf{1(b)}\)
$x^3 – 4x = -2$ at $r = 0.5391889$ and $s = 1.6751309$.
The area of the stated region is
\(
\int_r^s \left(-2 – \left(x^3 – 4x\right)\right) dx.
\)

\(\textbf{1(c)}\)
\( \text{Volume} = \int_0^2 \left(\sin(\pi x) – \left(x^3 – 4x\right)\right)^2 dx = 9.978. \)

\(\textbf{1(d)}\)
\( \text{Volume} = \int_0^2 \left(3 – x\right)\left(\sin(\pi x) – \left(x^3 – 4x\right)\right) dx = 8.369 \text{ or } 8.370. \)

Question 2

(a) Topic-2.3-Estimating Derivatives of a Function at a Point

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

(c) Topic-5.1-Using the Mean Value Theorem

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

Concert tickets went on sale at noon ($t = 0$) and were sold out within 9 hours. The number of people waiting in line to purchase tickets at time $t$ is modeled by a twice-differentiable function $L$ for $0 \leq t \leq 9$. Values of $L(t)$ at various times $t$ are shown in the table above.

(a) Use the data in the table to estimate the rate at which the number of people waiting in line was changing at 5:30 P.M. ($t = 5.5$). Show the computations that lead to your answer. Indicate units of measure.
(b) Use a trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first 4 hours that tickets were on sale.
(c) For $0 \leq t \leq 9$, what is the fewest number of times at which $L'(t)$ must equal $0$? Give a reason for your answer.
(d) The rate at which tickets were sold for $0 \leq t \leq 9$ is modeled by $r(t) = 550t e^{-t/2}$ tickets per hour. Based on the model, how many tickets were sold by 3 P.M. ($t = 3$), to the nearest whole number?

▶️Answer/Explanation

\(\textbf{2(a)}\)
$L'(5.5) \approx \frac{L(7) – L(4)}{7 – 4} = \frac{150 – 126}{3} = 8$ people per hour.

\(\textbf{2(b)}\)
The average number of people waiting in line during the first 4 hours is approximately:
\( \frac{1}{4} \left( L(0) + L(1) \cdot (1 – 0) + \frac{L(1) + L(3)}{2} \cdot (3 – 1) + \frac{L(3) + L(4)}{2} \cdot (4 – 3) \right) = 155.25 \text{ people}\).

\(\textbf{2(c)}\)
$L$ is differentiable on $[0, 9]$, so the Mean Value Theorem implies $L'(t) > 0$ for some $t$ in $(1, 3)$ and some $t$ in $(4, 7)$.
Similarly, $L'(t) < 0$ for some $t$ in $(3, 4)$ and some $t$ in $(7, 8)$.
Then, since $L’$ is continuous on $[0, 9]$, the Intermediate Value Theorem implies that $L'(t) = 0$ for at least three values of $t$ in $[0, 9]$.
OR:
The continuity of $L$ on $[1, 4]$ implies that $L$ attains a maximum value there. Since $L(3) > L(1)$ and $L(3) > L(4)$, this maximum occurs on $(1, 4)$.
Similarly, $L$ attains a minimum on $(3, 7)$ and a maximum on $(4, 8)$.
$L$ is differentiable, so $L'(t) = 0$ at each relative extreme point on $(0, 9)$.
Therefore, $L'(t) = 0$ at least three values of $t$ on $(0, 9)$. Note: There is a function $L$ that satisfies the given conditions with $L'(t) = 0$ for exactly three values of $t$.

\(\textbf{2(d)}\)
$\int_{0}^{3} r(t) \, dt = 972.784$
There were approximately $973$ tickets sold by 3 P.M.

Question 3

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

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

(c) Topic-10.12-Lagrange Error Bound


Let $h$ be a function having derivatives of all orders for $x > 0$. Selected values of $h$ and its first four derivatives are indicated in the table above. The function $h$ and these four derivatives are increasing on the interval $1 \leq x \leq 3$.
(a) Write the first-degree Taylor polynomial for $h$ about $x = 2$ and use it to approximate $h(1.9)$.
Is this approximation greater than or less than $h(1.9)$? Explain your reasoning.
(b) Write the third-degree Taylor polynomial for $h$ about $x = 2$ and use it to approximate $h(1.9)$.
(c) Use the Lagrange error bound to show that the third-degree Taylor polynomial for $h$ about $x = 2$ approximates $h(1.9)$ with error less than $3 \times 10^{-4}$.

▶️Answer/Explanation

\(\textbf{3(a)}\)
The first-degree Taylor polynomial for $h$ about $x = 2$ is given by:
\(
P_1(x) = 80 + 128(x – 2).
\)
Using this to approximate $h(1.9)$:
\(
h(1.9) \approx P_1(1.9) = 80 + 128(1.9 – 2) = 67.2.
\)
Since $h’$ is increasing on the interval $1 \leq x \leq 3$, we conclude:
\(
P_1(1.9) < h(1.9).
\)

\(\textbf{3(b)}\)
The third-degree Taylor polynomial for $h$ about $x = 2$ is given by:
\(
P_3(x) = 80 + 128(x – 2) + \frac{488}{6}(x – 2)^2 + \frac{448}{18}(x – 2)^3.
\)
Using this to approximate $h(1.9)$:
\(
h(1.9) \approx P_3(1.9) = 67.988.
\)

\(\textbf{3(c)}\)
The fourth derivative of $h$ is increasing on the interval $1 \leq x \leq 3$. Thus:
\(
\max_{1.9 \leq x \leq 2} \left| h^{(4)}(x) \right| = \frac{584}{9}.
\)
Using the Lagrange error bound, the error is:
\(
|h(1.9) – P_3(1.9)| \leq \frac{\frac{584}{9} \cdot |1.9 – 2|^4}{4!}.
\)
Simplifying:
\(
|h(1.9) – P_3(1.9)| \leq \frac{\frac{584}{9} \cdot (0.1)^4}{24} = 2.7037 \times 10^{-4}.
\)
Since $2.7037 \times 10^{-4} < 3 \times 10^{-4}$, the approximation satisfies the error bound.

Question 4

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

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

(c) Topic-5.3-Determining Intervals on Which a Function Is Increasing or Decreasing

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


A particle moves along the $x$-axis so that its velocity at time $t$, for $0 \leq t \leq 6$, is given by a differentiable function $v$ whose graph is shown above.
The velocity is $0$ at $t = 0$, $t = 3$, and $t = 5$, and the graph has horizontal tangents at $t = 1$ and $t = 4$.
The areas of the regions bounded by the $t$-axis and the graph of $v$ on the intervals $[0, 3]$, $[3, 5]$, and $[5, 6]$ are $8$, $3$, and $2$, respectively.
At time $t = 0$, the particle is at $x = -2$.
a) For $0 \leq t \leq 6$, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer.
(b) For how many values of $t$, where $0 \leq t \leq 6$, is the particle at $x = -8$? Explain your reasoning.
(c) On the interval $2 < t < 3$, is the speed of the particle increasing or decreasing? Give a reason for your answer.
(d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.

▶️Answer/Explanation

\(\textbf{4(a)}\)
Since $v(t) < 0$ for $0 < t < 3$ and $5 < t < 6$, and $v(t) > 0$ for $3 < t < 5$, we consider $t = 3$ and $t = 6$.
\(
x(3) = -2 + \int_{0}^{3} v(t) \, dt = -2 – 8 = -10
\)
\(
x(6) = -2 + \int_{0}^{6} v(t) \, dt = -2 – 8 + 3 – 2 = -9
\)
Therefore, the particle is farthest left at time $t = 3$ when its position is $x(3) = -10$.

\(\textbf{4(b)}\)
The particle moves continuously and monotonically from $x(0) = -2$ to $x(3) = -10$. Similarly, the particle moves continuously and monotonically from $x(3) = -10$ to $x(5) = -7$ and also from $x(5) = -7$ to $x(6) = -9$. \\ By the Intermediate Value Theorem, there are three values of $t$ for which the particle is at $x(t) = -8$.

\(\textbf{4(c)}\)
The speed is decreasing on the interval $2 < t < 3$ since on this interval $v < 0$ and $v$ is increasing.

\(\textbf{4(d)}\)
The acceleration is negative on the intervals $0 < t < 1$ and $4 < t < 6$ since velocity is decreasing on these intervals.

Question 5

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

(b) Topic-5.9-Connecting a Function Its First Derivative and Its Second Derivative

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

The derivative of a function \( f \) is given by \( f'(x) = (x – 3)e^x \) for \( x > 0 \), and \( f(1) = 7 \).
(a) The function \( f \) has a critical point at \( x = 3 \). At this point, does \( f \) have a relative minimum, a relative maximum, or neither? Justify your answer.
(b) On what intervals, if any, is the graph of \( f \) both decreasing and concave up? Explain your reasoning.
(c) Find the value of \( f(3) \).

▶️Answer/Explanation

\(\textbf{5(a)}\)
\( f'(x) < 0 \) for \( 0 < x < 3 \) and \( f'(x) > 0 \) for \( x > 3 \).
Therefore, \( f \) has a relative minimum at \( x = 3 \).

\(\textbf{5(b)}\)
\(
f”(x) = e^x + (x – 3)e^x = (x – 2)e^x
\)
\( f”(x) > 0 \) for \( x > 2 \).
\( f'(x) < 0 \) for \( 0 < x < 3 \).
Therefore, the graph of \( f \) is both decreasing and concave up on the interval \( 2 < x < 3 \).

\(\textbf{5(c)}\)
\(
f(3) = f(1) + \int_{1}^{3} f'(x) \, dx = 7 + \int_{1}^{3} (x – 3)e^x \, dx
\)
Let
\(
u = x – 3, \quad dv = e^x \, dx, \quad du = dx, \quad v = e^x.
\)
\(
f(3) = 7 + (x – 3)e^x \Big|_{1}^{3} – \int_{1}^{3} e^x \, dx
\)
\(
f(3) = 7 + \left((x – 3)e^x – e^x\right) \Big|_{1}^{3}
\)
\(
f(3) = 7 + 3e – e^3.
\)

Question 6

(a) Topic-7.3-Sketching Slope Fields

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

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

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

Consider the logistic differential equation \( \frac{dy}{dt} = \frac{y}{8}(6 – y). \)
Let \( y = f(t) \) be the particular solution to the differential equation with \( f(0) = 8 \).
(a) A slope field for this differential equation is given below. Sketch possible solution curves through the points \( (3, 2) \) and \( (0, 8) \).
(Note: Use the axes provided in the exam booklet).

(b) Use Euler’s method, starting at \( t = 0 \) with two steps of equal size, to approximate \( f(1) \).
(c) Write the second-degree Taylor polynomial for \( f \) about \( t = 0 \), and use it to approximate \( f(1) \).
(d) What is the range of \( f \) for \( t \geq 0 \)?

▶️Answer/Explanation

\(\textbf{6(a)}\)

\(\textbf{6(b)}\)
\( f\left(\frac{1}{2}\right) \approx 8 + (-2)\left(\frac{1}{2}\right) = 7 \)
\( f(1) \approx 7 + \left(-\frac{7}{8}\right)\left(\frac{1}{2}\right) = \frac{105}{16} \)

\(\textbf{6(c)}\)
\( \frac{d^2 y}{dt^2} = \frac{1}{8} \frac{dy}{dt}(6 – y) + \frac{y}{8} \left(\frac{dy}{dt}\right) \)
\( f(0) = 8, \quad f'(0) = \frac{dy}{dt}\bigg|_{t=0} = \frac{8}{8}(6 – 8) = -2; \)
\( f”(0) = \frac{d^2 y}{dt^2}\bigg|_{t=0} = \frac{1}{8}(-2)(6 – 8) + \frac{8}{8}(2) = \frac{5}{2} \)
The second-degree Taylor polynomial for \( f \) about \( t = 0 \) is \( P_2(t) = 8 – 2t + \frac{5}{4}t^2. \)
\( f(1) \approx P_2(1) = \frac{29}{4} \)

\(\textbf{6(d)}\)
The range of \( f \) for \( t \geq 0 \) is \( 6 < y \leq 8 \).

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