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Question 1

(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-8.1-Finding the Average Value of a Function on an Interval

(d) Topic-2.1-Defining Average and Instantaneous Rates of Change at a Point

Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute.
The traffic flow at a particular intersection is modeled by the function $F$ defined by
\(
F(t) = 82 + 4 \sin\left(\frac{t}{2}\right) \quad \text{for } 0 \leq t \leq 30,
\)
where $F(t)$ is measured in cars per minute and $t$ is measured in minutes.
(a) To the nearest whole number, how many cars pass through the intersection over the 30-minute period?
(b) Is the traffic flow increasing or decreasing at $t = 7$? Give a reason for your answer.
(c) What is the average value of the traffic flow over the time interval $10 \leq t \leq 15$? Indicate units of measure.
(d) What is the average rate of change of the traffic flow over the time interval $10 \leq t \leq 15$? Indicate units of measure.

▶️Answer/Explanation

\(\textbf{1(a)}\)
$\int_0^{30} F(t) \, dt = 2474 \, \text{cars}$

\(\textbf{1(b)}\)
$F'(7) = -1.872 \text{ or } -1.873$
Since $F'(7) < 0$, the traffic flow is decreasing at $t = 7$.

\(\textbf{1(c)}\)
$\frac{1}{5} \int_{10}^{15} F(t) \, dt = 81.899 \, \text{cars/min}$

\(\textbf{1(d)}\)
$\frac{F(15) – F(10)}{15 – 10} = 1.517 \text{ or } 1.518 \, \text{cars/min}^2$
Units of $\text{cars/min}$ in (c) and $\text{cars/min}^2$ in (d).

Question 2

(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 $f$ and $g$ be the functions given by $f(x) = 2x(1-x)$ and $g(x) = 3(x-1)\sqrt{x}$ for $0 \leq x \leq 1$. The graphs of $f$ and $g$ are shown in the figure above.

(a) Find the area of the shaded region enclosed by the graphs of $f$ and $g$.

(b) Find the volume of the solid generated when the shaded region enclosed by the graphs of $f$ and $g$ is revolved about the horizontal line $y = 2$.

(c) Let $h$ be the function given by $h(x) = kx(1-x)$ for $0 \leq x \leq 1$.
For each $k > 0$, the region (not shown) enclosed by the graphs of $h$ and $g$ is the base of a solid with square cross sections perpendicular to the $x$-axis.
There is a value of $k$ for which the volume of this solid is equal to 15.
Write, but do not solve, an equation involving an integral expression that could be used to find the value of $k$.

▶️Answer/Explanation

\(\textbf{2(a)}\)
\( \text{Area} = \int_0^1 \left( f(x) – g(x) \right) \, dx \)
\( = \int_0^1 \left( 2x(1-x) – 3(x-1)\sqrt{x} \right) \, dx = 1.133 \)

\(\textbf{2(b)}\)
\( \text{Volume} = \pi \int_0^1 \left[ (2 – g(x))^2 – (2 – f(x))^2 \right] \, dx \)
\( = \pi \int_0^1 \left[ (2 – 3(x-1)\sqrt{x})^2 – (2 – 2x(1-x))^2 \right] \, dx \)
\( = 16.179 \)

\(\textbf{2(c)}\)
\( \text{Volume} = \int_0^1 \left( h(x) – g(x) \right)^2 \, dx \)
\( \int_0^1 \left( kx(1-x) – 3(x-1)\sqrt{x} \right)^2 \, dx = 15 \)

Question 3

(a) Topic-9.4-Defining and Differentiating Vector-Valued Functions

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

(c) Topic-9.6-Solving Motion Problems Using Parametric and Vector-Valued Functions

(d) Topic-9.4-Defining and Differentiating Vector-Valued Functions

An object moving along a curve in the $xy$-plane has position $(x(t), y(t))$ at time $t \geq 0$ with
\(
\frac{dx}{dt} = 3 + \cos(t^2).
\)
The derivative $\frac{dy}{dt}$ is not explicitly given. At time $t = 2$, the object is at position $(1, 8)$.

(a) Find the $x$-coordinate of the position of the object at time $t = 4$.
(b) At time $t = 2$, the value of $\frac{dy}{dt}$ is $-7$. Write an equation for the line tangent to the curve at the point $(x(2), y(2))$.
(c) Find the speed of the object at time $t = 2$.
(d) For $t \geq 3$, the line tangent to the curve at $(x(t), y(t))$ has a slope of $2t + 1$. Find the acceleration vector of the object at time $t = 4$.

▶️Answer/Explanation

\(\textbf{3(a)}\)
\( x(4) = x(2) + \int_2^4 \left(3 + \cos(t^2)\right) \, dt \)
\( = 1 + \int_2^4 \left(3 + \cos(t^2)\right) \, dt = 7.132 \text{ or } 7.133. \)

\(\textbf{3(b)}\)
\( \left. \frac{dy}{dx} \right|_{t=2} = \frac{\left. \frac{dy}{dt} \right|_{t=2}}{\left. \frac{dx}{dt} \right|_{t=2}} = \frac{-7}{3 + \cos(4)} = -2.983. \)
\( y – 8 = -2.983(x – 1). \)

\(\textbf{3(c)}\)
\( \text{The speed of the object at time } t = 2 \text{ is } \sqrt{\left(x'(2)\right)^2 + \left(y'(2)\right)^2} = 7.382 \text{ or } 7.383. \)

\(\textbf{3(d)}\)
\(
x”(4) = 2.303
\)
\(
y'(t) = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (2t + 1)(3 + \cos(t^2)).
\)
\(
y”(4) = 24.813 \text{ or } 24.814.
\)
\(
\text{The acceleration vector at } t = 4 \text{ is } \langle 2.303, 24.813 \rangle \text{ or } \langle 2.303, 24.814 \rangle.
\)

Question 4

(a) Topic-3.2-Implicit Differentiation

(b) Topic-5.2-Extreme Value Theorem Global Versus Local Extrema and Critical Points

(c) Topic-5.7-Using the Second Derivative Test to Determine Extrema

Consider the curve given by $x^2 + 4y^2 = 7 + 3xy$.

(a) Show that $\frac{dy}{dx} = \frac{3y – 2x}{8y – 3x}$.
(b) Show that there is a point $P$ with $x$-coordinate 3 at which the line tangent to the curve at $P$ is horizontal. Find the $y$-coordinate of $P$.
(c) Find the value of $\frac{d^2y}{dx^2}$ at the point $P$ found in part (b).
Does the curve have a local maximum, a local minimum, or neither at the point $P$? Justify your answer.

▶️Answer/Explanation

\(\textbf{4(a)}\)
\(
2x + 8y y’ = 3y + 3xy’
\)
\(
(8y – 3x)y’ = 3y – 2x
\)
\(
y’ = \frac{3y – 2x}{8y – 3x}
\)

\(\textbf{4(b)}\)
\(
\frac{3y – 2x}{8y – 3x} = 0; \quad 3y – 2x = 0
\)
When $x = 3$, $3y = 6$, so $y = 2$.
\(
3^2 + 4 \cdot 2^2 = 25 \quad \text{and} \quad 7 + 3 \cdot 3 \cdot 2 = 25
\)
Therefore, $P = (3, 2)$ is on the curve and the slope is $0$ at this point.

\(\textbf{4(c)}\)
\(
\frac{d^2y}{dx^2} = \frac{(8y – 3x)(3y’ – 2) – (3y – 2x)(8y’ – 3)}{(8y – 3x)^2}
\)
At $P = (3, 2)$,
\(
\frac{d^2y}{dx^2} = \frac{(16 – 9)(-2) – (16 – 9)(-3)}{(16 – 9)^2} = \frac{-14}{49} = -\frac{2}{7}.
\)
Since $y’ = 0$ and $y” < 0$ at $P$, the curve has a local maximum at $P$.

Question 5

(a) Topic-7.9-Logistic Models with Differential Equations BC Only

(b) Topic-7.9-Logistic Models with Differential Equations BC Only

(c) Topic-7.7-Finding Particular Solutions Using Initial Conditions and Separation of Variables

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

A population is modeled by a function $P$ that satisfies the logistic differential equation
\(
\frac{dP}{dt} = \frac{P}{5} \left(1 – \frac{P}{12}\right).
\)
(a) If $P(0) = 3$, what is $\lim_{t \to \infty} P(t)$?
If $P(0) = 20$, what is $\lim_{t \to \infty} P(t)$?
(b) If $P(0) = 3$, for what value of $P$ is the population growing the fastest?
(c) A different population is modeled by a function $Y$ that satisfies the separable differential equation
\(
\frac{dY}{dt} = \frac{Y}{5} \left(1 – \frac{t}{12}\right).
\)
Find $Y(t)$ if $Y(0) = 3$.
(d) For the function $Y$ found in part (c), what is $\lim_{t \to \infty} Y(t)$?

▶️Answer/Explanation

\(\textbf{5(a)}\)
For this logistic differential equation, the carrying capacity is 12.
If $P(0) = 3$, $\lim_{t \to \infty} P(t) = 12$.
If $P(0) = 20$, $\lim_{t \to \infty} P(t) = 12$.

\(\textbf{5(b)}\)
The population is growing the fastest when $P$ is half the carrying capacity. Therefore, $P$ is growing the fastest when $P = 6$.

\(\textbf{5(c)}\)
\(
\frac{1}{Y} dY = \frac{1}{5} \left(1 – \frac{t}{12}\right) dt = \left(\frac{1}{5} – \frac{t}{60}\right) dt
\)
\(
\ln|Y| = \frac{t}{5} – \frac{t^2}{120} + C
\)
\(
Y(t) = K e^{\frac{t}{5} – \frac{t^2}{120}}
\)
Given $K = 3$,
\(
Y(t) = 3e^{\frac{t}{5} – \frac{t^2}{120}}.
\)

\(\textbf{5(d)}\)
$\lim_{t \to \infty} Y(t) = 0$

Question 6

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

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

(c) Topic-10.12-Lagrange Error Bound

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

Let $f$ be the function given by $f(x) = \sin\left(5x + \frac{\pi}{4}\right)$, and let $P(x)$ be the third-degree Taylor polynomial for $f$ about $x = 0$.
(a) Find $P(x)$.
(b) Find the coefficient of $x^{22}$ in the Taylor series for $f$ about $x = 0$.
(c) Use the Lagrange error bound to show that
\(
\left| f\left(\frac{1}{10}\right) – P\left(\frac{1}{10}\right) \right| < \frac{1}{100}.
\)
(d) Let $G$ be the function given by $G(x) = \int_0^x f(t) \, dt$. Write the third-degree Taylor polynomial for $G$ about $x = 0$.

▶️Answer/Explanation

\(\textbf{6(a)}\)
\(
f(0) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\)
\(
f'(0) = 5 \cos\left(\frac{\pi}{4}\right) = 5 \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}
\)
\(
f”(0) = -25 \sin\left(\frac{\pi}{4}\right) = -25 \cdot \frac{\sqrt{2}}{2} = \frac{-25\sqrt{2}}{2}
\)
\(
f”'(0) = -125 \cos\left(\frac{\pi}{4}\right) = -125 \cdot \frac{\sqrt{2}}{2} = \frac{-125\sqrt{2}}{2}
\)
\(
P(x) = \frac{\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}x – \frac{25\sqrt{2}}{2(2!)}x^2 – \frac{125\sqrt{2}}{2(3!)}x^3
\)

\(\textbf{6(b)}\)
$\frac{-5^{22}\sqrt{2}}{2(22!)}$

\(\textbf{6(c)}\)
\(
\left| f\left(\frac{1}{10}\right) – P\left(\frac{1}{10}\right) \right| \leq \max_{0 \leq c \leq \frac{1}{10}} \left| f^{(4)}(c) \right| \cdot \frac{\left(\frac{1}{10}\right)^4}{4!}
\)
\(
\leq \frac{625}{4!} \left(\frac{1}{10}\right)^4 = \frac{1}{384} < \frac{1}{100}
\)

\(\textbf{6(d)}\)
The third-degree Taylor polynomial for $G$ about $x = 0$ is
\(
\int_0^x \left(\frac{\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}t – \frac{25\sqrt{2}}{4}t^2\right) dt
\)
\(
= \frac{\sqrt{2}}{2}x + \frac{5\sqrt{2}}{4}x^2 – \frac{25\sqrt{2}}{12}x^3.
\)

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