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Question 1

(a) Topic-9.1 Defining and Differentiating Parametric Equations

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

(c) Topic-9.3 Finding Arc Lengths of Curves Given by Parametric Equations

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

An object moving along a curve in the $xy$-plane has position $(x(t), y(t))$ at time $t$ with
$\frac{dx}{dt} = \cos(t^3)$ and $\frac{dy}{dt} = 3 \sin(t^2)$ for $0 \leq t \leq 3$. At time $t = 2$, the object is at position $(4, 5)$.

(a) Write an equation for the line tangent to the curve at $(4, 5)$.
(b) Find the speed of the object at time $t = 2$.
(c) Find the total distance traveled by the object over the time interval $0 \leq t \leq 1$.
(d) Find the position of the object at time $t = 3$.

▶️Answer/Explanation

\(\textbf{1(a)}\)
$\frac{dy}{dx} = \frac{3 \sin(t^2)}{\cos(t^3)}$
$\left.\frac{dy}{dx}\right|_{t=2} = \frac{3 \sin(2^2)}{\cos(2^3)} = 15.604$
$y – 5 = 15.604(x – 4)$

\(\textbf{1(b)}\)
$\text{Speed} = \sqrt{\cos^2(8) + 9 \sin^2(4)} = 2.275$

\(\textbf{1(c)}\)
$\text{Distance} = \int_0^1 \sqrt{\cos^2(t^3) + 9 \sin^2(t^2)} , dt = 1.458$

\(\textbf{1(d)}\)
$x(3) = 4 + \int_2^3 \cos(t^3) , dt = 3.953 , \text{or} , 3.954$
$y(3) = 5 + \int_2^3 3 \sin(t^2) , dt = 4.906$

Question 2

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

(b) Topic-6.2 Approximating Areas with Riemann Sums

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

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



The temperature, in degrees Celsius (°C), of the water in a pond is a differentiable function $W$ of time $t$. The table above shows the water temperature as recorded every 3 days over a 15-day period.

(a) Use data from the table to find an approximation for $W'(12)$. Show the computations that lead to your answer. Indicate units of measure.
(b) Approximate the average temperature, in degrees Celsius, of the water over the time interval $0 \leq t \leq 15$ days by using a trapezoidal approximation with subintervals of length $\Delta t = 3$ days.
(c) A student proposes the function $P$, given by $P(t) = 20 + 10te^{-t/3}$, as a model for the temperature of the water in the pond at time $t$, where $t$ is measured in days and $P(t)$ is measured in degrees Celsius. Find $P'(12)$. Using appropriate units, explain the meaning of your answer in terms of water temperature.
(d) Use the function $P$ defined in part (c) to find the average value, in degrees Celsius, of $P(t)$ over the time interval $0 \leq t \leq 15$ days.

▶️Answer/Explanation

\(\textbf{2(a)}\)
Difference quotient; e.g.,
$W'(12) \approx \frac{W(15) – W(12)}{15 – 12} = -\frac{1}{3} , \text{°C/day or}$
$W'(12) \approx \frac{W(12) – W(9)}{12 – 9} = -\frac{2}{3} , \text{°C/day or}$
$W'(12) \approx \frac{W(15) – W(9)}{15 – 9} = -\frac{1}{2} , \text{°C/day}$

\(\textbf{2(b)}\)
$\frac{3}{2} \big(20 + 2(31) + 2(28) + 2(24) + 2(22) + 21\big) = 376.5$
Average temperature $\approx \frac{1}{15}(376.5) = 25.1 , \text{°C}$

\(\textbf{2(c)}\)
$P'(12) = 10e^{-t/3} – \frac{10}{3} te^{-t/3} \Big|_{t=12}$
$= -30e^{-4} = -0.549 , \text{°C/day}$
This means that the temperature is decreasing at the rate of $0.549 , \text{°C/day}$ when $t = 12$ days.

\(\textbf{2(d)}\)
$\frac{1}{15} \int_0^{15} \big(20 + 10te^{-t/3}\big) dt = 25.757 , \text{°C}$

Question 3

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

(b) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(d) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration



A car is traveling on a straight road with velocity $55 , \text{ft/sec}$ at time $t = 0$. For $0 \leq t \leq 18$ seconds, the car’s acceleration $a(t)$, in $\text{ft/sec}^2$, is the piecewise linear function defined by the graph above.

(a) Is the velocity of the car increasing at $t = 2$ seconds? Why or why not?
(b) At what time in the interval $0 \leq t \leq 18$, other than $t = 0$, is the velocity of the car $55 , \text{ft/sec}$? Why?
(c) On the time interval $0 \leq t \leq 18$, what is the car’s absolute maximum velocity, in $\text{ft/sec}$, and at what time does it occur? Justify your answer.
(d) At what times in the interval $0 \leq t \leq 18$, if any, is the car’s velocity equal to zero? Justify your answer.

▶️Answer/Explanation

\(\textbf{3(a)}\)
Since $v'(2) = a(2)$ and $a(2) = 15 > 0$, the velocity is increasing at $t = 2$.

\(\textbf{3(b)}\)
At time $t = 12$ because $ v(12) – v(0) = \int_0^{12} a(t) \, dt = 0$.

\(\textbf{3(c)}\)
The absolute maximum velocity is $115 \, \text{ft/sec}$ at $t = 6$.
The absolute maximum must occur at $t = 6$ or at an endpoint.
$ v(6) = 55 + \int_0^6 a(t) \, dt $
$ = 55 + 2(15) + \frac{1}{2}(4)(15) = 115 > v(0)$.
$\int_6^{18} a(t) \, dt < 0 \text{ so } v(18) < v(6)$.

\(\textbf{3(d)}\)
The car’s velocity is never equal to $0$. The absolute minimum occurs at $t = 16$ where
$ v(16) = 115 + \int_6^{16} a(t) \, dt = 115 – 105 = 10 > 0$.

Question 4

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

(b) Topic-5.6 Determining Concavity of Functions Over Their Domains

(c) Topic-4.6 Approximating Values of a Function Using Local Linearity and Linearization

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

Let $h$ be a function defined for all $x \neq 0$ such that $h(4) = -3$ and the derivative of $h$ is given by \[ h'(x) = \frac{x^2 – 2}{x} \quad \text{for all } x \neq 0. \]
(a) Find all values of $x$ for which the graph of $h$ has a horizontal tangent, and
determine whether $h$ has a local maximum, a local minimum, or neither at each of these values. Justify your answers.
(b) On what intervals, if any, is the graph of $h$ concave up? Justify your answer.
(c) Write an equation for the line tangent to the graph of $h$ at $x = 4$.
(d) Does the line tangent to the graph of $h$ at $x = 4$ lie above or below the graph of $h$ for $x > 4$? Why?

▶️Answer/Explanation

\(\textbf{4(a)}\)
$h'(x) = 0$ at $x = \pm \sqrt{2}$

Local minima at $x = -\sqrt{2}$ and $x = \sqrt{2}$.

\(\textbf{4(b)}\)
$h”(x) = 1 + \frac{2}{x^2} > 0$ for all $x \neq 0$.
Therefore, the graph of $h$ is concave up for all $x \neq 0$.

\(\textbf{4(c)}\)
$h'(4) = \frac{16 – 2}{4} = \frac{7}{2}$
$y + 3 = \frac{7}{2}(x – 4)$

\(\textbf{4(d)}\)
The tangent line is below the graph because the graph of $h$ is concave up for $x > 4$.

Question 5

(a) Topic-6.13 Evaluating Improper Integrals (BC Only)

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

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

Let $f$ be the function satisfying $f'(x) = -3x f(x)$, for all real numbers $x$, with $f(1) = 4$ and $\lim_{x \to \infty} f(x) = 0$.
(a) Evaluate $\int_{1}^{\infty} -3x f(x) \, dx$. Show the work that leads to your answer.
(b) Use Euler’s method, starting at $x = 1$ with a step size of $0.5$, to approximate $f(2)$.
(c) Write an expression for $y = f(x)$ by solving the differential equation $\frac{dy}{dx} = -3xy$ with the initial condition $f(1) = 4$.

▶️Answer/Explanation

\(\textbf{5(a)}\)
$\int_{1}^{\infty} -3x f(x) \, dx = \int_{1}^{\infty} f'(x) \, dx $
$= \lim_{b \to \infty} \int_{1}^{b} f'(x) \, dx$
$= \lim_{b \to \infty} \big[ f(x) \big]_{1}^{b}$,
which simplifies to $\lim_{b \to \infty} f(b) – f(1) = 0 – 4 = -4$.

\(\textbf{5(b)}\)
$f(1.5) \approx f(1) + f'(1)(0.5)$
$= 4 – 3(1)(4)(0.5)$
$= -2$.
Then, $f(2) \approx f(1.5) + f'(1.5)(0.5) \approx -2 – 3(1.5)(-2)(0.5) = 2.5$.

\(\textbf{5(c)}\)
Starting from $\frac{1}{y} \, dy = -3x \, dx$,
integrating gives $\ln y = -\frac{3}{2}x^2 + k$.
Thus, $y = Ce^{-\frac{3}{2}x^2}$.
Using $f(1) = 4$, $4 = Ce^{-\frac{3}{2}(1)^2}$, so $C = 4e^{\frac{3}{2}}$.
The final solution is $y = 4e^{\frac{3}{2}}e^{-\frac{3}{2}x^2}$.

Question 6

(a) Topic-10.13 Radius and Interval of Convergence of Power Series

(b) Topic-1.15 Connecting Limits at Infinity and Horizontal Asymptotes

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

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

A function $f$ is defined by

$f(x) = \frac{1}{3} + \frac{2}{3^2}x + \frac{3}{3^3}x^2 + \cdots + \frac{n+1}{3^{n+1}}x^n + \cdots$

for all $x$ in the interval of convergence of the given power series.

(a) Find the interval of convergence for this power series. Show the work that leads to your answer.
(b) Find $\lim_{x \to 0} \frac{f(x) – \frac{1}{3}}{x}$.
(c) Write the first three nonzero terms and the general term for an infinite series that represents $\int_{0}^{1} f(x) , dx$.
(d) Find the sum of the series determined in part (c).

▶️Answer/Explanation

\(\textbf{6(a)}\)
$\lim_{n \to \infty} \left| \frac{(n+2)x^{n+1}}{3^{n+2}} \cdot \frac{3^{n+1}}{(n+1)x^n} \right| = \lim_{n \to \infty} \left| \frac{(n+2)x}{(n+1)3} \right| = \left| \frac{x}{3} \right| < 1$
At $x = -3$, the series is $\sum_{n=0}^\infty (-1)^n \frac{n+1}{3}$, which diverges.
At $x = 3$, the series is $\sum_{n=0}^\infty \frac{n+1}{3}$, which diverges.
Therefore, the interval of convergence is $-3 < x < 3$.

\(\textbf{6(b)}\)
$f(x) – \frac{1}{3} = \lim_{x \to 0} \frac{2}{3^2}x + \frac{3}{3^3}x^2 + \frac{4}{3^4}x^3 + \cdots$
$\frac{2}{9}$

\(\textbf{6(c)}\)
$\int_0^1 f(x) , dx = \int_0^1 \left( \frac{1}{3} + \frac{2}{3^2}x + \frac{3}{3^3}x^2 + \cdots \right) , dx$
$= \frac{1}{3}x + \frac{1}{3^2}x^2 + \frac{1}{3^3}x^3 + \cdots + \frac{1}{3^{n+1}}x^{n+1} \Big|_0^1$
$= \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots$

\(\textbf{6(d)}\)
The series representing $\int_0^1 f(x) , dx$ is a geometric series.
$\int_0^1 f(x) , dx = \frac{\frac{1}{3}}{1 – \frac{1}{3}} = \frac{1}{2}$

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