Question 1
(a) Topic-8.4 Finding the Area Between Curves Expressed as Functions of x
(b) Topic-8.9 Volume with Disc Method Revolving Around the x-Axis
(c) Topic-8.7 Volumes with Cross Sections Squares and Rectangles
Let $R$ be the shaded region in the first quadrant enclosed by the graphs of $y = e^{-x^2}$, $y = 1 – \cos x$, and the $y$-axis, as shown in the figure above.
(a) Find the area of the region $R$.
(b) Find the volume of the solid generated when the region $R$ is revolved about the $x$-axis.
(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.
▶️Answer/Explanation
Region $R$
$e^{-x^2} = 1 – \cos x$ at $x = 0.941944 = A$
\(\textbf{1(a)}\)
Area $= \int_0^A \left( e^{-x^2} – (1 – \cos x) \right) dx$
$= 0.590 \text{ or } 0.591$
\(\textbf{1(b)}\)
Volume $= \pi \int_0^A \left( (e^{-x^2})^2 – (1 – \cos x)^2 \right) dx$
$= 0.55596\pi = 1.746 \text{ or } 1.747$
\(\textbf{1(c)}\)
Volume $= \int_0^A \left( e^{-x^2} – (1 – \cos x)^2 \right) dx$
$= 0.461$
Question 2
(a) Topic-4.1 Interpreting the Meaning of the Derivative in Context
(b) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(c) Topic-6.2 Approximating Areas with Riemann Sums
Two runners, $A$ and $B$, run on a straight racetrack for $0 \leq t \leq 10$ seconds. The graph above, which consists of two line segments, shows the velocity, in meters per second, of Runner $A$. The velocity, in meters per second, of Runner $B$ is given by the function $v$ defined by $v(t) = \frac{24t}{2t + 3}$.
(a) Find the velocity of Runner $A$ and the velocity of Runner $B$ at time $t = 2$ seconds. Indicate units of measure.
(b) Find the acceleration of Runner $A$ and the acceleration of Runner $B$ at time $t = 2$ seconds. Indicate units of measure.
(c) Find the total distance run by Runner $A$ and the total distance run by Runner $B$ over the time interval $0 \leq t \leq 10$ seconds. Indicate units of measure.
▶️Answer/Explanation
\(\textbf{2(a)}\)
Runner A: velocity $= \frac{10}{3} \cdot 2 = \frac{20}{3}$
$= 6.666 \text{ or } 6.667 , \text{meters/sec}$
Runner B: $v(2) = \frac{48}{7} = 6.857 , \text{meters/sec}$
\(\textbf{2(b)}\)
Runner A: acceleration $= \frac{10}{3} = 3.333 , \text{meters/sec}^2$
Runner B: $a(2) = v'(2) = \frac{72}{(2t + 3)^2} \Big|_{t=2}$
$= \frac{72}{49} = 1.469 , \text{meters/sec}^2$
\(\textbf{2(c)}\)
Runner A: distance $= \frac{1}{2} (3)(10) + 7(10) = 85 , \text{meters}$
Runner B: distance $= \int_0^{10} \frac{24t}{2t + 3} , dt = 83.336 , \text{meters}$
Question 3
(a) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
(b) Topic-10.13 Radius and Interval of Convergence of Power Series
(c) Topic-10.12 Lagrange Error Bound
The Taylor series about $x = 5$ for a certain function $f$ converges to $f(x)$ for all $x$ in the interval of convergence. The $n$th derivative of $f$ at $x = 5$ is given by $f^{(n)}(5) = \frac{(-1)^n n!}{2^n (n + 2)}$, and $f(5) = \frac{1}{2}$.
(a) Write the third-degree Taylor polynomial for $f$ about $x = 5$.
(b) Find the radius of convergence of the Taylor series for $f$ about $x = 5$.
(c) Show that the sixth-degree Taylor polynomial for $f$ about $x = 5$ approximates $f(6)$ with error less than $\frac{1}{1000}$.
▶️Answer/Explanation
\(\textbf{3(a)}\)
$f'(5) = \frac{-1!}{2(3)}, , f”(5) = \frac{2!}{4(4)}, , f”'(5) = \frac{-3!}{8(5)}$
$P_3(f, 5)(x) = \frac{1}{2} – \frac{1}{6}(x – 5) + \frac{1}{16}(x – 5)^2 – \frac{1}{40}(x – 5)^3$
\(\textbf{3(b)}\)
$a_n = \frac{f^{(n)}(5)}{n!} = \frac{(-1)^n}{2^n (n + 2)}$
$\lim_{n \to \infty} \left| \frac{(-1)^{n+1} (x – 5)^{n+1}}{2^{n+1} (n + 3)} \Bigg/ \frac{(-1)^n (x – 5)^n}{2^n (n + 2)} \right|$
$= \lim_{n \to \infty} \frac{1}{2} \frac{n + 2}{n + 3} |x – 5|$
$= \frac{|x – 5|}{2} < 1$
The radius of convergence is 2.
\(\textbf{3(c)}\)
The Taylor series about $x = 5$ for the function $f$, when evaluated at $x = 6$, is an alternating series with absolute value of terms decreasing to 0.
The error in approximating $f(6)$ with the 6th degree Taylor polynomial at $x = 6$ is less than the first omitted term in the series:
$\left| f(6) – P_6(f, 5)(6) \right| \leq \frac{1}{2^7 (9)} = \frac{1}{1152} < \frac{1}{1000}$
Question 4
(a) Topic-9.4 Defining and Differentiating Vector-Valued Functions
(b) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
(c) Topic-9.1 Defining and Differentiating Parametric Equations
(d) Topic-1.15 Connecting Limits at Infinity and Horizontal Asymptotes
A moving particle has position $(x(t), y(t))$ at time $t$. The position of the particle at time $t = 1$ is $(2, 6)$, and the velocity vector at any time $t > 0$ is given by $\left(1 – \frac{1}{t^2}, 2 + \frac{1}{t^2}\right)$.
(a) Find the acceleration vector at time $t = 3$.
(b) Find the position of the particle at time $t = 3$.
(c) For what time $t > 0$ does the line tangent to the path of the particle at $(x(t), y(t))$ have a slope of $8$?
(d) The particle approaches a line as $t \to \infty$. Find the slope of this line. Show the work that leads to your conclusion.
▶️Answer/Explanation
\(\textbf{4(a)}\)
Acceleration vector $= \left( x”(t), y”(t) \right) = \left( \frac{2}{t^3}, -\frac{2}{t^3} \right)$
$\left( x”(3), y”(3) \right) = \left( \frac{2}{27}, -\frac{2}{27} \right)$
\(\textbf{4(b)}\)
$(x(t), y(t)) = \left( t + \frac{1}{t} + C_1, 2t – \frac{1}{t} + C_2 \right)$
$(2, 6) = (x(1), y(1)) = (2 + C_1, 1 + C_2)$
$C_1 = 0, , C_2 = 5$
$x(3), y(3)) = \left( 3 + \frac{1}{3}, 6 – \frac{1}{3} + 5 \right) = \left( \frac{10}{3}, \frac{32}{3} \right)$
\(\textbf{4(c)}\)
$\frac{dy}{dx} = \frac{2 + \frac{1}{t^2}}{1 – \frac{1}{t^2}} = 8$
$2 + \frac{1}{t^2} = 8 \left( 1 – \frac{1}{t^2} \right); , t^2 = \frac{9}{6}$
$t = \frac{\sqrt{3}}{2}$
\(\textbf{4(d)}\)
$\lim_{t \to \infty} \frac{dy}{dx} = \lim_{t \to \infty} \frac{2 + \frac{1}{t^2}}{1 – \frac{1}{t^2}} = 2$
—or—
Since $x(t) \to \infty$ as $t \to \infty$, the slope of the line is:
$\lim_{t \to \infty} \frac{y(t)}{x(t)} = \lim_{t \to \infty} \frac{2t – \frac{1}{t} + 5}{t + \frac{1}{t}} = 2$
Question 5
(a) Topic-3.2 Implicit Differentiation
(b) Topic-3.1 The Chain Rule
(c) Topic-5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
Consider the curve given by $xy^2 – x^3y = 6$.
(a) Show that $\frac{dy}{dx} = \frac{3x^2y – y^2}{2xy – x^3}$.
(b) Find all points on the curve whose $x$-coordinate is 1, and write an equation for the tangent line at each of these points.
(c) Find the $x$-coordinate of each point on the curve where the tangent line is vertical.
▶️Answer/Explanation
\(\textbf{5(a)}\)
$y^2 + 2xy \frac{dy}{dx} – 3x^2 y – x^3 \frac{dy}{dx} = 0$
$\frac{dy}{dx} \left( 2xy – x^3 \right) = 3x^2 y – y^2$
$\frac{dy}{dx} = \frac{3x^2 y – y^2}{2xy – x^3}$
\(\textbf{5(b)}\)
When $x = 1$, $y^2 – y = 6$
$y^2 – y – 6 = 0$
$(y – 3)(y + 2) = 0$
$y = 3, , y = -2$
At $(1, 3)$, $\frac{dy}{dx} = \frac{9 – 9}{6 – 1} = 0$
Tangent line equation is $y = 3$
At $(1, -2)$, $\frac{dy}{dx} = \frac{-6 – 4}{-4 – 1} = \frac{-10}{-5} = 2$
Tangent line equation is $y + 2 = 2(x – 1)$
\(\textbf{5(c)}\)
Tangent line is vertical when $2xy – x^3 = 0$
$x \left( 2y – x^2 \right) = 0$ gives $x = 0$ or $y = \frac{1}{2}x^2$
There is no point on the curve with $x$-coordinate $0$.
When $y = \frac{1}{2}x^2$, $\frac{1}{4}x^5 – \frac{1}{2}x^5 = 6$
$-\frac{1}{4}x^5 = 6$
$x = \sqrt[5]{-24}$
Question 6
(a) Topic-7.3 Sketching Slope Fields
(b) Topic-7.3 Sketching Slope Fields
(c) Topic-7.6 Finding General Solutions Using Separation of Variables
(d) Topic-7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables
Consider the differential equation given by $\frac{dy}{dx} = x(y – 1)^2$.
(a) On the axes provided, sketch a slope field for the given differential equation at the eleven points indicated.
(b) Use the slope field for the given differential equation to explain why a solution could not have the graph shown below.
(c) Find the particular solution $y = f(x)$ to the given differential equation with the initial condition $f(0) = -1$.
(d) Find the range of the solution found in part (c).
▶️Answer/Explanation
\(\textbf{6(a)}\)
\(\textbf{6(b)}\)
The graph does not have slope $0$ where $y = 1$.
-or –
The slope field shown suggests that solutions are asymptotic to $y = 1$ from below, but the graph does not exhibit this behavior.
\(\textbf{6(c)}\)
$\frac{1}{(y – 1)^2} dy = x , dx$
$-\frac{1}{y – 1} = \frac{1}{2} x^2 + C$
$\frac{1}{2} = 0 + C; , C = \frac{1}{2}$
$-\frac{1}{y – 1} = \frac{1}{2}(x^2 + 1); , y = 1 – \frac{2}{x^2 + 1}$
\(\textbf{6(d)}\)
Range is $-1 \leq y < 1$.