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-8.11 Volume with Washer Method Revolving Around the x-Axis
Let $R$ be the region in the first quadrant bounded by the graph of $y = 8 – x^{\frac{3}{2}}$, the $x$-axis, and the $y$-axis.
(a) Find the area of the region $R$.
(b) Find the volume of the solid generated when $R$ is revolved about the $x$-axis.
(c) The vertical line $x = k$ divides the region $R$ into two regions such that when these two regions are revolved about the $x$-axis, they generate solids with equal volumes. Find the value of $k$.
▶️Answer/Explanation
\(\textbf{1(a)}\)
$A = \int_0^4 \left(8 – x^{3/2}\right) , dx$
$= \left[8x – \frac{2}{5}x^{5/2}\right]_0^4$
$= 32 – \frac{64}{5} – (0 – 0) = 32 – \frac{96}{5}$
$= 19.2$
\(\textbf{1(b)}\)
$V = \pi \int_0^4 \left(8 – x^{3/2}\right)^2 , dx$
$= \frac{576\pi}{5}$
$= 115.2\pi \approx 361.911$
\(\textbf{1(c)}\)
$\pi \int_0^k \left(8 – x^{3/2}\right)^2 , dx = \frac{115.2\pi}{2}$
or
$\pi \int_0^k \left(8 – x^{3/2}\right)^2 , dx = \pi \int_k^4 \left(8 – x^{3/2}\right)^2 , dx$
$\int_0^k \left(8 – x^{3/2}\right)^2 , dx = 57.6$
$\int_0^k \left(64 – 16x^{3/2} + x^3\right) , dx = 57.6$
$\left[64x – \frac{32}{5}x^{5/2} + \frac{x^4}{4}\right]_0^k = 57.6$
$64k – \frac{32}{5}k^{5/2} + \frac{k^4}{4} = 57.6$
$k \approx 0.995$ or $k \approx 0.994$
Question 2
(a) Topic-1.15 Connecting Limits at Infinity and Horizontal Asymptotes
(b) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
(c) Topic-5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
(d) Topic-5.11 Solving Optimization Problems
Let $f$ be the function given by $f(x) = 2xe^{2x}$.
(a) Find $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to \infty} f(x)$.
(b) Find the absolute minimum value of $f$. Justify that your answer is an absolute minimum.
(c) What is the range of $f$?
(d) Consider the family of functions defined by $y = bxe^{bx}$, where $b$ is a nonzero constant. Show that the absolute minimum value of $bxe^{bx}$ is the same for all nonzero values of $b$.
▶️Answer/Explanation
\(\textbf{2(a)}\)
$\lim_{x \to -\infty} 2xe^{2x} = 0$
$\lim_{x \to \infty} 2xe^{2x} = \infty$ or DNE
\(\textbf{2(b)}\)
$f'(x) = 2e^{2x} + 2x \cdot 2 \cdot e^{2x} = 2e^{2x}(1 + 2x) = 0$
If $x = -1/2$, then $f(-1/2) = -1/e$ or approximately $-0.368$ or $-0.367$.
$-1/e$ is an absolute minimum value because:
(i) $f'(x) < 0$ for all $x < -1/2$ and $f'(x) > 0$ for all $x > -1/2$,
or
(ii) $f'(x)$ changes from negative to positive at $x = -1/2$, which is the only critical number.
\(\textbf{2(c)}\)
The range of $f$ is $[-1/e, \infty)$ or $[-0.367, \infty)$ or $[-0.368, \infty)$.
\(\textbf{2(d)}\)
$y’ = be^{bx} + b^2xe^{bx} = be^{bx}(1 + bx) = 0$
If $x = -1/b$, then $y = -1/e$ at $x = -1/b$.
$y$ has an absolute minimum value of $-1/e$ for all nonzero $b$.
Question 3
(a) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
(b) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
(c) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
(d) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
Let $f$ be a function that has derivatives of all orders for all real numbers. Assume $f(0) = 5$, $f'(0) = -3$, $f^{”}(0) = 1$, and $f^{”’}(0) = 4$.
(a) Write the third-degree Taylor polynomial for $f$ about $x = 0$ and use it to approximate $f(0.2)$.
(b) Write the fourth-degree Taylor polynomial for $g$, where $g(x) = f(x^2)$, about $x = 0$.
(c) Write the third-degree Taylor polynomial for $h$, where $h(x) = \int_0^x f(t) , dt$, about $x = 0$.
(d) Let $h$ be defined as in part (c). Given that $f(1) = 3$, either find the exact value of $h(1)$ or explain why it cannot be determined.
▶️Answer/Explanation
\(\textbf{3(a)}\)
$P_3(f)(x) = 5 – 3x + \frac{1}{2}x^2 + \frac{2}{3}x^3$
$f(0.2) \approx P_3(f)(0.2) = 5 – 3(0.2) + \frac{0.04}{2} + \frac{2(0.008)}{3}$
$= 4.425$
\(\textbf{3(b)}\)
$P_4(g)(x) = P_2(f)(x^2) = 5 – 3x^2 + \frac{1}{2}x^4$
\(\textbf{3(c)}\)
$P_3(h)(x) = \int_0^x \left(5 – 3t + \frac{1}{2}t^2\right) dt$
$= \left[5t – 3\frac{t^2}{2} + \frac{1}{6}t^3\right]_0^x$
$= 5x – \frac{3}{2}x^2 + \frac{1}{6}x^3$
\(\textbf{3(d)}\)
$h(1) = \int_0^1 f(t) dt$
This cannot be determined because $f(t)$ is known only for $t = 0$ and $t = 1$.
Question 4
(a) Topic-7.3 Sketching Slope Fields
(b) Topic-7.5 Approximating Solutions Using Euler’s Method
(c) Topic-7.6 Finding General Solutions Using Separation of Variables
Consider the differential equation given by $\frac{dy}{dx} = \frac{xy}{2}$.
(a) On the axes provided below, sketch a slope field for the given differential equation at the nine points indicated.
(b) Let $y = f(x)$ be the particular solution to the given differential equation with the initial condition $f(0) = 3$. Use Euler’s method starting at $x = 0$, with a step size of 0.1, to approximate $f(0.2)$. Show the work that leads to your answer.
(c) Find the particular solution $y = f(x)$ to the given differential equation with the initial condition $f(0) = 3$. Use your solution to find $f(0.2)$.
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(\textbf{4(b)}\)
$f(0.1) \approx f(0) + f'(0)(0.1)$
$= 3 + \frac{1}{2}(0)(3)(0.1) = 3$
$f(0.2) \approx f(0.1) + f'(0.1)(0.1)$
$= 3 + \frac{1}{2}(0.1)(3)(0.1)$
$= 3 + \frac{0.03}{2} = 3.015$
\(\textbf{4(c)}\)
$\frac{dy}{dx} = \frac{xy}{2}$
$\int \frac{dy}{y} = \int \frac{x}{2} dx$
$\ln |y| = \frac{1}{4}x^2 + C_1$
$y = Ce^{x^2/4}$
$3 = Ce^0 \implies C = 3$
$y = 3e^{x^2/4}$
$f(0.2) = 3e^{0.04/4} = 3e^{0.01} \approx 3.030$
Question 5
(a) Topic-8.1 Finding the Average Value of a Function on an Interval
(b) Topic-8.1 Finding the Average Value of a Function on an Interval
(c) Topic-6.1 Exploring Accumulations of Change
(d) Topic-6.1 Exploring Accumulations of Change
The temperature outside a house during a 24-hour period is given by:
$F(t) = 80 – 10 \cos\left(\frac{\pi t}{12}\right), , 0 \leq t \leq 24$,
where $F(t)$ is measured in degrees Fahrenheit and $t$ is measured in hours.
(a) Sketch the graph of $F$ on the grid below.
(b) Find the average temperature, to the nearest degree Fahrenheit, between $t = 6$ and $t = 14$.
(c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of $t$ was the air conditioner cooling the house?
(d) The cost of cooling the house accumulates at the rate of $0.05$ per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?
▶️Answer/Explanation
\(\textbf{5(a)}\)
\(\textbf{5(b)}\)
$\text{Avg.} = \frac{1}{14 – 6} \int_{6}^{14} \left( 80 – 10 \cos\left(\frac{\pi t}{12}\right) \right) dt$
$= \frac{1}{8} (697.2957795)$
$= 87.162 , \text{or} , 87.161$
$\approx 87^\circ , \text{F}$
\(\textbf{5(c)}\)
$\left[ 80 – 10 \cos\left(\frac{\pi t}{12}\right) \right] – 78 \geq 0$
$\left[ 2 – 10 \cos\left(\frac{\pi t}{12}\right) \right] \geq 0$
$5.230 , \text{or} , 5.231 \leq t \leq 18.769 , \text{or} , 18.770$
\(\textbf{5(d)}\)
$C = 0.05 \int_{5.231}^{18.769} \left( \left[ 80 – 10 \cos\left(\frac{\pi t}{12}\right) \right] – 78 \right) dt$
$= 0.05 (101.92741) = 5.096 \approx 5.10$
Question 6
(a) Topic-7.6 Finding General Solutions Using Separation of Variables
(b) Topic-9.4 Defining and Differentiating Vector-Valued Functions
(c) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
A particle moves along the curve defined by the equation $y = x^3 – 3x$. The $x$-coordinate of the particle, $x(t)$, satisfies the equation $\frac{dx}{dt} = \frac{1}{\sqrt{2t + 1}}$, for $t \geq 0$ with initial condition $x(0) = -4$.
(a) Find $x(t)$ in terms of $t$.
(b) Find $\frac{dy}{dt}$ in terms of $t$.
(c) Find the location and speed of the particle at time $t = 4$.
▶️Answer/Explanation
\(\textbf{6(a)}\)
$x(t) = \int \frac{1}{\sqrt{2t + 1}} , dt$
$x(t) = \sqrt{2t + 1} + C$
$x(0) = -4 = 1 + C \implies C = -5$
$x(t) = \sqrt{2t + 1} – 5$
\(\textbf{6(b)}\)
$y = x^3 – 3x$
$\frac{dy}{dt} = 3x^2 \frac{dx}{dt} – 3 \frac{dx}{dt}$
$= \left( 3x^2 – 3 \right) \frac{dx}{dt}$
$= \left[ 3 \left( \sqrt{2t + 1} – 5 \right)^2 – 3 \right] \left[ \frac{1}{\sqrt{2t + 1}} \right]$
$x(4) = \sqrt{9} – 5 = -2$
$y(4) = (-2)^3 – 3(-2) = -2$
$\text{Location at } t = 4 \text{ is } (-2, -2)$
\(\textbf{6(c)}\)
$\frac{dx}{dt} \bigg|{t=4} = \frac{1}{3}$
$\frac{dy}{dt} \bigg|{t=4} = \frac{3 \left( 3 – 5 \right)^2 – 3}{3} = 3$
$\text{Speed} = \sqrt{\left( \frac{1}{3} \right)^2 + 3^2} = \sqrt{\frac{82}{9}} = 3.018$