Question 1
(a) Topic-9.1 Defining and Differentiating Parametric Equations
(b) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
(c) Topic-9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
A particle moves in the $xy$-plane so that its position at any time $t$, $0 \leq t \leq \pi$, is given by:
$x(t) = \frac{t^2}{2} – \ln(1 + t)$ and $y(t) = 3 \sin t$.
(a) Sketch the path of the particle in the $xy$-plane below. Indicate the direction of motion along the path.
(b) At what time $t$, $0 \leq t \leq \pi$, does $x(t)$ attain its minimum value? What is the position $(x(t), y(t))$ of the particle at this time?
(c) At what time $t$, $0 < t < \pi$, is the particle on the $y$-axis? Find the speed and the acceleration vector of the particle at this time.
▶️Answer/Explanation
\(\textbf{1(a)}\)
\(\textbf{1(b)}\)
$x'(t) = t – \frac{1}{1 + t} = 0$
$t^2 + t – 1 = 0$
$t = \frac{-1 \pm \sqrt{5}}{2}$ or $t = 0.618$ in $[0, \pi]$
$x(0.618) = -0.290$
$y(0.618) = 1.738$
\(\textbf{1(c)}\)
$x(t) = \frac{t^2}{2} – \ln(1 + t) = 0$
$t = 1.285$ or $1.286$
$x'(t) = t – \frac{1}{1 + t}$
$y'(t) = 3 \cos t$
$\text{speed} = \sqrt{\left( x'(1.286) \right)^2 + \left( y'(1.286) \right)^2} = 1.96$
$x”(t) = 1 + \frac{1}{(1 + t)^2}$
$y”(t) = -3 \sin t$
$\text{acceleration vector} = \langle x”(1.286), y”(1.286) \rangle = \langle 1.91, -2.879 \rangle$
Question 2
(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-8.10 Volume with Disc Method Revolving Around Other Axes
The shaded region, $R$, is bounded by the graph of $y = x^2$ and the line $y = 4$, as shown in the figure above.
(a) Find the area of $R$.
(b) Find the volume of the solid generated by revolving $R$ about the $x$-axis.
(c) There exists a number $k$, $k > 4$, such that when $R$ is revolved about the line $y = k$, the resulting solid has the same volume as the solid in part (b). Write, but do not solve, an equation involving an integral expression that can be used to find the value of $k$.
▶️Answer/Explanation
\(\textbf{2(a)}\)
$\text{Area} = \int_{-2}^2 (4 – x^2) , dx$
$= 2 \int_0^2 (4 – x^2) , dx$
$= 2 \left[ 4x – \frac{x^3}{3} \right]_0^2$
$= 2 \left[ \frac{32}{3} \right]$
$= \frac{32}{3} \approx 10.666 , \text{or} , 10.667$
\(\textbf{2(b)}\)
$\text{Volume} = \pi \int_{-2}^2 \left( 4^2 – (x^2)^2 \right) , dx$
$= 2\pi \int_0^2 (16 – x^4) , dx$
$= 2\pi \left[ 16x – \frac{x^5}{5} \right]_0^2$
$= 2\pi \left[ \frac{256}{5} \right]$
$= \frac{256\pi}{5} \approx 160.849 , \text{or} , 160.850$
\(\textbf{2(c)}\)
$\pi \int_{-2}^2 \left[ (k – x^2)^2 – (k – 4)^2 \right] , dx = \frac{256\pi}{5}$
Question 3
(a) Topic-6.2 Approximating Areas with Riemann Sums
(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
The rate at which water flows out of a pipe, in gallons per hour, is given by a differentiable function $R$ of time $t$.
The table above shows the rate as measured every 3 hours for a 24-hour period.
(a) Use a midpoint Riemann sum with 4 subdivisions of equal length to approximate $\int_0^{24} R(t) \, dt$.
Using correct units, explain the meaning of your answer in terms of water flow.
(b) Is there some time $t$, $0 < t < 24$, such that $R'(t) = 0$? Justify your answer.
(c) The rate of water flow $R(t)$ can be approximated by $Q(t) = \frac{1}{79} (768 + 23t – t^2)$. Use $Q(t)$ to approximate the average rate of water flow during the 24-hour time period. Indicate units of measure.
▶️Answer/Explanation
\(\textbf{3(a)}\)
$\int_0^{24} R(t) , dt \approx 6[R(3) + R(9) + R(15) + R(21)]$
$= 6[10.4 + 11.2 + 11.3 + 10.2]$
$= 258.6 , \text{gallons}$
This is an approximation to the total flow in gallons of water from the pipe in the 24-hour period.
\(\textbf{3(b)}\)
Yes;
Since $R(0) = R(24) = 9.6$, the Mean Value Theorem guarantees that there is a $t$, $0 < t < 24$, such that $R'(t) = 0$.
\(\textbf{3(c)}\)
Average rate of flow $\approx$ average value of $Q(t)$
$\frac{1}{24} \int_0^{24} \frac{1}{79} (768 + 23t – t^2) , dt$
$= 10.785 , \text{gal/hr or} , 10.784 , \text{gal/hr}$
Question 4
(a) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
(b) Topic-10.12 Lagrange Error Bound
(c) Topic-10.11 Finding Taylor Polynomial Approximations of Functions
The function $f$ has derivatives of all orders for all real numbers $x$. Assume $f(2) = -3$, $f'(2) = 5$, $f”(2) = 3$, and $f”'(2) = -8$.
(a) Write the third-degree Taylor polynomial for $f$ about $x = 2$ and use it to approximate $f(1.5)$.
(b) The fourth derivative of $f$ satisfies the inequality $|f^{(4)}(x)| \leq 3$ for all $x$ in the closed interval $[1.5, 2]$. Use the Lagrange error bound on the approximation to $f(1.5)$ found in part (a) to explain
why $f(1.5) \neq -5$.
(c) Write the fourth-degree Taylor polynomial, $P(x)$, for $g(x) = f(x^2 + 2)$ about $x = 0$. Use $P$ to explain why $g$ must have a relative minimum at $x = 0$.
▶️Answer/Explanation
\(\textbf{4(a)}\)
$T_3(f, 2)(x) = -3 + 5(x – 2) + \frac{3}{2}(x – 2)^2 – \frac{8}{6}(x – 2)^3$
$f(1.5) \approx T_3(f, 2)(1.5)$
$= -3 + 5(-0.5) + \frac{3}{2}(-0.5)^2 – \frac{4}{3}(-0.5)^3$
$= -4.9583 \approx -4.958$
\(\textbf{4(b)}\)
Lagrange Error Bound $= \frac{3}{4!} |1.5 – 2|^4 = 0.0078125$
$f(1.5) > -4.9583 – 0.0078125 = -4.966 > -5$
Therefore, $f(1.5) \neq -5$.
\(\textbf{4(c)}\)
$P(x) = T_4(g, 0)(x)$
$= T_2(f, 2)(x^2 + 2) = -3 + 5x^2 + \frac{3}{2}x^4$
The coefficient of $x$ in $P(x)$ is $g'(0)$. This coefficient is $0$, so $g'(0) = 0$.
The coefficient of $x^2$ in $P(x)$ is $\frac{g”(0)}{2!}$. This coefficient is $5$, so $g”(0) = 10$, which is greater than $0$.
Therefore, $g$ has a relative minimum at $x = 0$.
Question 5
(a) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
(b) Topic-5.9 Connecting a Function, Its First Derivative, and Its Second Derivative
(c) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
(d) Topic-5.6 Determining Concavity of Functions Over Their Domains
The graph of the function $f$, consisting of three line segments, is given above. Let $g(x) = \int_1^x f(t) , dt$.
(a) Compute $g(4)$ and $g(-2)$.
(b) Find the instantaneous rate of change of $g$, with respect to $x$, at $x = 1$.
(c) Find the absolute minimum value of $g$ on the closed interval $[-2, 4]$. Justify your answer.
(d) The second derivative of $g$ is not defined at $x = 1$ and $x = 2$. How many of these values are $x$-coordinates of points of inflection of the graph of $g$? Justify your answer.
▶️Answer/Explanation
\(\textbf{5(a)}\)
$g(4) = \int_1^4 f(t) , dt = \frac{3}{2} + 1 + \frac{1}{2} – 1 = \frac{5}{2}$
$g(-2) = \int_1^{-2} f(t) , dt = -\frac{1}{2}(12) = -6$
\(\textbf{5(b)}\)
$g'(1) = f(1) = 4$
\(\textbf{5(c)}\)
$g$ is increasing on $[-2, 3]$ and decreasing on $[3, 4]$.
Therefore, $g$ has an absolute minimum at an endpoint of $[-2, 4]$.
Since $g(-2) = -6$ and $g(4) = \frac{5}{2}$, the absolute minimum value is $-6$.
\(\textbf{5(d)}\)
One; $x = 1$
On $(-2, 1)$, $g”(x) = f'(x) > 0$
On $(1, 2)$, $g”(x) = f'(x) < 0$
On $(2, 4)$, $g”(x) = f'(x) < 0$
Therefore $(1, g(1))$ is a point of inflection and $(2, g(2))$ is not.
Question 6
(a) Topic-4.6 Approximating Values of a Function Using Local Linearity and Linearization
(b) Topic-7.5 Approximating Solutions Using Euler’s Method
(c) Topic-6.4 The Fundamental Theorem of Calculus and Accumulation Functions
Let $f$ be the function whose graph goes through the point $(3, 6)$ and whose derivative is given by $f'(x) = \frac{1 + e^x}{x^2}$.
(a) Write an equation of the line tangent to the graph of $f$ at $x = 3$ and use it to approximate $f(3.1)$.
(b) Use Euler’s method, starting at $x = 3$ with a step size of $0.05$, to approximate $f(3.1)$. Use $f”$ to explain why this approximation is less than $f(3.1)$.
(c) Use $\int_3^{3.1} f'(x) , dx$ to evaluate $f(3.1)$.
▶️Answer/Explanation
\(\textbf{1(a)}\)
$f'(3) = \frac{1 + e^3}{9} = 2.342 \text{ or } 2.343$
$y – 6 = \frac{1 + e^3}{9}(x – 3)$
$y = 6 + \frac{1 + e^3}{9}(x – 3)$
$f(3.1) \approx 6 + \frac{1 + e^3}{9}(0.1) = 6.234$
\(\textbf{1(b)}\)
$f(3.05) \approx f(3) + f'(3)(0.05)$
$= 6 + 0.11714 = 6.11714$
$f(3.1) \approx 6.11714 + f'(3.05)(0.05)$
$= 6.11714 + (2.37735)(0.05) = 6.236$
$f”(x) = \frac{x^2 e^x – 2x(1 + e^x)}{x^4} = \frac{(x – 2)e^x – 2}{x^3}$
For $x \geq 3$, $f”(x) > 0$ and the graph of $f$ is concave upward on $(3, 3.1)$. Therefore, the Euler approximation lines at 3 and 3.05 lie below the graph.
\(\textbf{1(c)}\)
$f(3.1) – f(3) = \int_3^{3.1} \frac{1 + e^x}{x^2} , dx$
$f(3.1) = 6 + 0.2378 = 6.237 \text{ or } 6.238$