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ap13_frq_calculus_bc

Question 1

(a) Topic-2.3 Estimating Derivatives of a Function at a Point

(b) Topic-6.9 Integrating Using Substitution

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

(d) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals

On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by \(G(t)=90+45 \cos\left ( \frac{t^{2}}{18} \right ),\)  where t is measured in hours and  0 ≤ t ≤ 8. At the beginning of the workday (t = 0), the plant has 500 tons of unprocessed gravel. During the hours of operation,0 ≤ t ≤ 8 the plant processes gravel at a constant rate of 100 tons per hour.

(a) Find G’(5). Using correct units, interpret your answer in the context of the problem.

(b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday.

(c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time t = 5 hours? Show the work that leads to your answer.

(d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer.

▶️Answer/Explanation

\(
\textbf{1(a) } G'(5) = -24.588 \text{ (or } -24.587 \text{)}
\)

\(
\text{The rate at which gravel is arriving is decreasing by 24.588 (or 24.587) tons per hour per hour at time } t = 5 \text{ hours.}
\)

\(
\textbf{2(b) } \int_{0}^{8} G(t) \, dt = 825.551 \text{ tons.}
\)

\(
\textbf{1(c) } G(5) = 98.140764 < 100
\)

\(
\text{At time } t = 5, \text{ the rate at which unprocessed gravel is arriving is less than the rate at which it is being processed.}
\)

\(
\text{Therefore, the amount of unprocessed gravel at the plant is decreasing at time } t = 5.
\)

\(
\textbf{1(d)} \text{The amount of unprocessed gravel at time } t \text{ is given by}
\)

\(
A(t) = 500 + \int_{0}^{t} (G(s) – 100) \, ds.
\)

\(
A'(t) = G(t) – 100 = 0 \implies t = 4.923480
\)

\(
\begin{array}{|c|c|}
\hline
t & A(t) \\
\hline
0 & 500 \\
4.92348 & 635.376123 \\
8 & 525.551089 \\
\hline
\end{array}
\)

\(
\text{The maximum amount of unprocessed gravel at the plant during this workday is 635.376 tons.}
\)

Question 2

(a) Topic-9.9-Finding the Area of the Region Bounded by Two Polar Curves

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

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

The graphs of the polar curves $r = 3$ and $r = 4 – 2\sin \theta$ are shown in the figure above. The curves intersect when $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
(a) Let $S$ be the shaded region that is inside the graph of $r = 3$ and also inside the graph of $r = 4 – 2\sin \theta$. Find the area of $S$.
(b) A particle moves along the polar curve $r = 4 – 2\sin \theta$ so that at time $t$ seconds, $\theta = t^2$. Find the time $t$ in the interval $1 \leq t \leq 2$ for which the $x$-coordinate of the particle’s position is $-1$.
(c) For the particle described in part (b), find the position vector in terms of $t$. Find the velocity vector at time $t = 1.5$.

▶️Answer/Explanation

\(\textbf{2(a)}\)
Area of the shaded region:
\(
\text{Area inside the circle} – \left(\text{Area inside } r_1 \text{ and outside } r_2 \text{ from } \frac{\pi}{6} \text{ to } \frac{5\pi}{6}\right)
\)
\(
= \pi(3)^2 – \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left(r_1^2 – r_2^2\right) \, d\theta = 24.709.
\)
\(\text{OR}\)
\(
\left(\frac{2}{3} \text{ of the area inside the circle}\right) + \left(\text{Area inside } r_2 \text{ from } \frac{\pi}{6} \text{ to } \frac{5\pi}{6}\right)
\)
\(
= \frac{2}{3}\left(\pi \cdot 3^2\right) + \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left(r_2^2\right) \, d\theta = 24.709.
\)
\(\text{Note: Make sure you are in the polar mode in your calculator.}\)
\(\textbf{2(b)}\)
If $\theta = t^2$, we want $t$ on $[1,2]$ where $x = -1$:
\(
x = r \cos \theta \implies -1 = (4 – 2 \sin t^2) \cos t^2.
\)
Let $\theta = t^2$ and set $0 = (4 – 2 \sin t^2) \cos t^2 + 1$.
Graph this in function mode on the calculator and find the zeros on $[1,2]$:
\(
t = 1.428.
\)
\(\textbf{2(c)}\)
Position vector:  $\langle x(t), y(t) \rangle$
\(
x = r \cos \theta = (4 – 2 \sin t^2) \cos t^2 \implies x(t) = (4 – 2 \sin t^2) \cos t^2.
\)
\(
y = r \sin \theta = (4 – 2 \sin t^2) \sin t^2 \implies y(t) = (4 – 2 \sin t^2) \sin t^2.
\)
Position vector = $\langle (4 – 2 \sin t^2) \cos t^2, \, (4 – 2 \sin t^2) \sin t^2 \rangle$.
Velocity vector at $t = 1.5$:
\(
\vec{v}(1.5) = \langle x'(1.5), y'(1.5) \rangle = \langle -8.072, -1.673 \rangle.
\)

Question 3

(a) Topic-2.3 Estimating Derivatives of a Function at a Point

(b) Topic-5.1 Using the Mean Value Theorem

(c) Topic-6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

(d) Topic-2.3 Estimating Derivatives of a Function at a Point

t
(minutes)
0123456
C (t)
(ounces)
05.38.811.212.813.814.5

Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, 0 ≤ t ≤ 6, is given by a differentiable function C, where t is measured in minutes. Selected values of C (t), measured in ounces, are given in the table above.
(a) Use the data in the table to approximate C'(3.5) . Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, 2 ≤ t ≤ 4 at which C’ (t)= 2 ? Justify your answer.
(c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of \(\frac{1}{6}\int_{0}^{6}C(t)dt.\) Using correct units, explain the meaning of \(\frac{1}{6}\int_{0}^{6}C(t)dt\) in the context of the problem.
(d) The amount of coffee in the cup, in ounces, is modeled by \(B(t)= 16-16e^{-0.4t}\). Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5.

▶️Answer/Explanation

\(
\textbf{3(a) } C'(3.5) \approx \frac{C(4) – C(3)}{4 – 3} = \frac{12.8 – 11.2}{1} = 1.6 \, \text{ounces/min}
\)

\(
\textbf{3(b) } C \text{ is differentiable } \implies C \text{ is continuous (on the closed interval)}
\)
\(
\frac{C(4) – C(2)}{4 – 2} = \frac{12.8 – 8.8}{2} = 2
\)
\(
\text{Therefore, by the Mean Value Theorem, there is at least one time } t, \, 2 < t < 4, \text{ for which } C'(t) = 2.
\)

\(
\textbf{3(c) } \frac{1}{6} \int_{0}^{6} C(t) \, dt \approx \frac{1}{6} [2 \cdot C(1) + 2 \cdot C(3) + 2 \cdot C(5)]
\)
\(
= \frac{1}{6} [2 \cdot 5.3 + 2 \cdot 11.2 + 2 \cdot 13.8]
\)
\(
= \frac{1}{6} (60.6) = 10.1 \, \text{ounces}
\)
\(
\frac{1}{6} \int_{0}^{6} C(t) \, dt \text{ is the average amount of coffee in the cup, in ounces, over the time interval } 0 \leq t \leq 6 \, \text{minutes.}
\)

\(
\textbf{3(d) } B'(t) = -16(-0.4)e^{-0.4t} = 6.4e^{-0.4t}
\)
\(
B'(5) = 6.4e^{-0.4(5)} = \frac{6.4}{e^{2}} \, \text{ounces/min}
\)

Question 4

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

(b) Topic-8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

(c) Topic-5.6 Determining Concavity of Functions over Their Domains

(d) Topic-2.5 Applying the Power Rule

The figure above shows the graph of f’, the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤ 8.  The graph of f’ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of f’ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f (8) = 4.
(a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer.
(b) Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤ 8.  Justify your answer.
(c) On what open intervals contained in 0 < x < 8  is the graph of f both concave down and increasing? Explain your reasoning.
(d) The function g is defined by \(g(x)=(f(x))^{3}\), if \(f(3)= -\frac{5}{2},\) find the slope of the line tangent to the graph of g at x = 3.

▶️Answer/Explanation

\(
\textbf{4(a) } x = 6 \text{ is the only critical point at which } f’ \text{ changes sign from negative to positive. Therefore, } f \text{ has a local minimum at } x = 6.
\)

\(
\textbf{4(b)} \text{From part (a), the absolute minimum occurs either at } x = 6 \text{ or at an endpoint.}
\)
\(
f(0) = f(8) + \int_{8}^{0} f'(x) \, dx
\)
\(
= f(8) – \int_{0}^{8} f'(x) \, dx = 4 – 12 = -8
\)
\(
f(6) = f(8) + \int_{8}^{6} f'(x) \, dx
\)
\(
= f(8) – \int_{6}^{8} f'(x) \, dx = 4 – 7 = -3
\)
\(
f(8) = 4
\)
\(
\text{The absolute minimum value of } f \text{ on the closed interval } [0, 8] \text{ is } -8.
\)

\(
\textbf{4(c)} \text{The graph of } f \text{ is concave down and increasing on } 0 < x < 1 \text{ and } 3 < x < 4, \text{ because } f’ \text{ is decreasing and positive on these intervals.}
\)

\(
\textbf{4(d) } g'(x) = 3[f(x)]^{2} \cdot f'(x)
\)
\(
g'(3) = 3[f(3)]^{2} \cdot f'(3) = 3 \left( -\frac{5}{2} \right)^{2} \cdot 4 = 75
\)

Question 5

(a) Topic-4.7-Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

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

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

Consider the differential equation
\(
\frac{dy}{dx} = y^2(2x + 2).
\)
Let $y = f(x)$ be the particular solution to the differential equation with initial condition $f(0) = -1$.

(a) Find
\(
\lim_{x \to 0} \frac{f(x) + 1}{\sin x}.
\)
Show the work that leads to your answer.

(b) Use Euler’s method, starting at $x = 0$ with two steps of equal size, to approximate $f\left(\frac{1}{2}\right)$.

(c) Find $y = f(x)$, the particular solution to the differential equation with initial condition $f(0) = -1$.

▶️Answer/Explanation

\(\textbf{5(a)}\)
\(
\lim_{x \to 0} \frac{f(x) + 1}{\sin x} = \lim_{x \to 0} \frac{-1 + 1}{0} = \frac{0}{0} \quad \text{so we can use L’Hopital’s Rule.}
\)
\(
\lim_{x \to 0} \frac{f(x) + 1}{\sin x} = \lim_{x \to 0} \frac{f'(x)}{\cos x} = \frac{f'(0)}{\cos 0} = \frac{(-1)^2(2(0) + 2)}{1} = 2.
\)
\(\textbf{5(b)}\)
Euler’s method with 2 steps $\left(\Delta x = \frac{1}{4}\right)$:
\(
f(0) = -1
\)
\(
f\left(\frac{1}{4}\right) \approx f(0) + f'(0) \left(\frac{1}{4}\right) = -1 + (-1)^2(2(0) + 2)\left(\frac{1}{4}\right) = -1 + \frac{1}{4} = -\frac{1}{2}.
\)
\(
f\left(\frac{1}{2}\right) \approx f\left(\frac{1}{4}\right) + f’\left(\frac{1}{4}\right) \left(\frac{1}{4}\right).
\)
\(
f’\left(\frac{1}{4}\right) = (-f(1/4))^2(2(1/4) + 2) = \left(-\frac{1}{2}\right)^2\left(2\left(\frac{1}{4}\right) + 2\right) = \frac{1}{4}(2.5) = \frac{5}{16}.
\)
\(
f\left(\frac{1}{2}\right) \approx -\frac{1}{2} + \frac{5}{16}\left(\frac{1}{4}\right) = -\frac{1}{2} + \frac{5}{64} = -\frac{32}{64} + \frac{5}{64} = -\frac{11}{32}.
\)
\(\textbf{5(c)}\)
\(
\frac{dy}{dx} = y^2(2x + 2) \quad \implies \quad \frac{dy}{y^2} = (2x + 2)dx.
\)
\(
\int \frac{dy}{y^2} = \int (2x + 2) dx
\)
\(
-\frac{1}{y} = x^2 + 2x + C \quad \text{Since } f(0) = -1 \implies -\frac{1}{-1} = 0 + 0 + C \implies C = 1.
\)
\(
-\frac{1}{y} = x^2 + 2x + 1.
\)
\(
y = -\frac{1}{x^2 + 2x + 1} \quad \text{OR} \quad y = \frac{1}{-x^2 – 2x – 1}.
\)
For $x > -1$ \(\quad\) (While important, points are usually not deducted if it is not stated.)

Question 6

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

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

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

A function $f$ has derivatives of all orders at $x = 0$. Let $P_n(x)$ denote the $n$th-degree Taylor polynomial for $f$ about $x = 0$.

(a) It is known that $f(0) = -4$ and that $P_1\left(\frac{1}{2}\right) = -3$. Show that $f'(0) = 2$.

(b) It is known that $f”(0) = -\frac{2}{3}$ and $f”'(0) = \frac{1}{3}$. Find $P_3(x)$.

(c) The function $h$ has first derivative given by $h'(x) = f(2x)$. It is known that $h(0) = 7$. Find the third-degree Taylor polynomial for $h$ about $x = 0$.

▶️Answer/Explanation

$P_n(x)$ is the $n$th-degree Taylor polynomial for $f$ about $x = 0$ (Maclaurin polynomial).

\(\textbf{6(a)}\)
\(
f(0) = -4 \quad \text{and} \quad P_1\left(\frac{1}{2}\right) = -3.
\)
\(
P_1(x) = f(0) + f'(0)x
\)
\(
P_1\left(\frac{1}{2}\right) = -4 + f'(0)\left(\frac{1}{2}\right) = -3 \implies f'(0) = \frac{-3 + 4}{1/2} = 2.
\)
\(\textbf{6(b)}\)
\(
f^{”}(0) = -\frac{2}{3} \quad \text{and} \quad f^{”’}(0) = \frac{1}{3}.
\)
\(
P_3(x) = f(0) + f'(0)x + \frac{f^{”}(0)}{2!}x^2 + \frac{f^{”’}(0)}{3!}x^3
\)
\(
= -4 + 2x + \frac{-2/3}{2!}x^2 + \frac{1/3}{3!}x^3 = -4 + 2x – \frac{1}{3}x^2 + \frac{1}{18}x^3.
\)
\(\textbf{6(c)}\)
\(
h'(x) = f(2x) \quad \text{and} \quad h(0) = 7.
\)
METHOD 1:
\(
h”(x) = 2f'(2x) \quad \text{and} \quad h^{”’}(x) = 4f^{”}(2x).
\)
The third-degree Taylor polynomial for $h$ about $x = 0$ is:
\(
h(0) + h'(0)x + \frac{h^{”}(0)}{2!}x^2 + \frac{h^{”’}(0)}{3!}x^3
\)
\(
= 7 + f(2(0))x + \frac{2f'(2(0))}{2!}x^2 + \frac{4f^{”}(2(0))}{3!}x^3.
\)
\(
= 7 + (-4)x + \frac{2(2)}{2!}x^2 + \frac{4(-2/3)}{3!}x^3.
\)
\(
= 7 – 4x + 2x^2 – \frac{4}{9}x^3.
\)
METHOD 2:
\(
h'(x) = f(2x) \approx P_3(2x) = -4 + 2(2x) + \frac{-2/3}{2!}(2x)^2 + \frac{1/3}{3!}(2x)^3
\)
\(
= -4 + 4x – \frac{4}{3}x^2 + \frac{4}{9}x^3.
\)
\(
h(x) \approx \int \left(-4 + 4x – \frac{4}{3}x^2 + \frac{4}{9}x^3\right)dx + C.
\)
\(
= -4x + 2x^2 – \frac{4}{9}x^3 + \frac{1}{9}x^4 + C.
\)
Since $h(0) = 7$,
\(
7 = 0 + 0 + 0 + C \implies C = 7.
\)
\(
h(x) = 7 – 4x + 2x^2 – \frac{4}{9}x^3.
\)

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