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ap13_frq_calculus_ab

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-1.1 Introducing Calculus: Can Change Occur at an Instant?

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

(c) Topic-6.1 Exploring Accumulations of Change

(d) Topic-2.1 Defining Average and Instantaneous Rates of Change at a Point

A particle moves along a straight line. For 0 ≤ t ≤ 5, the velocity of the particle is given by \(v(t)= -2 + (t^{2}+3t)^{6/5} – t^{3}\), and the position of the particle is given by s(t). It is known that s(0) = 10.
(a) Find all values of t in the interval 0 ≤ t ≤ 4 for which the speed of the particle is 2.
(b) Write an expression involving an integral that gives the position s(t). Use this expression to find the position of the particle at time t = 5.
(c) Find all times t in the interval 0 ≤ t ≤ 5 at which the particle changes direction. Justify your answer.
(d) Is the speed of the particle increasing or decreasing at time t = 4 ? Give a reason for your answer.

▶️Answer/Explanation

\(
\textbf{2(a)} \text{Solve } |v(t)| = 2 \text{ on } 2 \leq t \leq 4.
\)
\(
t = 3.128 \text{ (or 3.127) and } t = 3.473
\)

\(
\textbf{2(b) } s(t) = 10 + \int_{0}^{t} v(x) \, dx
\)
\(
s(5) = 10 + \int_{0}^{5} v(x) \, dx = -9.207
\)

\(
\textbf{2(c) } v(t) = 0 \text{ when } t = 0.536033, \, 3.317756
\)
\(
v(t) \text{ changes sign from negative to positive at time } t = 0.536033.
\)
\(
v(t) \text{ changes sign from positive to negative at time } t = 3.317756.
\)
\(
\text{Therefore, the particle changes direction at time } t = 0.536 \text{ and time } t = 3.318 \text{ (or 3.317).}
\)

\(
\textbf{2(d) } v(4) = -11.475758 < 0, \, a(4) = v'(4) = -22.295714 < 0
\)
\(
\text{The speed is increasing at time } t = 4 \text{ because velocity and acceleration have the same sign.}
\)

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-8.4 Finding the Area Between Curves Expressed as Functions of x

(b) Topic-8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

(c) Topic-8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

Let \(f(x)= 2x^{2}-6x+4\) and \(g(x)= 4 cos \left ( \frac{1}{4}\pi x \right ).\) Let R be the region bounded by the graphs of f and g, as shown in the figure above.
(a) Find the area of R.
(b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 4.
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square.
Write, but do not evaluate, an integral expression that gives the volume of the solid. 

▶️Answer/Explanation

\(
\textbf{5(a)} \text{Area } = \int_{0}^{2} [g(x) – f(x)] \, dx
\)
\(
= \int_{0}^{2} \left[ 4 \cos \left( \frac{\pi}{4} x \right) – \left( 2x^2 – 6x + 4 \right) \right] \, dx
\)
\(
= \left[ 4 \cdot \frac{4}{\pi} \sin \left( \frac{\pi}{4} x \right) – \left( \frac{2x^3}{3} – 3x^2 + 4x \right) \right]_{0}^{2}
\)
\(
= \frac{16}{\pi} – \left( \frac{16}{3} – 12 + 8 \right) = \frac{16}{\pi} – \frac{4}{3}
\)

\(
\textbf{5(b)} \text{Volume } = \pi \int_{0}^{2} \left[ (4 – f(x))^2 – (4 – g(x))^2 \right] \, dx
\)
\(
= \pi \int_{0}^{2} \left[ (4 – (2x^2 – 6x + 4))^2 – (4 – 4 \cos \left( \frac{\pi}{4} x \right))^2 \right] \, dx
\)

\(
\textbf{5(c)} \text{Volume } = \int_{0}^{2} [g(x) – f(x)]^2 \, dx
\)
\(
= \int_{0}^{2} \left[ 4 \cos \left( \frac{\pi}{4} x \right) – (2x^2 – 6x + 4) \right]^2 \, dx
\)

Question 6

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

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

Consider the differential equation \(\frac{dy}{dx}=e^{y}(3x^{2}-6x).\) Let y = f (x) be the particular solution to the differential equation that passes through (1 , 0) .
(a) Write an equation for the line tangent to the graph of f at the point (1, 0) . Use the tangent line to approximate f (1, 2).
(b) Find y = f(x), the particular solution to the differential equation that passes through (1, 0). 

▶️Answer/Explanation

\(
\textbf{6(a) } \frac{dy}{dx} \Big|_{(x, y)=(1, 0)} = e^0 \left(3 \cdot 1^2 – 6 \cdot 1\right) = -3
\)
\(
\text{An equation for the tangent line is } y = -3(x – 1).
\)
\(
f(1.2) \approx -3(1.2 – 1) = -0.6
\)

\(
\textbf{6(b) } \frac{dy}{e^y} = \left(3x^2 – 6x\right) \, dx
\)
\(
\int \frac{dy}{e^y} = \int \left(3x^2 – 6x\right) \, dx
\)
\(
-e^{-y} = x^3 – 3x^2 + C
\)
\(
-e^{-0} = 1^3 – 3 \cdot 1^2 + C \implies C = 1
\)
\(
-e^{-y} = x^3 – 3x^2 + 1
\)
\(
e^{-y} = -x^3 + 3x^2 – 1
\)
\(
-y = \ln(-x^3 + 3x^2 – 1)
\)
\(
y = -\ln(-x^3 + 3x^2 – 1)
\)

\(\text{Note: This solution is valid on an interval containing } x = 1 \text{ for which } -x^3 + 3x^2 – 1 > 0.\)

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