AP Calculus BC Unit 5: Analytical Applications

Practice AP Calculus BC Unit 5 questions on Mean and Extreme Value Theorems, derivative tests, concavity, and optimization.

Covered syllabus points

5.1 Using the Mean Value Theorem 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema 5.6 Determining Concavity of Functions over Their Domains 5.7 Using the Second Derivative Test to Determine Extrema 5.8 Sketching Graphs of Functions and Their Derivatives 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative 5.10 Introduction to Optimization Problems 5.11 Solving Optimization Problems 5.12 Exploring Behaviors of Implicit Relations

[Maximum number: 5]

An invasive species of plant appears in a fruit grove at time t=0 and begins to spread. The

function C defined by $C(t)=7.6 \arctan (0.2 t)$ models the number of acres in the fruit grove

affected by the species t weeks after the species appears. It can be shown that $C^{\prime}(t)=\frac{38}{25+t^{2}}$ .

(Note: Your calculator should be in radian mode.)

(a)

Find the time t when the instantaneous rate of change of C equals the average rate of change

of C over the time interval $0 \leq t \leq 4$ . Show the setup for your calculations.

[ 2 ]

B Find the time t when the instantaneous rate of change of C equals the average rate of change of C over

the time interval $0 \leq t \leq 4$ . Show the setup for your calculations.

$\frac{C(4)-C(0)}{4-0}=1.282008$	Uses average rate of change	Point 3 (P3)
$C^{\prime}(t)=\frac{38}{25+t^{2}}=1.282008 \Rightarrow t=2.154298$ <br>The instantaneous rate of change of C equals the average rate of change of C over the interval $0 \leq t \leq 4$ at time t=2.154.	Answer with supporting work	Point 4 (P4)

Scoring Notes for Part B

- P3 may be earned by presenting the expression or value for the average rate of change. Note that

because C(0)=0 and the interval is $0 \leq t \leq 4$ , any of the following will earn $\mathbf{P 3}: \frac{\int_{0}^{4} C^{\prime}(t) d t}{4}$ ,

$\frac{C(4)-C(0)}{4-0}, \frac{C(4)}{4}, \frac{5.128-0}{4-0}, \frac{5.128}{4}$ , or 1.282. However, neither P3 nor P4 is earned by just

presenting t=1.282.

- P4 is earned for the correct answer supported by the appropriate equation. A reported answer should

be accurate to three places after the decimal point, rounded or truncated. An inappropriately rounded

answer does not earn the point, unless an earlier point was not earned due to inappropriate rounding.

The following response, for example, earns both $\mathbf{P 3}$ and $\mathbf{P 4}: C^{\prime}(t)=\frac{C(4)-C(0)}{4}$ when t=2.154.

(b)

At time t=4 weeks after the invasive species appears in the fruit grove, measures are taken

to counter the spread of the species. The function A, defined by $A(t)=C(t)-\int_{4}^{t} 0.1 \cdot \ln (x) d x$ ,

models the number of acres affected by the species over the time interval $4 \leq t \leq 36$ . At what

time t, for $4 \leq t \leq 36, \operatorname{does} A$ attain its maximum value? Justify your answer.

[ 3 ]

D At time t=4 weeks after the invasive species appears in the fruit grove, measures are taken to counter

the spread of the species. The function A, defined by $A(t)=C(t)-\int_{4}^{t} 0.1 \cdot \ln (x) d x$ , models the number

of acres affected by the species over the time interval $4 \leq t \leq 36$ . At what time t, for $4 \leq t \leq 36$ , does

A attain its maximum value? Justify your answer.

$A^{\prime}(t)=C^{\prime}(t)-0.1 \cdot \ln t$	Considers $A^{\prime}(t)=0 \quad$ Point 7 (P7)
For $4 \leq t \leq 36$ , the maximum value of A(t) occurs when $A^{\prime}(t)=0$ or at an endpoint. $A^{\prime}(t)=C^{\prime}(t)-0.1 \cdot \ln t=0 \Rightarrow C^{\prime}(t)=0.1 \cdot \ln t$
$\Rightarrow t=11.441700$ t	A(t)<br>4	5.128031<br>11.441700	7.316978<br>36	1.743056	Justification Point 8 (P8)
Therefore, the number of acres affected by the species is a maximum at time t=11.442 (or 11.441) weeks.	Answer with supporting work

Scoring Notes for

[Maximum number: 2]

The graph of the differentiable function f, shown for $-6 \leq x \leq 7$ , has a horizontal tangent at x=-2 and is linear for $0 \leq x \leq 7$ . Let R be the region in the second quadrant bounded by the graph of f, the vertical line x=-6, and the x - and y-axes. Region R has area 12.

For the function g defined in part (a), find all values of x in the interval $0 \leq x \leq 6$ at which the graph of g has a critical point. Give a reason for your answer.

$g^{\prime}(x)=f(x)$

Fundamental
Theorem of Calculus

1 point

$g^{\prime}(x)=f(x)=0 \Rightarrow x=4$

Answer with reason

1 point

Therefore, the graph of g has a critical point at x=4.

Scoring notes:

- The first point is earned for explicitly making the connection $g^{\prime}=f$ in this part.

○ A response that writes $g^{\prime \prime}=f^{\prime}$ earns the first point but can only earn the second point by

reasoning from f=0.

- A response that does not earn the first point is eligible to earn the second point with an implied

application of the FTC (e.g., "Because $g^{\prime}(4)=0, x=4$ is a critical point").

- A response that reports any additional critical points in 0<x<6 does not earn the second point.

○ Any presented critical point outside the interval 0<x<6 will not affect scoring.

Total for part (b) 2 points

[Maximum number: 6]

Let f be a differentiable function with f(4)=3. On the interval $0 \leq x \leq 7$ , the graph of $f^{\prime}$ , the derivative of f, consists of a semicircle and two line segments, as shown in the figure above.

(a)

Find the x-coordinates of all points of inflection of the graph of f for 0<x<7. Justify your answer.

[ 2 ]

The graph of f has a point of inflection at each of x=2 and

Answer

1 point

x=6, because $f^{\prime}(x)$ changes from decreasing to increasing at
x=2 and from increasing to decreasing at x=6.

Justification

1 point

Scoring notes:

- A response that gives only one of x=2 or x=6, along with a correct justification, earns 1 of the

2 points.

- A response that claims that there is a point of inflection at any value other than x=2 or x=6

earns neither point.

- To earn the second point a response must use correct reasoning based on the graph of $f^{\prime}$ . Examples

of correct reasoning include:

○ Correctly discussing the signs of the slopes of the graph of $f^{\prime}$

○ Citing x=2 and x=6 as the locations of local extrema on the graph of $f^{\prime}$

- Examples of reasoning not (sufficiently) connected to the graph of $f^{\prime}$ include:

○ Reasoning based on sign changes in $f^{\prime \prime}$ unless the connection is made between the sign of $f^{\prime \prime}$

and the slopes of the graph of $f^{\prime}$

○ Reasoning based only on the concavity of the graph of f

- The second point cannot be earned by use of vague or undefined terms such as "it" or "the function"

or "the derivative."

- Responses that report inflection points as ordered pairs must report the points $(2,3+\pi)$ and (6,5)

in order to earn the first point. If the y-coordinates are reported incorrectly, the response remains

eligible for the second point.

Total for part (b) 2 points

(b)

Let g be the function defined by g(x)=f(x)-x. On what intervals, if any, is g decreasing for $0 \leq x \leq 7$ ? Show the analysis that leads to your answer.

[ 2 ]

$g^{\prime}(x)=f^{\prime}(x)-1$ $g^{\prime}(x)=f^{\prime}(x)-1$

1 point

$f^{\prime}(x)-1 \leq 0 \Rightarrow f^{\prime}(x) \leq 1$

The graph of g is decreasing on the interval $0 \leq x \leq 5$ because $g^{\prime}(x) \leq 0$ on this interval.

Interval with reason

1 point

Scoring notes:

- The first point can be earned for $f^{\prime}(x) \leq 1$ or the equivalent, in words or symbols.

- Endpoints do not need to be included in the interval to be eligible for the second point.

Total for part (c) 2 points

(c)

For the function g defined in part (c), find the absolute minimum value on the interval $0 \leq x \leq 7$ . Justify your answer.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

[ 2 ]

For 0<x<5, $g^{\prime}(x)<0$ , and for 5<x<7, $g^{\prime}(x)>0$ . Therefore, the absolute minimum occurs at x=5. Since $g(5)=f(5)-5=\frac{7}{2}-5=-\frac{3}{2}$ , the absolute minimum value of g is $-\frac{3}{2}$ .

Scoring notes:

- A justification using a local argument must state that $g^{\prime}$ changes from negative to positive at x=5 and that x=5 is the only critical point.
- A response that explicitly isolates x=5 as the only critical number in (0,7) earns the critical-point component.
- Using a candidates test, the relevant values are $g(0)=3+2\pi$ , $g(5)=-\frac{3}{2}$ , and $g(7)=-\frac{1}{2}$ ; the minimum is $-\frac{3}{2}$ .

Total for part (d)

for question 3

9 points

[Maximum number: 3]

Curve C is defined by the polar equation $r(\theta)=2 \sin ^{2} \theta$ for $0 \leq \theta \leq \pi$ . Curve C and the semicircle

$r=\frac{1}{2}$ for $0 \leq \theta \leq \pi$ are shown in the x y-plane.

Figure for Question Unit 5: Analytical Applications of Differentiation question 4 — AP Calculus BC

(Note: Your calculator should be in radian mode.)

It can be shown that $\frac{d x}{d \theta}=4 \sin \theta \cos ^{2} \theta-2 \sin ^{3} \theta$ for curve C. For $0 \leq \theta \leq \frac{\pi}{2}$ , find the value

of $\theta$ that corresponds to the point on curve C that is farthest from the y-axis. Justify your

answer.

C It can be shown that $\frac{d x}{d \theta}=4 \sin \theta \cos ^{2} \theta-2 \sin ^{3} \theta$ for curve C. For $0 \leq \theta \leq \frac{\pi}{2}$ , find the value of $\theta$ that

corresponds to the point on curve C that is farthest from the y-axis. Justify your answer.

For $0 \leq \theta \leq \frac{\pi}{2}$ , the curve C is in the first quadrant. Thus, a point on the curve will be farthest away from the y-axis when the x-coordinate attains its maximum value. This will either occur when $\frac{d x}{d \theta}=0$ or at an endpoint of the interval $0 \leq \theta \leq \frac{\pi}{2}.$ $\frac{d x}{d \theta}=0$	Considers $\frac{d x}{d \theta}=0$	Point 5 (P5)
$\Rightarrow \theta=0.955317$ $\theta$	$x(\theta)=r(\theta) \cos \theta$ <br>0	0<br>0.955317	0.769800<br> $\frac{\pi}{2}$	0	Justification	Point 6 (P6)
Therefore, the value of $\theta$ for which the point on the curve is farthest from the y-axis is 0.955.	Answer with supporting work	Point 7 (P7)

Scoring Notes for Part C

- P5 is earned for considering $\frac{d x}{d \theta}=0. \mathbf{P 5}$ is not earned by just presenting $\theta=0.955317$ .

A response that discusses the sign of $\frac{d x}{d \theta}$ changing or uses the phrase "critical points of $x(\theta)$ " also

earns P5.

- The value $\theta=0.955317$ might be presented as $\arccos \left(\frac{1}{\sqrt{3}}\right), \arcsin \left(\sqrt{\frac{2}{3}}\right)$ , or $\arctan (\sqrt{2})$ .

- To earn P6 using a candidates test, a response must make a global argument by correctly evaluating

$x(\theta)$ at $\theta=0, \theta=0.955317$ , and $\theta=\frac{\pi}{2}$ . The evaluations must be correct to the first digit after

the decimal, rounded or truncated.

- Alternate justifications:

○ $\frac{d x}{d \theta}>0$ for $0<\theta<0.955$ , and $\frac{d x}{d \theta}<0$ for $0.955<\theta<\frac{\pi}{2}$ . Therefore, $\theta=0.955$ is the

location of the absolute maximum for $x(\theta)$ on the interval $0 \leq \theta \leq \frac{\pi}{2}$ .

○ Because $\frac{d x}{d \theta}$ changes sign from positive to negative at $\theta=0.955$ (this might be presented as

" $\frac{d x}{d \theta}>0$ for $\theta<0.955$ , and $\frac{d x}{d \theta}<0$ for $\theta>0.955$ "), it is the location of a relative maximum

for $x(\theta)$ . And because $\theta=0.955$ is the only critical point of $x(\theta)$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$ ,

it is the location of the absolute maximum for $x(\theta)$ on the interval.

○ Because $\left.\frac{d x}{d \theta}\right|_{\theta=0.955}=0$ and $\left.\frac{d^{2} x}{d \theta^{2}}\right|_{\theta=0.955}<0, \theta=0.955$ is the location of a relative

maximum for $x(\theta)$ . And because $\theta=0.955$ is the only critical point of $x(\theta)$ in the interval

$0 \leq \theta \leq \frac{\pi}{2}$ , it is the location of the absolute maximum for $x(\theta)$ on the interval.

- A response that presents only a local argument (such as a First Derivative Test or a Second

Derivative Test) or an incorrect global argument does not earn P6 but is eligible for P7 with the

correct answer. A reported answer should be accurate to three places after the decimal point,

rounded or truncated. An inappropriately rounded answer does not earn the point, unless an earlier

point was not earned due to inappropriate rounding.

All question bank results loaded

Covered syllabus points

Unit 5: Analytical Applications of Differentiation question 1

Question (a)

Question (b)

Unit 5: Analytical Applications of Differentiation question 2

Unit 5: Analytical Applications of Differentiation question 3

Question (a)

Question (b)

Question (c)

Unit 5: Analytical Applications of Differentiation question 4