AP Calculus AB Unit 5: Analytical Applications

Explore AP Calculus Unit 5 questions on the Mean Value Theorem, critical points, extrema, concavity, and derivative-based function analysis.

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 Part D

- P7 is earned for considering $A^{\prime}(t)=0, C^{\prime}(t)-0.1 \cdot \ln t=0$ , or $C^{\prime}(t)=0.1 \cdot \ln t$ . P7 is not earned

by just presenting t=11.441700.

A response that discusses the sign of $A^{\prime}(t)$ changing or uses the phrase "critical points of A " also

earns P7.

- To earn P8 using a candidates test, a response must make a global argument by correctly evaluating

A(t) at t=4, t=11.441700, and t=36. The evaluations must be correct to the first digit after

the decimal, rounded or truncated.

- Alternate justifications:

○ $A^{\prime}(t)>0$ for 4<t<11.442, and $A^{\prime}(t)<0$ for 11.442<t<36. Therefore, t=11.442 is the

location of the absolute maximum for A on the interval $4 \leq t \leq 36$ .

○ Because $A^{\prime}(t)$ changes sign from positive to negative at t=11.442 (this might be presented as

" $A^{\prime}(t)>0$ for t<11.442, and $A^{\prime}(t)<0$ for t>11.442 "), it is the location of a relative

maximum for A. And because t=11.442 is the only critical point of A in the interval

$4 \leq t \leq 36$ , it is the location of the absolute maximum for A on the interval.

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

Test) or an incorrect global argument does not earn P8 but is eligible for P9 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.

Part A (AB): Graphing calculator required

[Maximum number: 2]

Let f be the function defined above.

Must there be a value of x at which f(x) attains an absolute maximum on the closed interval

$-3 \leq x \leq 4 ?$ Justify your answer.

Figure for Question Unit 5: Analytical Applications of Differentiation question 2 — AP Calculus AB

Graph of f

$\lim _{x \rightarrow 0^{-}} f(x)=f(0)=3$ and $\lim _{x \rightarrow 0^{+}} f(x)=3$ , so f is continuous at

x=0.

Because f is continuous on [-3, 4], the Extreme Value Theorem

guarantees that f attains an absolute maximum on [-3, 4].

1: answer

$2:\left\{\begin{array}{l}1: f^{\prime}(3) \\ 1: \text { equation }\end{array}\right.$

4: $\left\{\begin{array}{l}1 \text {: integrals of } f \text { over } \\ \quad-3 \leq x \leq 0 \text { and } 0 \leq x \leq 4 \\ 1 \text {: value of } \int_{-3}^{0} \sqrt{9-x^{2}} d x \\ 1 \text {: antiderivative of } \\ \quad-x+3 \cos \left(\frac{\pi x}{2}\right) \\ 1 \text {: answer }\end{array}\right.$ $2:\left\{\begin{array}{l}1: \text { continuity at } x=0 \\ 1: \text { answer with justification }\end{array}\right.$

Question 4

[Maximum number: 6]

The functions f and g are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of x.

(a)

Let k be a differentiable function such that $k^{\prime}(x)=(f(x))^{2} \cdot g(x)$ . Is the graph of k concave up or concave down at the point where x=4 ? Give a reason for your answer.

[ 3 ]

Let k be a differentiable function such that $k^{\prime}(x)=(f(x))^{2} \cdot g(x)$ . Is the graph of k concave up or concave down at the point where x=4 ? Give a reason for your answer.

k^{\prime \prime}(x)=2 f(x) \cdot f^{\prime}(x) \cdot g(x)+(f(x))^{2} \cdot g^{\prime}(x)

Product or chain rule 1 point

$k^{\prime \prime}(4)=2 f(4) \cdot f^{\prime}(4) \cdot g(4)+(f(4))^{2} \cdot g^{\prime}(4)$ $=2 \cdot 4 \cdot 3 \cdot(-3)+4^{2} \cdot 2=-72+32=-40$ $k^{\prime \prime}(4) \quad 1$ point

The graph of k is concave down at the point where x=4
because $k^{\prime \prime}(4)<0$ and $k^{\prime \prime}$ is continuous.

Answer with reason 1 point

Scoring notes:

- The first point is earned for either $k^{\prime \prime}(x)=2 f(x) \cdot f^{\prime}(x) \cdot g(x)+(f(x))^{2} \cdot g^{\prime}(x)$ or $k^{\prime \prime}(4)=2 f(4) \cdot f^{\prime}(4) \cdot g(4)+(f(4))^{2} \cdot g^{\prime}(4)$ .

- The first point is also earned by any of the following incorrect expressions, each of which has a single error in the application of the product rule or the chain rule:

$2 f(x) \cdot g(x)+(f(x))^{2} \cdot g^{\prime}(x)$ or $2 f(4) \cdot g(4)+(f(4))^{2} \cdot g^{\prime}(4)$ $2 f^{\prime}(x) \cdot g(x)+(f(x))^{2} \cdot g^{\prime}(x)$ or $2 f^{\prime}(4) \cdot g(4)+(f(4))^{2} \cdot g^{\prime}(4)$ $f^{\prime}(x) \cdot g(x)+(f(x))^{2} \cdot g^{\prime}(x)$ or $f^{\prime}(4) \cdot g(4)+(f(4))^{2} \cdot g^{\prime}(4)$ $2 f(x) \cdot f^{\prime}(x) \cdot g^{\prime}(x)$ or $2 f(4) \cdot f^{\prime}(4) \cdot g^{\prime}(4)$

Note: A response that presents one of these expressions cannot earn the second point.

- To earn the second point a response must correctly find $k^{\prime \prime}(4)=-40$ (or equivalent) with supporting work.

- The third point is earned for an answer and reason that are consistent with any declared nonzero value of $k^{\prime \prime}(4)$ .

Total for part (b) 3 points

(b)

Is the function m defined in part (c) increasing, decreasing, or neither at x=2 ? 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.

[ 3 ]

[Maximum number: 3]

The depth of seawater at a location can be modeled by the function H that satisfies the differential equation $\frac{d H}{d t}=\frac{1}{2}(H-1) \cos \left(\frac{t}{2}\right)$ , where H(t) is measured in feet and t is measured in hours after noon ( t=0 ). It is known that H(0)=4.

For 0<t<5, it can be shown that H(t)>1. Find the value of t, for 0<t<5, at which H has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.

Because H(t)>1, then $\frac{dH}{dt}=0$ implies $\cos \left(\frac{t}{2}\right)=0$ .
This implies that $t=\pi$ is a critical point.}

Considers sign of
$\frac{d H}{d t}$

1 point

Identifies $t=\pi$

1 point

For $0<t<\pi, \frac{d H}{d t}>0$ and for $\pi<t<5, \frac{d H}{d t}<0$ . Therefore,
$t=\pi$ is the location of a relative maximum value of H.

Answer with justification

1 point

Scoring notes:

- The first point is earned for considering $\frac{d H}{d t}=0, \frac{d H}{d t}>0, \frac{d H}{d t}<0, \cos \left(\frac{t}{2}\right)=0, \cos \left(\frac{t}{2}\right)>0$ ,

or $\cos \left(\frac{t}{2}\right)<0$ .

- The second point is earned for identifying $t=\pi$ , with or without supporting work. A response

may consider H=1 or t=1 as potential critical points without penalty.

- The third point cannot be earned without the first point. The third point is earned only for a correct

justification and a correct answer of "relative maximum."

- The justification can be shown by determining the sign of $\frac{d H}{d t}$ (or $\cos \left(\frac{t}{2}\right)$ ) at a single value in

$0<t<\pi$ and at a single value in $\pi<t<5$ . It is not necessary to state that $\frac{d H}{d t}$ does not change

sign on these intervals.

- The third point can also be earned by using the Second Derivative Test. For example:

Therefore, $t=\pi$ is the location of a relative maximum value of H.

Total for part (b) 3 points

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)

Unit 5: Analytical Applications of Differentiation question 4