AP Calculus BC Unit 4: Contextual Applications

Practice AP Calculus BC Unit 4 questions on interpreting derivatives, solving motion and related-rate problems, and using linearization.

Covered syllabus points

4.1 Interpreting the Meaning of the Derivative in Context 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration 4.3 Rates of Change in Applied Contexts Other Than Motion 4.4 Introduction to Related Rates 4.5 Solving Related Rates Problems 4.6 Approximating Values of a Function Using Local Linearity and Linearization 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

[Maximum number: 2]

A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function f, where f(t) is measured in gallons per second and t is measured in seconds since pumping began. Selected values of f(t) are given in the table.

Using the model g defined in part (c), find the value of $g^{\prime}(140)$ . Interpret the meaning of your answer in the context of the problem.

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.

Figure for Question Unit 4: Contextual Applications of Differentiation question 1 — AP Calculus BC

$\begin{aligned} g^{\prime}(140) \approx-0.004908 g^{\prime}(140) =-0.005(\text { or }-0.004) \end{aligned}$ $g^{\prime}(140)$

1 point

The rate at which gasoline is flowing into the tank is decreasing at
a rate of 0.005 (or 0.004 ) gallon per second per second at time
t=140 seconds.

Interpretation

1 point

Scoring notes:

- The exact value of $g^{\prime}(140)$ is $\frac{1}{500} \cos \left(\frac{49}{36}\right)-\frac{49}{9000} \sin \left(\frac{49}{36}\right)$ .

- The value of $g^{\prime}(140)$ may appear only in the interpretation.

- To be eligible for the second point a response must present some numerical value for $g^{\prime}(140)$ .

- To earn the second point the interpretation must include "the rate of flow of gasoline is changing

at a rate of [the declared value of $g^{\prime}(140)$ ]" and "at t=140 " (or equivalent).

- An interpretation of "decreasing at a rate of -0.005" or "increasing at a rate of 0.005" does not

earn the second point.

- Degree mode: In degree mode, $g^{\prime}(140)=0.001997$ or 0.00187.

Total for part (d) for question 19 points

Part A (BC): Graphing calculator required

[Maximum number: 2]

Johanna jogs along a straight path. For $0 \leq t \leq 40$ , Johanna's velocity is given by a differentiable function v. Selected values of v(t), where t is measured in minutes and v(t) is measured in meters per minute, are given in the table above.

Bob is riding his bicycle along the same path. For $0 \leq t \leq 10$ , Bob's velocity is modeled by $B(t)=t^{3}-6 t^{2}+300$ , where t is measured in minutes and B(t) is measured in meters per minute. Find Bob's acceleration at time t=5.

Bob's acceleration is $B^{\prime}(t)=3 t^{2}-12 t$ .

B^{\prime}(5)=3(25)-12(5)=15 \text { meters } / \mathrm{min}^{2}

[Maximum number: 2]

From 5 A.M. to 10 A.M., the rate at which vehicles arrive at a certain toll plaza is given by
$A(t)=450 \sqrt{\sin (0.62 t)}$ , where t is the number of hours after 5 A.M. and A(t) is measured in vehicles per hour. Traffic is flowing smoothly at 5 A.M. with no vehicles waiting in line.

Is the rate at which vehicles arrive at the toll plaza at 6 A.M. ( t=1 ) increasing or decreasing? Give a reason for your answer.

$A^{\prime}(1)=148.947272$

Considers $A^{\prime}(1)$

1 point

Because $A^{\prime}(1)>0$ , the rate at which the vehicles arrive at the toll
plaza is increasing.

Answer with reason

1 point

Scoring notes:

- The response need not present the value of $A^{\prime}(1)$ . The second line of the model solution earns both

points.

- An incorrect value assigned to $A^{\prime}(1)$ earns the first point (but will not earn the second point).

- Without a reference to t=1, the first point is earned by any of the following:

○ 148.947 accurate to the number of decimals presented, with zero up to three decimal places

(i.e., 149, 148, 148.9, 148.95, or 148.94)

○ $A^{\prime}(t)=148.947$ by itself

- To be eligible for the second point, the first point must be earned.

- To earn the second point, there must be a reference to t=1.

- Degree mode: $A^{\prime}(1)=23.404311$

Total for part (c)

2 points

[Maximum number: 3]

An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function r, where r(t) is measured in centimeters and t is measured in days. The table above gives selected values of $r^{\prime}(t)$ , the rate of change of the radius, over the time interval $0 \leq t \leq 12$ .

The height of the cone decreases at a rate of 2 centimeters per day. At time t=3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time t=3 days. (The volume V of a cone with radius r and height h is $V=\frac{1}{3} \pi r^{2} h$ .)

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.

Figure 1

Figure 2

$\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^{2}\frac{dh}{dt}$

Product rule

1 point

Chain rule

1 point

$\left.\frac{dV}{dt}\right|_{t=3}=\frac{2}{3}\pi(100)(50)(-5)+\frac{1}{3}\pi(100)^{2}(-2)=-\frac{70,000\pi}{3}$

The rate of change of the volume of the sculpture at t=3 is $-\frac{70,000\pi}{3}$ cubic centimeters per day.

Answer

1 point

Scoring notes:

- The first 2 points could be earned in either order.
- A completely correct product rule with the correct differentials earns the product rule and chain rule points.
- The third point is earned for the correct substitution and answer $-\frac{70,000\pi}{3}$ .
- Units are not required or read in this part.

Total for part (d)

for question 4

9 points

All question bank results loaded

Covered syllabus points

Unit 4: Contextual Applications of Differentiation question 1

Unit 4: Contextual Applications of Differentiation question 2

Unit 4: Contextual Applications of Differentiation question 3

Unit 4: Contextual Applications of Differentiation question 4