AP Calculus BC : 4.1 Interpreting the Meaning of the Derivative in Context- Exam Style questions with Answer- MCQ

Question

The rate at which water leaks from a tank, in gallons per hour, is modeled by \( R \), a differentiable function of the number of hours after the leak is discovered. Which of the following is the best interpretation of \( R'(3) \)?

A) The amount of water, in gallons, that has leaked out of the tank during the first three hours after the leak is discovered

B) The amount of change, in gallons per hour, in the rate at which water is leaking during the three hours after the leak is discovered

C) The rate at which water leaks from the tank, in gallons per hour, three hours after the leak is discovered

D) The rate of change of the rate at which water leaks from the tank, in gallons per hour, three hours after the leak is discovered

▶️ Answer/Explanation

Solution

\( R'(3) \) is the rate of change of the leakage rate at \( t = 3 \).

A) Total water leaked is \( \int R(t) \, dt \), not \( R'(3) \).

B) Change over an interval, not instantaneous at \( t = 3 \).

C) \( R(3) \) is the rate, not its derivative.

D) Matches \( R'(3) \) as the rate of change of the rate.

✅ Answer: D)

Question

A weight suspended by a spring vibrates vertically according to the function \( D \) given by:

\[ D(t) = 2\sin\left(4\pi\left(t + \frac{1}{8}\right)\right) \]

where \( D(t) \), in centimeters, is the directed distance of the weight from its central position \( t \) seconds after the start of the motion. Assume the positive direction is upward. What is the instantaneous rate of change of the weight’s position, in centimeters per second, at the moment the weight is first 1 centimeter above its central position?

A) \( -157.914 \)
B) \( -21.765 \)
C) \( -12 \)
D) \( 0 \)

▶️ Answer/Explanation

Solution

Step 1: Find when \( D(t) = 1 \) cm (first occurrence)

\[ 1 = 2\sin\left(4\pi\left(t + \frac{1}{8}\right)\right) \] \[ \sin\left(4\pi\left(t + \frac{1}{8}\right)\right) = \frac{1}{2} \] \[ 4\pi\left(t + \frac{1}{8}\right) = \frac{\pi}{6} \text{ (first positive solution)} \] \[ t + \frac{1}{8} = \frac{1}{24} \] \[ t = \frac{1}{24} – \frac{1}{8} = -\frac{1}{12} \text{ seconds} \]

Since time can’t be negative, we take the next period solution:

\[ 4\pi\left(t + \frac{1}{8}\right) = \frac{\pi}{6} + 2\pi \] \[ t + \frac{1}{8} = \frac{1}{24} + \frac{1}{2} = \frac{13}{24} \] \[ t = \frac{13}{24} – \frac{3}{24} = \frac{10}{24} = \frac{5}{12} \text{ seconds} \]

Step 2: Compute the derivative \( D'(t) \)

\[ D'(t) = 2\cos\left(4\pi\left(t + \frac{1}{8}\right)\right) \cdot 4\pi \] \[ D'(t) = 8\pi\cos\left(4\pi\left(t + \frac{1}{8}\right)\right) \]

Step 3: Evaluate at \( t = \frac{5}{12} \)

\[ D’\left(\frac{5}{12}\right) = 8\pi\cos\left(4\pi\left(\frac{5}{12} + \frac{1}{8}\right)\right) \] \[ = 8\pi\cos\left(4\pi\left(\frac{10}{24} + \frac{3}{24}\right)\right) \] \[ = 8\pi\cos\left(4\pi \cdot \frac{13}{24}\right) \] \[ = 8\pi\cos\left(\frac{13\pi}{6}\right) \] \[ = 8\pi\cos\left(2\pi + \frac{\pi}{6}\right) \] \[ = 8\pi\cos\left(\frac{\pi}{6}\right) \] \[ = 8\pi \cdot \frac{\sqrt{3}}{2} \] \[ = 4\pi\sqrt{3} \approx 21.765 \text{ cm/s} \]

Since the weight is moving downward through this position (as it’s the first time reaching +1cm after starting at \( D(0) = \sqrt{2} \) cm), we take the negative value.

✅ Answer: B) \( -21.765 \)

Question

A city planner developed the function \( P \) to model the population of a city over the next 25 years, where \( P(t) \) is the projected population of the city \( t \) years from now. According to the model, which of the following gives the rate at which the population of the city will be changing at time \( t = 5 \)?

A) \( P(5) \)
B) \( P(6) – P(4) \)
C) \( \frac{P(5)}{5} \)
D) \( P'(5) \)

▶️ Answer/Explanation

Solution

Understanding the Problem:

We need to identify which expression represents the instantaneous rate of change of the population at exactly \( t = 5 \) years.

Analyzing Each Option:

A) \( P(5) \):
This gives the actual population at \( t = 5 \) years, not the rate of change.

B) \( P(6) – P(4) \):
This represents the average rate of change between \( t = 4 \) and \( t = 6 \) years (a 2-year period centered at \( t = 5 \)), not the instantaneous rate at exactly \( t = 5 \).

C) \( \frac{P(5)}{5} \):
This incorrectly divides the population by time, giving a meaningless value that doesn’t represent any rate of change.

D) \( P'(5) \):
This is the derivative of \( P \) evaluated at \( t = 5 \), which by definition gives the instantaneous rate of change of the population at exactly 5 years.

Conclusion:
The derivative \( P'(t) \) represents the instantaneous rate of change of the population at time \( t \). Therefore, \( P'(5) \) correctly gives the rate of change at exactly 5 years.

✅ Answer: D) \( P'(5) \)

Question

The figure below shows the graph of the differentiable function \( f \) for \( 1 \leq x \leq 8 \), and the secant line through the points \( (1, f(1)) \) and \( (8, f(8)) \).

Graph of function and secant line

For how many values of \( x \) in the closed interval \([1, 8]\) does the instantaneous rate of change of \( f \) at \( x \) equal the average rate of change of \( f \) over that interval?

A) Zero

B) Three

C) Four

D) Five

▶️ Answer/Explanation

Solution

The average rate of change of \( f \) on the interval \([1, 8]\) is the slope of the secant line between the points \( (1, f(1)) \) and \( (8, f(8)) \).

The instantaneous rate of change is given by the slope of the tangent line to the curve at a point.

We need to count how many times the slope of the tangent to the curve matches the slope of the secant line.

From the graph, the tangent to the curve appears to be parallel to the secant line in four distinct places.

This is a direct application of the Mean Value Theorem, which ensures at least one such point exists for a differentiable function, and in this case we observe 4 such tangent points.

✅ Answer: C

AP Calculus BC : 4.1 Interpreting the Meaning of the Derivative in Context- Exam Style questions with Answer- MCQ

Resources

Members

Company