AP Calculus BC : 4.3 Rates of Change in Applied Contexts Other Than Motion- Exam Style questions with Answer- MCQ

Question

The area of a circular region is increasing at a rate of 96π square meters per second. When the area of the region is 64π square meters, how fast, in meters per second, is the radius of the region increasing?

(A) 6

(B) 8

(D) \(\sqrt[4]{3}\)

(E) \(\sqrt[12]{3}\)

▶️ Answer/Explanation

Detailed Solution

Given: \( A = \pi r^2 \), \( \frac{dA}{dt} = 96\pi \), \( A = 64\pi \).

Differentiate: \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).

Find \( r \): \( 64\pi = \pi r^2 \), so \( r = 8 \).

Substitute: \( 96\pi = 2\pi \cdot 8 \cdot \frac{dr}{dt} \), solve: \( \frac{dr}{dt} = \frac{96\pi}{16\pi} = 6 \).

✅ Answer: A)

Question

A spherical balloon is being inflated. What is the volume of the sphere at the instant when the rate of increase of the surface area is four times the rate of increase of the radius of the sphere?

(A) \(\frac{1}{2\pi}\) cubic units

(B) \(\frac{3\pi r^{2}}{2}\) cubic units

(D) \(\frac{1}{6\pi^{2}}\) cubic units

▶️ Answer/Explanation

Solution

Surface area: \( S = 4 \pi r^2 \), so \( \frac{dS}{dt} = 8 \pi r \frac{dr}{dt} \).

Given: \( \frac{dS}{dt} = 4 \frac{dr}{dt} \), substitute: \( 4 \frac{dr}{dt} = 8 \pi r \frac{dr}{dt} \), solve: \( r = \frac{1}{2 \pi} \).

Volume: \( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( \frac{1}{2 \pi} \right)^3 = \frac{1}{6 \pi^2} \).

✅ Answer: D)

Question

A conical funnel has a base diameter of 4 cm and a height of 5 cm. The funnel is initially full, but water is draining at a constant rate of \( 2 \text{ cm}^3/\text{s} \). How fast is the water level falling when the water is 2.5 cm high?

A) \( \frac{1}{2\pi} \text{ cm/s} \)
B) \( -\frac{1}{2\pi} \text{ cm/s} \)
C) \( \frac{2}{\pi} \text{ cm/s} \)
D) \( -\frac{2}{\pi} \text{ cm/s} \)

▶️ Answer/Explanation

Solution

Step 1: Establish the relationship between radius and height

Using similar triangles:

\[ \frac{r}{h} = \frac{2}{5} \Rightarrow r = \frac{2}{5}h \]

Step 2: Express volume in terms of height

\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{2}{5}h\right)^2 h = \frac{4}{75}\pi h^3 \]

Step 3: Differentiate volume with respect to time

\[ \frac{dV}{dt} = \frac{4}{25}\pi h^2 \frac{dh}{dt} \]

Step 4: Substitute known values

Given: \[ \frac{dV}{dt} = -2 \text{ cm}^3/\text{s} \quad \text{(negative because volume is decreasing)} \] \[ h = 2.5 \text{ cm} \]

Substitute into the equation:

\[ -2 = \frac{4}{25}\pi (2.5)^2 \frac{dh}{dt} \] \[ -2 = \frac{4}{25}\pi (6.25) \frac{dh}{dt} \] \[ -2 = \pi \frac{dh}{dt} \] \[ \frac{dh}{dt} = -\frac{2}{\pi} \text{ cm/s} \]

✅ Answer: D) \( -\frac{2}{\pi} \text{ cm/s} \)

Question

A light on the ground 50 m from a tall building shines on a 1-m tall child. The child is walking away from the light at a velocity of 1 m/s. How fast is the shadow on the building decreasing when the child is 10 m away from the building?
(A) \(-\frac{5}{4}m/s\)
(B) \(-\frac{1}{50}m/s\)
(C) \(-\frac{1}{32}m/s\)
(D)\(-\frac{1}{20}m/s\)

▶️ Answer/Explanation

Solution

Let \( x \) be the distance from the light to the child, and \( S \) be the shadow’s height on the building. When the child is 10 m from the building, \( x = 50 – 10 = 40 \) m. The child moves away from the light at \( \frac{dx}{dt} = 1 \) m/s.

The following diagram illustrates the setup using similar triangles:

Using similar triangles (child’s height 1 m, shadow height \( S \), bases \( x \) and 50):

\[ \frac{1}{S} = \frac{x}{50} \]

\[ S = \frac{50}{x} = 50x^{-1} \]

Differentiate with respect to time:

\[ \frac{dS}{dt} = -50x^{-2} \frac{dx}{dt} \]

Since \( \frac{dx}{dt} = 1 \):

\[ \frac{dS}{dt} = -\frac{50}{x^2} \]

At \( x = 40 \):

\[ \frac{dS}{dt} = -\frac{50}{40^2} = -\frac{50}{1600} = -\frac{1}{32} \text{ m/s} \]

The negative sign indicates the shadow’s height is decreasing, matching the question’s context.

✅ Answer: C)

AP Calculus BC : 4.3 Rates of Change in Applied Contexts Other Than Motion- Exam Style questions with Answer- MCQ

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