AP Calculus AB Practice Quiz: Solving Related Rates Problems

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

The radius of a circle is increasing at a constant rate of 3 centimeters per second. At the instant when the radius of the circle is 10 centimeters, what is the rate of change of the area of the circle?

A)30π cm²/sB)60π cm²/sC)90π cm²/sD)100π cm²/s

All Questions (7)

A) 30π cm²/s

B) 60π cm²/s

C) 90π cm²/s

D) 100π cm²/s

Correct Answer: B

Let A be the area and r be the radius of the circle. The formula relating them is A = πr². We are given dr/dt = 3 cm/s and we want to find dA/dt when r = 10 cm. Differentiating the area formula with respect to time t gives dA/dt = 2πr(dr/dt). Substituting the given values: dA/dt = 2π(10)(3) = 60π cm²/s.

A 13-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 5 feet per second. How fast is the top of the ladder sliding down the wall at the instant the bottom of the ladder is 12 feet from the wall?

A) -12 ft/s

B) 12 ft/s

C) -5/12 ft/s

D) -12/5 ft/s

Correct Answer: A

Let x be the distance from the base of the wall to the bottom of the ladder, and y be the height of the top of the ladder on the wall. By the Pythagorean theorem, x² + y² = 13². We are given dx/dt = 5 ft/s and we want to find dy/dt when x = 12 ft. First, find y when x = 12: 12² + y² = 13², so 144 + y² = 169, which gives y = 5 ft. Differentiate the equation with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0. Substitute the known values: 2(12)(5) + 2(5)(dy/dt) = 0. This simplifies to 120 + 10(dy/dt) = 0. Solving for dy/dt gives dy/dt = -120/10 = -12 ft/s. The negative sign indicates the top of the ladder is sliding down.

Water is being pumped into an inverted conical tank at a rate of 2 cubic meters per minute. The tank has a height of 10 meters and a radius at the top of 5 meters. What is the rate of change of the water's height when the water is 6 meters deep?

A) 1/(18π) m/min

B) 8/(9π) m/min

C) 1/(9π) m/min

D) 2/(9π) m/min

Correct Answer: D

Let V be the volume, r be the radius, and h be the height of the water in the cone. The volume is V = (1/3)πr²h. By similar triangles, the ratio of the radius to the height is constant: r/h = 5/10, so r = h/2. Substitute this into the volume formula to get V in terms of h only: V = (1/3)π(h/2)²h = (π/12)h³. We are given dV/dt = 2 m³/min and we want to find dh/dt when h = 6 m. Differentiate the volume formula with respect to time t: dV/dt = (π/12) * 3h²(dh/dt) = (π/4)h²(dh/dt). Substitute the known values: 2 = (π/4)(6)²(dh/dt). This simplifies to 2 = (π/4)(36)(dh/dt), or 2 = 9π(dh/dt). Solving for dh/dt gives dh/dt = 2/(9π) m/min.

Car A is traveling east at 60 mph and Car B is traveling north at 45 mph. Both are headed for the same intersection. At what rate is the distance between the cars changing when Car A is 120 miles from the intersection and Car B is 90 miles from the intersection?

A) 105 mph

B) 75 mph

C) 68 mph

D) 50 mph

Correct Answer: B

Let x be the distance of Car A from the intersection, y be the distance of Car B, and z be the distance between them. The relationship is x² + y² = z². We are given dx/dt = -60 mph and dy/dt = -45 mph (negative because the distances are decreasing). We want to find dz/dt when x = 120 and y = 90. First, find z at this instant: z² = 120² + 90² = 14400 + 8100 = 22500, so z = 150 miles. Differentiate the equation with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt). Substitute the known values: 2(120)(-60) + 2(90)(-45) = 2(150)(dz/dt). This simplifies to -14400 - 8100 = 300(dz/dt), so -22500 = 300(dz/dt). Solving for dz/dt gives -75 mph. The rate of change is 75 mph (the distance is decreasing).

The side length of a cube is increasing at a rate of 2 inches per minute. What is the rate of change of the volume of the cube at the instant when the side length is 5 inches?

A) 30 in³/min

B) 60 in³/min

C) 150 in³/min

D) 300 in³/min

Correct Answer: C

Let V be the volume and s be the side length of the cube. The formula for the volume is V = s³. We are given ds/dt = 2 in/min and we want to find dV/dt when s = 5 in. Differentiating the volume formula with respect to time t gives dV/dt = 3s²(ds/dt). Substitute the given values: dV/dt = 3(5)²(2) = 3(25)(2) = 150 in³/min.

A 6-foot-tall person is walking away from a 15-foot-tall lamppost at a constant rate of 5 feet per second. At what rate is the tip of the person's shadow moving away from the lamppost?

A) 5 ft/s

B) 10/3 ft/s

C) 25/3 ft/s

D) 15/2 ft/s

Correct Answer: C

Let x be the person's distance from the lamppost and s be the length of the shadow. The position of the tip of the shadow from the lamppost is P = x + s. We are given dx/dt = 5 ft/s and we want to find dP/dt. By similar triangles, (x+s)/15 = s/6. Cross-multiplying gives 6(x+s) = 15s, which simplifies to 6x + 6s = 15s, or 6x = 9s. Solving for s gives s = (2/3)x. Now substitute this into the position formula: P = x + (2/3)x = (5/3)x. Differentiate with respect to time t: dP/dt = (5/3)(dx/dt). Substitute the given rate: dP/dt = (5/3)(5) = 25/3 ft/s. This rate is constant and does not depend on the person's distance from the lamppost.

Air is being pumped into a spherical balloon at a rate of 100 cubic centimeters per second. What is the rate of change of the radius of the balloon when its radius is 5 centimeters?

A) 1/π cm/s

B) π cm/s

C) 1/(25π) cm/s

D) 25π cm/s

Correct Answer: A

Let V be the volume and r be the radius of the sphere. The formula for the volume is V = (4/3)πr³. We are given dV/dt = 100 cm³/s and we want to find dr/dt when r = 5 cm. Differentiating the volume formula with respect to time t gives dV/dt = (4/3)π * 3r²(dr/dt) = 4πr²(dr/dt). Substitute the known values: 100 = 4π(5)²(dr/dt). This simplifies to 100 = 4π(25)(dr/dt), or 100 = 100π(dr/dt). Solving for dr/dt gives dr/dt = 100/(100π) = 1/π cm/s.