← Back to the Related Rates worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution states the equation used, the differentiation step and the units. If your answer differs, check whether values were substituted before differentiating, which is the most common error in related rates.
Standards: AP Calculus AB Unit 4 · Level: Grade 12 AP Calculus AB
Problem 1 Interpreting marginal cost
When 200 cupcakes are being baked, the cost is increasing at a rate of $1.85 per cupcake. That is, baking the 201st cupcake adds about $1.85 to the total cost.
Problem 2 Velocity and acceleration at an instant
v(t) = 3t2 − 18t + 15, so v(2) = 12 − 36 + 15 = −9 ft/s. a(t) = 6t − 18, so a(2) = −6 ft/s2. The speed is |−9| = 9 ft/s. Both are negative, so the cart is speeding up.
Problem 3 Direction of motion
v(t) = 3t2 − 12t + 9 = 3(t − 1)(t − 3), which is zero at t = 1 and t = 3. Testing a point in each interval gives v(0) = 9 > 0, v(2) = −3 < 0 and v(4) = 9 > 0. The particle moves right on [0, 1) and (3, ∞) and left on (1, 3).
Problem 4 Acceleration from a table
The nearest times around 3.5 are 2 and 5: 16 − 75 − 2 = 9/3 = 3 m/s2.
Problem 5 Speeding up or slowing down
a(t) = 2t − 8. At t = 1: v = 5 and a = −6, opposite signs, so the particle is slowing down. At t = 7: v = 5 and a = 6, same sign, so it is speeding up.
Problem 6 Linearization of a cube root
f(27) = 3 and f′(x) = 1/(3x2/3), so f′(27) = 1/27. The linearization is L(x) = 3 + (x − 27)/27. Then ∛29 ≈ 3 + 2/27 ≈ 3.0741. The cube root graph is concave down, so the tangent line lies above it and the estimate is too high.
Problem 7 Linear approximation of an exponential
g(0) = 1 and g′(x) = 5e5x, so g′(0) = 5. Then L(x) = 1 + 5x and g(0.02) ≈ 1.1. Since g″ = 25e5x > 0, the graph is concave up and the tangent line lies below it: an underestimate (actual ≈ 1.1052).
Problem 8 L’Hôpital’s Rule once
Direct substitution gives 0/0. Differentiate top and bottom: 4e4x6 cos 6x. At x = 0 this is 4/6. The limit is 2/3.
Problem 9 L’Hôpital’s Rule twice
The form is ∞/∞. First pass: 2x3e3x, still ∞/∞. Second pass: 29e3x, which tends to 0. The limit is 0.
Problem 10 Growing circle
A = πr2, so dA/dt = 2πr dr/dt = 2π(15)(0.4) = 12π cm2/s (about 37.7 cm2/s).
Problem 11 Inflating a balloon
dV/dt = 4πr2 dr/dt. Substitute: 90 = 4π(9) dr/dt = 36π dr/dt. So dr/dt = 90/(36π) = 5/(2π) in/s ≈ 0.80 in/s.
Problem 12 Sliding ladder
x2 + y2 = 289. When x = 8, y = √(289 − 64) = 15. Differentiate: x dxdt + y dydt = 0, so 8(1.5) + 15 dydt = 0 and dydt = −12/15 = −0.8 ft/s. The top slides down at 0.8 ft/s.
Problem 13 Marginal cost and actual cost
C′(x) = 0.04x + 4, so C′(100) = $8 per skateboard. It estimates the extra cost of the 101st skateboard. The actual cost is C(101) − C(100) = 8.02, so the estimate is within $0.02.
Problem 14 Filling a conical tank
By similar triangles r = h/2, so V = (1/3)π(h/2)2h = πh3/12. Differentiate: dV/dt = (πh2/4) dh/dt. Substitute: 20 = (16π/4) dh/dt = 4π dh/dt, so dh/dt = 5/π ft/min ≈ 1.59 ft/min.
Problem 15 Relative growth of a population
P′(t) = 320e0.04t. At t = 5, P′(5) = 320e0.2 ≈ 390.85 people per year. The relative rate is 320e0.04t/(8,000e0.04t) = 0.04, so the population grows 4% per year at all times.
Problem 16 Shadow of a person walking
Let T be the distance from the post to the shadow tip. Similar triangles give T − x6 = T15. So 15T − 15x = 6T, 9T = 15x, and T = (5/3)x. Then dT/dt = (5/3)(4) = 20/3 ft/s. The shadow’s length is T − x, which grows at 20/3 − 4 = 8/3 ft/s.
Problem 17 Distance traveled by a particle
v(t) = 3(t − 2)(t − 6): the particle moves right on [0, 2], left on [2, 6], and right on [6, 7]. Positions: s(0) = 0, s(2) = 32, s(6) = 0, s(7) = 7. Distances: 32 + 32 + 7 = 71. The net displacement is 7 − 0 = 7.
Problem 18 Trigonometric limit with two passes
The first form is 0/0. Differentiate: 5 sin 5x2x, still 0/0. Differentiate again: 25 cos 5x2. At x = 0 this is 25/2. The limit is 25/2.
Problem 19 Intervals of speeding up
v(t) = 6(t − 2)(t − 3) and a(t) = 12t − 30, which is zero at t = 2.5. Signs: on (0, 2), v > 0, a < 0 (slowing); on (2, 2.5), v < 0, a < 0 (speeding up); on (2.5, 3), v < 0, a > 0 (slowing); on (3, 4), v > 0, a > 0 (speeding up). The particle speeds up on (2, 2.5) and (3, 4).
Problem 20 Distance from a radar station
Let x be the ground distance and D the straight-line distance. Then D2 = x2 + 16, so 2D dD/dt = 2x dxdt. When x = 3, D = 5. Therefore dD/dt = (3)(500)/5 = 300 mph.
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