AP Calculus AB Quiz: Introduction To Related Rates

What this quiz covers

This quiz focuses on Introduction To Related Rates, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $3$
2. $-3$ (correct answer)
3. $-6$
4. $6$
5. $0$

Explanation: This problem demonstrates constraint relationships in perimeter. Given $P = 2x + 2y$ and $\dfrac{dP}{dt} = 0$ (constant perimeter), taking the derivative: $2\dfrac{dx}{dt} + 2\dfrac{dy}{dt} = 0$. With $\dfrac{dx}{dt} = 3$, we have $2(3) + 2\dfrac{dy}{dt} = 0$, so $6 + 2\dfrac{dy}{dt} = 0$. Therefore, $\dfrac{dy}{dt} = -3$. The negative sign indicates that as one dimension increases, the other must decrease to maintain constant perimeter.

Question 2

1. $-1.2\pi\text{ cm}^3/\text{hr}$ (correct answer)
2. $1.2\pi\text{ cm}^3/\text{hr}$
3. $-0.3\pi\text{ cm}^3/\text{hr}$
4. $-4\pi\text{ cm}^3/\text{hr}$
5. $-12\pi\text{ cm}^3/\text{hr}$

Explanation: This problem demonstrates volume rate change for a cylinder with decreasing height. Given $V = \pi r^2 h$ with $r = 2$ cm (constant) and $dh/dt = -0.3$ cm/hr (negative for decreasing), taking the derivative: $dV/dt = \pi r^2 (dh/dt) = \pi (2)^2 (-0.3) = \pi (4) (-0.3) = -1.2 \pi \text{ cm}^3/\text{hr}$. The negative sign indicates volume is decreasing as the candle burns down, which matches physical expectation.

Question 3

1. $\dfrac{32}{9}\text{ ft/min}$ (correct answer)
2. $-\dfrac{32}{9}\text{ ft/min}$
3. $\dfrac{2}{9}\text{ ft/min}$
4. $-\dfrac{2}{9}\text{ ft/min}$
5. $16\text{ ft/min}$

Explanation: This problem uses constant area constraint for a rectangle. Given $A = lw = 144$ (constant), when $l = 9$ and $\frac{dl}{dt} = -2$ (negative for decreasing), we find $w = \frac{144}{9} = 16$. Taking the derivative: $l \frac{dw}{dt} + w \frac{dl}{dt} = 0$. Substituting: $9 \frac{dw}{dt} + 16(-2) = 0$, so $9 \frac{dw}{dt} = 32$. Therefore, $\frac{dw}{dt} = \frac{32}{9} \text{ ft/min}$. The positive result indicates width increases as length decreases to maintain constant area.

Question 4

1. $120 \, \text{cm}^2 / \text{min}$ (correct answer)
2. $60 \, \text{cm}^2 / \text{min}$
3. $30 \, \text{cm}^2 / \text{min}$
4. $300 \, \text{cm}^2 / \text{min}$
5. $150 \, \text{cm}^2 / \text{min}$

Explanation: This problem connects volume and surface area rates for a cube. Given $V = s^3$ and $SA = 6s^2$, with $\frac{dV}{dt} = 150 \, \text{cm}^3/\text{min}$ when $s = 5 \, \text{cm}$. First, find $ds/dt$: $150 = 3s^2 (ds/dt) = 3(25)(ds/dt) = 75(ds/dt)$, so $ds/dt = 2 \, \text{cm/min}$. Then $dSA/dt = 12s(ds/dt) = 12(5)(2) = 120 \, \text{cm}^2/\text{min}$. This shows how volume changes translate to surface area changes through the common rate of side length change.

Question 5

1. $-2\pi\text{ cm}^2/\text{s}$ (correct answer)
2. $2\pi\text{ cm}^2/\text{s}$
3. $-0.05\pi\text{ cm}^2/\text{s}$
4. $-20\pi\text{ cm}^2/\text{s}$
5. $-4\pi\text{ cm}^2/\text{s}$

Explanation: This problem demonstrates area rate change for a decreasing circular radius. Given A = πr² and dr/dt = -0.05 cm/s (negative for decreasing) when r = 20 cm, taking the derivative: dA/dt = 2πr(dr/dt) = 2π(20)(-0.05) = -2π cm²/s. The negative sign correctly indicates the area is decreasing as the bracelet contracts. Students might forget the negative sign or miscalculate the coefficient.

Question 6

1. $\dfrac{3}{\sqrt{2}}\text{ cm/min}$
2. $3\sqrt{2}\text{ cm/min}$ (correct answer)
3. $6\text{ cm/min}$
4. $\sqrt{2}\text{ cm/min}$
5. $\dfrac{\sqrt{2}}{3}\text{ cm/min}$

Explanation: This problem applies the relationship between square side length and diagonal. Given $d = s\sqrt{2}$, taking the derivative: $\dfrac{dd}{dt} = \sqrt{2} \dfrac{ds}{dt}$. With $\dfrac{ds}{dt} = 3$ cm/min, we have $\dfrac{dd}{dt} = \sqrt{2}(3) = 3\sqrt{2}$ cm/min. This demonstrates how geometric relationships create proportional rate relationships. Students might forget the $\sqrt{2}$ factor or confuse the direction of the relationship.

Question 7

1. $1\text{ ft/min}$
2. $-1\text{ ft/min}$ (correct answer)
3. $-2\text{ ft/min}$
4. $2\text{ ft/min}$
5. $0\text{ ft/min}$

Explanation: This problem involves related rates with constant perimeter constraint. For a rectangle with perimeter P = 2l + 2w = 40, we have l + w = 20. Taking the derivative: dl/dt + dw/dt = 0. Given l = 12 and dl/dt = 1 ft/min, we substitute: 1 + dw/dt = 0. Solving: dw/dt = -1 ft/min. The negative sign correctly indicates that as length increases, width must decrease to maintain constant perimeter. A common error would be forgetting the constraint or the negative relationship.

Question 8

1. $200\ln(1.5)$ (correct answer)
2. $\ln(1.5)$
3. $300$
4. $200(1.5)$
5. $\dfrac{1}{200\ln(1.5)}$

Explanation: This problem requires differentiating an exponential function. Given P = 200(1.5)^t, taking the derivative: dP/dt = 200(1.5)^t × ln(1.5) using the chain rule for exponential functions. At t = 0, dP/dt = 200(1.5)^0 × ln(1.5) = 200(1) × ln(1.5) = 200ln(1.5). Students might forget the natural logarithm factor or confuse the exponential differentiation rule.

Question 9

1. $0\text{ m}^2/\text{day}$ (correct answer)
2. $4\pi\text{ m}^2/\text{day}$
3. $40\pi\text{ m}^2/\text{day}$
4. $-4\pi\text{ m}^2/\text{day}$
5. $400\pi\text{ m}^2/\text{day}$

Explanation: This problem involves a constant area constraint where the area is unchanging. Given A = πr² = 400π (constant), taking the derivative: dA/dt = 2πr(dr/dt) = 0. Since area is constant, its rate of change must be zero regardless of how radius might be changing. This demonstrates that constraints can override individual parameter changes. Students might mistakenly calculate a non-zero rate based on the radius change.

Question 10

1. $4\pi\text{ mm}^2/\text{hr}$ (correct answer)
2. $2\pi\text{ mm}^2/\text{hr}$
3. $20\pi\text{ mm}^2/\text{hr}$
4. $0.2\pi\text{ mm}^2/\text{hr}$
5. $40\pi\text{ mm}^2/\text{hr}$

Explanation: This problem applies the basic formula for the rate of change of circular area. Given $A = \pi r^2$ and $\dfrac{dr}{dt} = 0.2$ mm/hr when $r = 10$ mm, taking the derivative: $\dfrac{dA}{dt} = 2\pi r (\dfrac{dr}{dt}) = 2\pi(10)(0.2) = 4\pi$ mm$^2$/hr. This is a direct application of the chain rule to the area formula. Students might forget the coefficient 2 from differentiating r$^2$ or miscalculate the substitution.

Question 11

1. $50\pi\text{ m/hr}$
2. $\dfrac{2}{\pi}\text{ m/hr}$
3. $\dfrac{1}{2}\text{ m/hr}$
4. $2\text{ m/hr}$ (correct answer)
5. $\dfrac{1}{2\pi}\text{ m/hr}$

Explanation: This problem demonstrates interpreting related rates for a cylindrical tank with fixed dimensions. For a cylinder, $V = \pi r^2 h$ where r is constant at 5 m. Taking the derivative: $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ since r is constant. Given $\frac{dV}{dt} = 50\pi$ m$^3$/hr and r = 5 m, we substitute:$50\pi = \pi(5)^2 \frac{dh}{dt} = 25\pi \frac{dh}{dt}$. Solving:$\frac{dh}{dt} = \frac{50\pi}{25\pi} = 2$m/hr. Students might incorrectly include dr/dt terms even though radius is fixed, or miscalculate the area$\pi r^2$.

Question 12

1. $200\text{ g/hr}$
2. $600\text{ g/hr}$ (correct answer)
3. $\dfrac{200}{3}\text{ g/hr}$
4. $3\text{ g/hr}$
5. $203\text{ g/hr}$

Explanation: This problem applies linear related rates to a biological context. Given $B = 3n$, taking the derivative: $\dfrac{dB}{dt} = 3 \dfrac{dn}{dt}$. Since $\dfrac{dn}{dt} = 200$ cells/hr, we have $\dfrac{dB}{dt} = 3(200) = 600$ g/hr. This is a straightforward application of the constant multiple rule for derivatives. Students might overthink this problem or forget to apply the coefficient 3 correctly.

Question 13

1. $\dfrac{4}{3}\text{ cm/min}$
2. $\dfrac{2}{3}\text{ cm/min}$
3. $4\text{ cm/min}$ (correct answer)
4. $24\pi\text{ cm/min}$
5. $\dfrac{8}{3}\text{ cm/min}$

Explanation: This problem involves the total surface area of a cylinder including top and bottom. For a cylinder, SA = 2πr² + 2πrh = 2πr(r + h). Taking the derivative with r = 3 constant: dSA/dt = 2π(3)(dh/dt) = 6π(dh/dt). Given dSA/dt = 24π cm²/min, we solve: 24π = 6π(dh/dt), so dh/dt = 4 cm/min. Students might forget to include both circular ends or incorrectly differentiate the surface area formula.

Question 14

1. $-8\pi$
2. $8\pi$
3. $10\pi$ (correct answer)
4. $-10\pi$
5. $18\pi$

Explanation: This problem demonstrates the product rule applied to lateral surface area of a cylinder. Given L = 2πrh with r = 2, h = 9, dr/dt = 1, and dh/dt = -2, taking the derivative: dL/dt = 2π[h(dr/dt) + r(dh/dt)] = 2π[9(1) + 2(-2)] = 2π[9 - 4] = 2π(5) = 10π. The positive result shows lateral area is increasing despite height decreasing, because radius increase dominates.

Question 15

1. $-\dfrac{1}{4}$ (correct answer)
2. $\dfrac{1}{4}$
3. $-5$
4. $5$
5. $-\dfrac{1}{20}$

Explanation: This problem uses the constraint relationship y = 100/x for a hyperbolic function. Taking the derivative: dy/dt = -100/x² × dx/dt. When x = 20 and dx/dt = 1, we have dy/dt = -100/(20)² × 1 = -100/400 = -1/4. The negative sign indicates that as x increases, y decreases to maintain the constant product relationship. Students might forget the negative sign from the chain rule.

Question 16

1. $2\pi\text{ cm}^2/\text{s}$
2. $4\pi\text{ cm}^2/\text{s}$ (correct answer)
3. $8\pi\text{ cm}^2/\text{s}$
4. $20\pi\text{ cm}^2/\text{s}$
5. $0.8\pi\text{ cm}^2/\text{s}$

Explanation: This problem applies area rate change using diameter for a circle. Given A = π(d/2)² = πd²/4 and dd/dt = 0.8 cm/s when d = 10 cm, taking the derivative: dA/dt = π(d/2)(dd/dt) = π(10/2)(0.8) = π(5)(0.8) = 4π cm²/s. This demonstrates how diameter-based formulations affect the derivative calculation compared to radius-based approaches.

Question 17

1. $500\text{ dollars/month}$
2. $20\text{ dollars/month}$ (correct answer)
3. $520\text{ dollars/month}$
4. $\dfrac{1}{20}\text{ months/dollar}$
5. $0\text{ dollars/month}$

Explanation: This problem involves interpreting the rate of change in a linear function. Given $B = 500 + 20t$, taking the derivative with respect to time: $\frac{dB}{dt} = 20$ dollars/month. This represents the constant rate at which the balance increases each month. The constant term 500 disappears upon differentiation, leaving only the coefficient of t. Students might mistakenly include the constant term or confuse the units.

Question 18

1. $-10\pi \text{ cm}^3/\text{min}$ (correct answer)
2. $10\pi \text{ cm}^3/\text{min}$
3. $-\dfrac{5}{2}\pi \text{ cm}^3/\text{min}$
4. $-0.1\pi \text{ cm}^3/\text{min}$
5. $-100\pi \text{ cm}^3/\text{min}$

Explanation: This problem demonstrates related rates with decreasing radius for spherical volume. Given $V = \frac{4}{3} \pi r^3$ and $\frac{dr}{dt} = -0.1 \, \text{cm/min}$ (negative for decreasing) when $r = 5$ cm, taking the derivative: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} = 4 \pi (5)^2 (-0.1) = 4 \pi (25) (-0.1) = -10 \pi \, \text{cm}^3/\text{min}$. The negative sign correctly indicates volume is decreasing as radius decreases. Students might forget the negative sign or miscalculate the derivative.

Question 19

1. $15\text{ cm}^2/\text{s}$
2. $30\text{ cm}^2/\text{s}$ (correct answer)
3. $60\text{ cm}^2/\text{s}$
4. $3\text{ cm}^2/\text{s}$
5. $45\text{ cm}^2/\text{s}$

Explanation: This problem connects diagonal and area rates for a square. For a square with diagonal $d$, the side length is $s = \frac{d}{\sqrt{2}}$, so area $A = s^2 = \frac{d^2}{2}$. Taking the derivative: $\frac{dA}{dt} = d \frac{dd}{dt}$. When $d = 10$ cm and $\frac{dd}{dt} = 3$ cm/s, we have $\frac{dA}{dt} = 10(3) = 30$ cm²/s. Students might incorrectly use the side length relationship or forget to properly convert between diagonal and area formulas.

Question 20

1. $-20\text{ dollars/day}$ (correct answer)
2. $20\text{ dollars/day}$
3. $-0.4\text{ dollars/day}$
4. $-50\text{ dollars/day}$
5. $50\text{ dollars/day}$

Explanation: This problem demonstrates interpreting rates in a linear business context. Given $R = 50p$, taking the derivative: $\dfrac{dR}{dt} = 50 \dfrac{dp}{dt}$. Since price decreases at 0.4 dollars/day, $\dfrac{dp}{dt} = -0.4$. Therefore, $\dfrac{dR}{dt} = 50(-0.4) = -20$ dollars/day. The negative sign indicates revenue is decreasing as price decreases. A common error would be forgetting the negative sign or confusing the direction of change.