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AP Calculus AB Quiz

AP Calculus AB Quiz: Solving Related Rates Problems

Practice Solving Related Rates Problems in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 13

0 of 13 answered

A circle’s radius decreases at 111 cm/s; what is dAdt\dfrac{dA}{dt}dtdA​ when r=7r=7r=7, given A=πr2A=\pi r^2A=πr2?

Select an answer to continue

What this quiz covers

This quiz focuses on Solving Related Rates Problems, 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.

All questions

Question 1

A circle’s radius decreases at 111 cm/s; what is dAdt\dfrac{dA}{dt}dtdA​ when r=7r=7r=7, given A=πr2A=\pi r^2A=πr2?

  1. −14π-14\pi−14π cm2^22/s (correct answer)
  2. 14π14\pi14π cm2^22/s
  3. −7π-7\pi−7π cm2^22/s
  4. −49π-49\pi−49π cm2^22/s
  5. 49π49\pi49π cm2^22/s

Explanation: Related rates in AP Calculus AB for areas like circle A=πr2A=\pi r^2A=πr2 with dr/dt=−1dr/dt=-1dr/dt=−1, dA/dt=2πr⋅dr/dtdA/dt=2\pi r \cdot dr/dtdA/dt=2πr⋅dr/dt. At r=7, dA/dt=2π⋅7⋅(−1)=−14πdA/dt=2\pi \cdot 7 \cdot (-1)=-14\pidA/dt=2π⋅7⋅(−1)=−14π. Indicates decreasing area. A common error is using πr⋅dr/dt\pi r \cdot dr/dtπr⋅dr/dt, getting −7π-7\pi−7π. Include the 2 from power rule. The correct answer is −14π-14\pi−14π cm2^22/s. A transferable strategy is to differentiate power functions carefully and interpret negative rates as decreases.

Question 2

A sphere’s radius increases at 222 cm/s; what is dVdt\dfrac{dV}{dt}dtdV​ when r=3r=3r=3 cm, given V=43πr3V=\frac{4}{3}\pi r^3V=34​πr3?

  1. 72π72\pi72π cm3^33/s (correct answer)
  2. 24π24\pi24π cm3^33/s
  3. 18π18\pi18π cm3^33/s
  4. 36π36\pi36π cm3^33/s
  5. 8π8\pi8π cm3^33/s

Explanation: This problem involves related rates, where the volume of a sphere changes as its radius increases. The volume is given by V = (4/3)πr³, and we differentiate both sides with respect to time t to get dV/dt = 4πr² dr/dt. Given dr/dt = 2 cm/s and r = 3 cm, we substitute to find dV/dt = 4π(3)²(2) = 72π cm³/s. A tempting misinterpretation is to plug in the values before differentiating, which would incorrectly ignore the chain rule. Instead, always differentiate first and then substitute the known values. The correct reasoning confirms the rate of volume increase at that instant. A transferable strategy is to identify the relating equation, differentiate implicitly with respect to time, solve for the desired rate, and evaluate at the given point.

Question 3

For a rectangle with constant perimeter P=60P=60P=60, length increases at 222 cm/s; find dAdt\frac{dA}{dt}dtdA​ when L=20L=20L=20.

  1. −20-20−20 cm2^22/s (correct answer)
  2. 404040 cm2^22/s
  3. −40-40−40 cm2^22/s
  4. 202020 cm2^22/s
  5. 000 cm2^22/s

Explanation: This related rates problem deals with a rectangle of constant perimeter P = 60 = 2L + 2W, so W = 30 - L. The area A = L W = L(30 - L), and dA/dt = (30 - 2L) dL/dt. At L = 20 and dL/dt = 2 cm/s, dA/dt = (30 - 40)(2) = -20 cm²/s. A common mistake is to assume area increases when length does, ignoring the decreasing width. The negative rate indicates area decrease past the maximum. A transferable strategy is to express dependent variables in terms of one, differentiate, and substitute rates and values carefully.

Question 4

A rectangle has diagonal 101010 with sides x,yx,yx,y so x2+y2=100x^2+y^2=100x2+y2=100; if dxdt=1\frac{dx}{dt}=1dtdx​=1, find dydt\frac{dy}{dt}dtdy​ at x=8x=8x=8.

  1. −86-\frac{8}{6}−68​
  2. −68-\frac{6}{8}−86​
  3. 68\frac{6}{8}86​
  4. −43-\frac{4}{3}−34​ (correct answer)
  5. 43\frac{4}{3}34​

Explanation: Related rates in AP Calculus AB connect changing quantities through implicit differentiation with respect to time. For a rectangle with diagonal 10, so x2+y2=100x^2 + y^2 = 100x2+y2=100 and dxdt=1\frac{dx}{dt} = 1dtdx​=1, differentiate to get 2xdxdt+2ydydt=02x \frac{dx}{dt} + 2y \frac{dy}{dt} = 02xdtdx​+2ydtdy​=0, solving for dydt=−(xy)dxdt\frac{dy}{dt} = -\left(\frac{x}{y}\right) \frac{dx}{dt}dtdy​=−(yx​)dtdx​. At x=8x = 8x=8, y=6y = 6y=6, dydt=−86×1=−43\frac{dy}{dt} = -\frac{8}{6} \times 1 = -\frac{4}{3}dtdy​=−68​×1=−34​. A tempting error is using y=100−64=6y = \sqrt{100 - 64} = 6y=100−64​=6 but forgetting the negative sign, getting 43\frac{4}{3}34​. Remember rates can be negative indicating decrease. The correct answer is −43-\frac{4}{3}−34​. A transferable strategy is to solve for the unknown rate after differentiating and consider signs based on context.

Question 5

A right triangle has legs xxx and yyy with hypotenuse 131313; if dxdt=5\frac{dx}{dt}=5dtdx​=5, find dydt\frac{dy}{dt}dtdy​ when x=5x=5x=5.

  1. −2512-\frac{25}{12}−1225​ (correct answer)
  2. 2512\frac{25}{12}1225​
  3. −1225-\frac{12}{25}−2512​
  4. −512-\frac{5}{12}−125​
  5. 512\frac{5}{12}125​

Explanation: In this related rates problem, a right triangle has legs x, y and hypotenuse 13, so x² + y² = 169. Differentiate: 2x dx/dt + 2y dy/dt = 0, dy/dt = -(x/y) dx/dt. At x = 5, y = 12, dx/dt = 5, dy/dt = -(5/12)(5) = -25/12. A common mistake is to use the hypotenuse in the rate formula incorrectly. The negative indicates y decreases as x increases. A transferable strategy is to use the Pythagorean theorem, differentiate, and plug in after solving for the rate.

Question 6

A point moves on x2−y2=15x^2 - y^2 = 15x2−y2=15; if dxdt=2\frac{dx}{dt} = 2dtdx​=2, find dydt\frac{dy}{dt}dtdy​ at (4,1)(4,1)(4,1).

  1. 444
  2. −4-4−4
  3. −8-8−8
  4. 888 (correct answer)
  5. −2-2−2

Explanation: For hyperbolas in AP Calculus AB related rates, x2−y2=15x^2 - y^2 = 15x2−y2=15 with dxdt=2\frac{dx}{dt} = 2dtdx​=2, differentiate to 2xdxdt−2ydydt=02x \frac{dx}{dt} - 2y \frac{dy}{dt} = 02xdtdx​−2ydtdy​=0, so dydt=xydxdt\frac{dy}{dt} = \frac{x}{y} \frac{dx}{dt}dtdy​=yx​dtdx​. At (4,1), dydt=(4/1)∗2=8\frac{dy}{dt} = (4/1)*2 = 8dtdy​=(4/1)∗2=8. This shows the relation. A mistake might be switching signs, getting -8 for y2−x2y^2 - x^2y2−x2. Check the equation form. The correct answer is 8. A transferable strategy is to differentiate conic sections implicitly and isolate the desired rate using point values.

Question 7

A kite string length sss satisfies s2=x2+144s^2=x^2+144s2=x2+144; if dxdt=2\frac{dx}{dt}=2dtdx​=2, find dsdt\frac{ds}{dt}dtds​ when x=5x=5x=5.

  1. 513\frac{5}{13}135​
  2. 213\frac{2}{13}132​
  3. 135\frac{13}{5}513​
  4. 1013\frac{10}{13}1310​ (correct answer)
  5. 113\frac{1}{13}131​

Explanation: Related rates in AP Calculus AB help analyze how rates interconnect in geometric contexts like distances. For kite string s² = x² + 144 with dx/dt = 2, differentiate to 2s ds/dt = 2x dx/dt, so ds/dt = (x/s) dx/dt. At x = 5, s = 13, ds/dt = (5/13)*2 = 10/13. A mistake might be using ds/dt = dx/dt without the ratio, getting 2. Always include the chain rule factors. The correct answer is 10/13. A transferable strategy is to differentiate squared relations carefully and solve for the target rate using current values.

Question 8

Water fills a cone with V=13πr2hV=\frac{1}{3}\pi r^2hV=31​πr2h and r=12hr=\frac{1}{2}hr=21​h; if dhdt=3\frac{dh}{dt}=3dtdh​=3, find dVdt\frac{dV}{dt}dtdV​ at h=4h=4h=4.

  1. 16π16\pi16π units3^33/s
  2. 8π8\pi8π units3^33/s
  3. 12π12\pi12π units3^33/s (correct answer)
  4. 4π4\pi4π units3^33/s
  5. 24π24\pi24π units3^33/s

Explanation: This related rates problem concerns the volume of a cone where the radius r is half the height h, so r = h/2. Substitute into V = (1/3)πr²h to get V = (1/3)π(h/2)²h = (π/12)h³. Differentiate with respect to t: dV/dt = (π/4)h² dh/dt. At h = 4 and dh/dt = 3, dV/dt = (π/4)(16)(3) = 12π units³/s. A common mistake is to differentiate without substituting r in terms of h, leading to extra variables. The correct approach yields 12π by expressing everything in terms of one variable before differentiating. A transferable strategy is to express the quantity in terms of a single changing variable when possible, differentiate, and plug in values.

Question 9

A rectangle has constant area A=100A=100A=100 with A=LWA=LWA=LW; if dLdt=2\frac{dL}{dt}=2dtdL​=2, find dWdt\frac{dW}{dt}dtdW​ when L=20L=20L=20.

  1. −12-\frac{1}{2}−21​ (correct answer)
  2. 12\frac{1}{2}21​
  3. −2-2−2
  4. −15-\frac{1}{5}−51​
  5. 15\frac{1}{5}51​

Explanation: Related rates in AP Calculus AB for constant areas like A=100=LW with dL/dt=2, differentiate L dW/dt + W dL/dt=0, dW/dt=-(W/L) dL/dt. At L=20, W=5, dW/dt=-(5/20)*2=-1/2. Shows inverse change. A mistake might be using dW/dt = -dL/dt, getting -2. Include the ratio. The correct answer is -1/2. A transferable strategy is to set dA/dt=0 for constants and solve for compensatory rates.

Question 10

A triangle has area A=12xyA=\frac{1}{2}x yA=21​xy; if dxdt=4\frac{dx}{dt}=4dtdx​=4 and y=10y=10y=10 constant, find dAdt\frac{dA}{dt}dtdA​.

  1. 404040
  2. 202020 (correct answer)
  3. 101010
  4. 808080
  5. 444

Explanation: Related rates in AP Calculus AB apply to areas like triangle A=(1/2)xy with dx/dt=4 and y=10 constant. Then dA/dt = (1/2)(y dx/dt) since dy/dt=0, giving (1/2)104=20. No x value needed as it's independent. A common mistake is unnecessarily including x or forgetting the 1/2. Verify which variables change. The correct answer is 20. A transferable strategy is to set rates of constants to zero and simplify the differentiated equation.

Question 11

A balloon’s surface area is S=4πr2S=4\pi r^2S=4πr2; if drdt=12\frac{dr}{dt}=\frac{1}{2}dtdr​=21​, find dSdt\frac{dS}{dt}dtdS​ when r=6r=6r=6.

  1. 24π units2/s24\pi \text{ units}^2 / \text{s}24π units2/s (correct answer)
  2. 12π units2/s12\pi \text{ units}^2 / \text{s}12π units2/s
  3. 48π units2/s48\pi \text{ units}^2 / \text{s}48π units2/s
  4. 6π units2/s6\pi \text{ units}^2 / \text{s}6π units2/s
  5. 72π units2/s72\pi \text{ units}^2 / \text{s}72π units2/s

Explanation: This problem applies related rates to the surface area of a balloon S=4πr2S = 4\pi r^2S=4πr2. Differentiate: dSdt=8πrdrdt\frac{dS}{dt} = 8\pi r \frac{dr}{dt}dtdS​=8πrdtdr​. At r=6r = 6r=6 and drdt=12\frac{dr}{dt} = \frac{1}{2}dtdr​=21​, dSdt=8π(6)(12)=24π\frac{dS}{dt} = 8\pi(6)(\frac{1}{2}) = 24\pidtdS​=8π(6)(21​)=24π units2^22/s. A common mistake is to use the volume derivative instead, leading to incorrect powers. The rate reflects faster area growth for larger radii. A transferable strategy is to differentiate surface or volume formulas, incorporate given rates, and evaluate at specified points.

Question 12

A circle expands with drdt=4\frac{dr}{dt}=4dtdr​=4; what is dAdt\frac{dA}{dt}dtdA​ when r=2r=2r=2, given A=πr2A=\pi r^2A=πr2?

  1. 8π8\pi8π
  2. 16π16\pi16π (correct answer)
  3. 32π32\pi32π
  4. 4π4\pi4π
  5. 64π64\pi64π

Explanation: This related rates problem involves circle area A=πr2A = \pi r^2A=πr2 with drdt=4\frac{dr}{dt} = 4dtdr​=4. Differentiate: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}dtdA​=2πrdtdr​. At r=2r=2r=2, dAdt=2π(2)(4)=16π\frac{dA}{dt} = 2\pi(2)(4) = 16\pidtdA​=2π(2)(4)=16π. A tempting misinterpretation is using the circumference derivative. The rate doubles with radius. A transferable strategy is to apply the power rule with chain rule for geometric rates.

Question 13

A cube’s surface area is S=6s2S=6s^2S=6s2; if dsdt=2\frac{ds}{dt}=2dtds​=2, find dSdt\frac{dS}{dt}dtdS​ when s=3s=3s=3.

  1. 242424
  2. 727272 (correct answer)
  3. 363636
  4. 121212
  5. 108108108

Explanation: This related rates problem finds the surface area change for a cube S=6s². Differentiate: dS/dt=12s ds/dt. At s=3, ds/dt=2, dS/dt=12(3)(2)=72. A tempting misinterpretation is using volume instead. The rate increases linearly with s. A transferable strategy is to differentiate polynomial expressions and evaluate step-by-step.