AP Calculus AB Related Rates Worksheet Solutions (PDF)
This related rates answers for AP Calculus AB covers related rates (Common Core AP CALC AB 4.4), with a free PDF to print.
This answer key gives a full worked solution for every problem in the related rates worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.
AP Calculus AB · Common Core State Standards: AP CALC AB 4.4
Related rates answers: exercises 1 to 10
Exercise 1 – Growing ripple
\(A=\pi r^{2}\), so \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\).
Substitute \(r=10\) and \(\frac{dr}{dt}=3\): \(2\pi(10)(3)=60\pi\).
Final answer: \(60\pi\approx 188.5\text{ ft}^{2}\text{/s}\).
Exercise 2 – Melting ice cube
\(V=x^{3}\), so \(\frac{dV}{dt}=3x^{2}\frac{dx}{dt}\). With \(-6=3(9)\frac{dx}{dt}\): \(\frac{dx}{dt}=-\frac{2}{9}\).
\(S=6x^{2}\), so \(\frac{dS}{dt}=12x\frac{dx}{dt}=12(3)\left(-\frac{2}{9}\right)=-8\).
Final answer: \(\frac{dx}{dt}=-\frac{2}{9}\text{ cm/min;}\ \frac{dS}{dt}=-8\text{ cm}^{2}\text{/min}\).
Exercise 3 – Growing cube
\(V=x^{3}\) gives \(\frac{dV}{dt}=3x^{2}\frac{dx}{dt}=3(9)(2)=54\).
\(S=6x^{2}\) gives \(\frac{dS}{dt}=12x\frac{dx}{dt}=12(3)(2)=72\).
Final answer: \(54\text{ cm}^{3}\text{/s and }72\text{ cm}^{2}\text{/s}\).
Exercise 4 – Sliding ladder
\(x^{2}+y^{2}=169\). Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\).
When \(x=5\), \(y=12\). So \(10(2)+24\frac{dy}{dt}=0\) and \(\frac{dy}{dt}=-\frac{5}{6}\).
Final answer: \(-\frac{5}{6}\text{ ft/s (the top slides down)}\).
Exercise 5 – Rectangle
\(A=LW\), so \(\frac{dA}{dt}=L^{\prime}W+LW^{\prime}=3(5)+8(-2)=-1\).
\(P=2L+2W\), so \(\frac{dP}{dt}=2(3)+2(-2)=2\).
Final answer: \(\frac{dA}{dt}=-1\text{ in}^{2}\text{/s},\ \frac{dP}{dt}=2\text{ in/s}\).
Exercise 6 – Two cars
After 0.5 h: \(y=30\) and \(x=20\) miles, so \(D=\sqrt{20^{2}+30^{2}}=\sqrt{1300}=10\sqrt{13}\).
\(D^{2}=x^{2}+y^{2}\) gives \(D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\), so \(\frac{dD}{dt}=\frac{20(40)+30(60)}{10\sqrt{13}}=\frac{2600}{10\sqrt{13}}=\frac{260}{\sqrt{13}}=20\sqrt{13}\).
Final answer: \(20\sqrt{13}\approx 72.1\text{ mph}\).
Exercise 7 – Cone-shaped cup
Similar triangles: \(\frac{r}{h}=\frac{3}{6}\), so \(r=\frac{h}{2}\). Then \(V=\frac{1}{3}\pi\left(\frac{h}{2}\right)^{2}h=\frac{\pi}{12}h^{3}\).
\(\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}\). With \(h=4\): \(2=\frac{\pi}{4}(16)\frac{dh}{dt}=4\pi\frac{dh}{dt}\).
Final answer: \(\frac{dh}{dt}=\frac{1}{2\pi}\approx 0.159\text{ in/s}\).
Exercise 8 – Shadow length
Let \(x\) be her distance from the light and \(T\) the distance to the tip of the shadow. Similar triangles: \(\frac{T}{15}=\frac{T-x}{6}\), so \(6T=15T-15x\) and \(T=\frac{5}{3}x\).
\(\frac{dT}{dt}=\frac{5}{3}(4)=\frac{20}{3}\). The shadow length is \(T-x=\frac{2}{3}x\), so it grows at \(\frac{2}{3}(4)=\frac{8}{3}\).
Final answer: \(\frac{20}{3}\approx 6.67\text{ ft/s and }\frac{8}{3}\approx 2.67\text{ ft/s}\).
Exercise 9 – Angle of elevation
\(\tan\theta=\frac{h}{100}\), so \(\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{100}\frac{dh}{dt}\).
At \(h=100\), \(\theta=\frac{\pi}{4}\) and \(\sec^{2}\theta=2\). So \(2\frac{d\theta}{dt}=\frac{5}{100}\).
Final answer: \(\frac{d\theta}{dt}=0.025\text{ rad/s}\).
Exercise 10 – Sand pile
\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi h^{3}\). So \(\frac{dV}{dt}=\pi h^{2}\frac{dh}{dt}\).
\(10=\pi(25)\frac{dh}{dt}\), so \(\frac{dh}{dt}=\frac{2}{5\pi}\).
Final answer: \(\frac{2}{5\pi}\approx 0.127\text{ ft/min}\).
Exercises 11 and up
Exercise 11 – Melting sphere
\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=4\pi(16)(-0.5)=-32\pi\).
Final answer: \(-32\pi\approx -100.5\text{ cm}^{3}\text{/min}\).
Exercise 12 – Circle with circumference
\(C=2\pi r\), so \(\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{2}{\pi}\). At \(C=20\pi\), \(r=10\).
\(\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi(10)\frac{2}{\pi}=40\).
Final answer: \(40\text{ cm}^{2}\text{/s}\).
Exercise 13 – Gas under pressure
At \(V=10\), \(P=20\). Differentiate: \(\frac{dP}{dt}V+P\frac{dV}{dt}=0\).
\(\frac{dP}{dt}(10)+20(-3)=0\), so \(\frac{dP}{dt}=6\).
Final answer: \(6\text{ units per second (the pressure is rising)}\).
Exercise 14 – Draining cylinder
\(V=\pi r^{2}h=4\pi h\), so \(\frac{dV}{dt}=4\pi\frac{dh}{dt}\).
\(-12=4\pi\frac{dh}{dt}\) gives \(\frac{dh}{dt}=-\frac{3}{\pi}\).
Final answer: \(-\frac{3}{\pi}\approx -0.955\text{ ft/min}\).
Exercise 15 – Kite
\(\ell^{2}=x^{2}+80^{2}\). At \(\ell=100\), \(x=60\).
\(\ell\frac{d\ell}{dt}=x\frac{dx}{dt}\), so \(\frac{d\ell}{dt}=\frac{60(5)}{100}=3\).
Final answer: \(3\text{ ft/s}\).
Exercise 16 – Boat and dock
\(x^{2}+64=L^{2}\). At \(L=10\), \(x=6\).
\(x\frac{dx}{dt}=L\frac{dL}{dt}\), so \(6\frac{dx}{dt}=10(-1.5)\) and \(\frac{dx}{dt}=-2.5\).
Final answer: \(2.5\text{ ft/s toward the dock}\).
Exercise 17 – Runner on a baseball diamond
\(D^{2}=90^{2}+s^{2}\). At \(s=30\), \(D=\sqrt{9000}=30\sqrt{10}\).
\(D\frac{dD}{dt}=s\frac{ds}{dt}\), so \(\frac{dD}{dt}=\frac{30(20)}{30\sqrt{10}}=\frac{20}{\sqrt{10}}=2\sqrt{10}\).
Final answer: \(2\sqrt{10}\approx 6.32\text{ ft/s}\).
Exercise 18 – Approaching plane
\(z^{2}=x^{2}+9\). At \(x=4\), \(z=5\).
\(z\frac{dz}{dt}=x\frac{dx}{dt}\), so \(\frac{dz}{dt}=\frac{4(-480)}{5}=-384\).
Final answer: \(-384\text{ mph}\).
Exercise 19 – Triangle with changing angle
\(A=\frac{1}{2}(6)(8)\sin\theta=24\sin\theta\), so \(\frac{dA}{dt}=24\cos\theta\frac{d\theta}{dt}\).
At \(\theta=\frac{\pi}{3}\): \(24\cdot\frac{1}{2}\cdot 0.1=1.2\).
Final answer: \(1.2\text{ cm}^{2}\text{/s}\).
Exercise 20 – Oil spill
\(\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi(50)(2)=200\pi\).
Final answer: \(200\pi\approx 628.3\text{ ft}^{2}\text{/min}\).
Exercise 21 – Triangular trough
At depth \(h\), the width of the water surface is \(2h\) (similar triangles: \(\frac{w}{h}=\frac{4}{2}\)). The cross-section area is \(\frac{1}{2}(2h)(h)=h^{2}\), so \(V=10h^{2}\).
\(\frac{dV}{dt}=20h\frac{dh}{dt}\). At \(h=1\): \(3=20\frac{dh}{dt}\).
Final answer: \(\frac{dh}{dt}=\frac{3}{20}=0.15\text{ ft/min}\).
Exercise 22 – Volume and surface area
\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=4\pi(36)(0.4)=57.6\pi\). \(S=4\pi r^{2}\) gives \(\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(6)(0.4)=19.2\pi\).
Ratio: \(\frac{57.6\pi}{19.2\pi}=3=\frac{6}{2}\).
Final answer: \(57.6\pi\approx 181\text{ cm}^{3}\text{/s;}\ 19.2\pi\approx 60.3\text{ cm}^{2}\text{/s}\).
Key facts behind this related rates answers
- Differentiate the relating equation with respect to time t, using the chain rule on every variable.
- Substitute instant values only after differentiating; constants such as a ladder length stay constant.
- A quantity that is decreasing has a negative rate.
- Use similar triangles or a given ratio to reduce a formula to one variable before differentiating.
How to use this related rates answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
Download the related rates answers as a PDF with the button below and keep it with your notes.
Download the related rates answers as a PDF with the button below and keep it with your notes.
Download the related rates answers as a PDF with the button below and keep it with your notes.
Download the related rates answers as a PDF with the button below and keep it with your notes.
Download the related rates answers as a PDF with the button below and keep it with your notes.
Download the related rates answers as a PDF with the button below and keep it with your notes.
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