AP Calculus AB · Lesson 2

Related rates

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In a related-rates problem, two or more quantities change with time and are tied together by an equation. You know some rates and want another.

The method: 1. Name every changing quantity and write down the rates you know and the rate you want. 2. Write an equation that links the quantities (geometry, the Pythagorean theorem, similar triangles, a trig ratio). 3. Differentiate both sides with respect to tt; each changing variable picks up its own rate, such as dxdt\frac{dx}{dt}, by the chain rule. 4. Substitute the values for the moment asked about and solve. 5. Answer with units, and check the sign: a quantity that is shrinking has a negative rate.

If the equation has one variable too many, use a fixed relationship (for example, a shape whose radius is always half its height) to remove it before you differentiate.

Trap Do not put in the moment’s values too early

Values like "when the radius is 10 cm" are true only at one instant. If you substitute r=10r = 10 into A=πr2A = \pi r^2 first, you get A=100πA = 100\pi, a constant, and its derivative is 0. Differentiate first, then substitute.

The edges of a cube of ice shrink at 0.2 centimeter per minute as it melts. How fast is the volume changing when each edge is 10 centimeters long?

  1. -200 cubic centimeters per minute
  2. -60 cubic centimeters per minuteAnswer
  3. -20 cubic centimeters per minute
  4. -6 cubic centimeters per minute

How to solve it

  1. Link the quantities: V=s3V = s^3, with dsdt=−0.2\frac{ds}{dt} = -0.2 (negative, because the edges shrink).
  2. Differentiate with respect to tt: dVdt=3s2dsdt\frac{dV}{dt} = 3s^2 \frac{ds}{dt}.
  3. Substitute s=10s = 10: 3(100)(−0.2)=−603(100)(-0.2) = -60 cubic centimeters per minute.

Why each choice is right or wrong

  • A. Incorrect. -200 uses s3dsdt=1000(−0.2)s^3 \frac{ds}{dt} = 1000(-0.2); the derivative of s3s^3 is 3s23s^2, not s3s^3.
  • B. Correct. V=s3V = s^3, so dVdt=3s2dsdt=3(10)2(−0.2)=−60\frac{dV}{dt} = 3s^2 \frac{ds}{dt} = 3(10)^2(-0.2) = -60 cubic centimeters per minute.
  • C. Incorrect. -20 is s2dsdt=100(−0.2)s^2 \frac{ds}{dt} = 100(-0.2), which drops the factor 3 from the power rule.
  • D. Incorrect. -6 is 3sdsdt=30(−0.2)3s \frac{ds}{dt} = 30(-0.2), which uses 3s3s in place of 3s23s^2.

Explain Explain it

Task verbs: calculate, explain

Right triangles are the most common link. If a horizontal distance xx and a fixed height hh make a straight-line distance DD, then D2=x2+h2D^2 = x^2 + h^2 and 2DdDdt=2xdxdt2D\frac{dD}{dt} = 2x\frac{dx}{dt}, because hh is constant.

Explain your setup in words: which quantities change, which stay fixed, and why a rate is negative. A reader should see the equation before any numbers.

A drone flies horizontally at a constant height of 30 meters, moving at 4 meters per second away from a point directly above a person standing on the ground. How fast is the distance between the person and the drone increasing when the drone is 40 meters from that point, measured horizontally?

  1. 2.4 meters per second
  2. 3.2 meters per secondAnswer
  3. 4 meters per second
  4. 163\frac{16}{3} meters per second

How to solve it

  1. Let xx be the horizontal distance and DD the distance to the person: D2=x2+302D^2 = x^2 + 30^2.
  2. Differentiate: 2DdDdt=2xdxdt2D\frac{dD}{dt} = 2x\frac{dx}{dt}, so dDdt=xDdxdt\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt}.
  3. When x=40x = 40, D=402+302=50D = \sqrt{40^2 + 30^2} = 50, so dDdt=4050(4)=3.2\frac{dD}{dt} = \frac{40}{50}(4) = 3.2 meters per second.

Why each choice is right or wrong

  • A. Incorrect. 2.4 is 3050(4)\frac{30}{50}(4): it uses the height, 30, where the horizontal distance, 40, belongs.
  • B. Correct. D2=x2+302D^2 = x^2 + 30^2, so dDdt=xDdxdt=4050(4)=3.2\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt} = \frac{40}{50}(4) = 3.2 meters per second.
  • C. Incorrect. 4 is the drone’s horizontal speed; the straight-line distance grows more slowly, because part of it is the fixed height.
  • D. Incorrect. 163\frac{16}{3} is 4030(4)\frac{40}{30}(4), which divides by the height instead of by the distance D=50D = 50.

The side of a square photo on a screen grows at 0.5 centimeter per second. Find how fast the photo’s area is growing when the side is 8 centimeters. Then say why the area does not grow at the same rate at every moment.

Compare with a model answer

A full answer does this

  • Differentiates A=s2A = s^2 with respect to time to get dAdt=2sdsdt\frac{dA}{dt} = 2s\frac{ds}{dt}
  • Finds that the area grows at 8 square centimeters per second
  • Says the area’s rate depends on the side length, which keeps changing

Model answer

With A=s2A = s^2, differentiating with respect to time gives dAdt=2sdsdt\frac{dA}{dt} = 2s\frac{ds}{dt}. When s=8s = 8, dAdt=2(8)(0.5)=8\frac{dA}{dt} = 2(8)(0.5) = 8 square centimeters per second. The rate is not the same at every moment, because 2sdsdt2s\frac{ds}{dt} depends on the side length ss: as the side grows, the area grows faster.

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