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Related rates: write the equation, differentiate in time, then plug in

In a related rates problem, two or more quantities change together, you know how fast one is changing, and you need how fast another is. The method is always the same three steps, and the most common mistake is doing the third step before the second.

Why it works

Every quantity in the problem is a function of time, even if the problem never writes t. If an equation like x² + y² = 100 is true at every moment, then both sides change at the same rate, so their derivatives with respect to t are equal. The chain rule is what turns the derivative of x² into 2x · dx/dt: x depends on t, so its own rate comes along. That new equation connects the rates, the same way the original one connected the quantities.

The numbers in the question describe one instant. If you plug them in before differentiating, you're treating a changing quantity as a constant, and its rate disappears from the equation.

Example 1: a growing circle

A ripple spreads in a pond, and its radius grows at 3 cm/s. How fast is the area inside it growing when the radius is 10 cm?

1. A = πr²True at every moment.
2. dAdt = 2πr drdtDifferentiate in t. The chain rule brings in dr/dt.
3. = 2π · 10 · 3Now plug in r = 10 and dr/dt = 3.
4. 60π ≈ 188.5 cm²/sThe units are area per time.

Example 2: a sliding ladder

A 10 ft ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?

1. x² + y² = 100x is the bottom's distance from the wall, y is the top's height. The ladder's length doesn't change.
2. 2x dxdt + 2y dydt = 0Differentiate in t. The constant 100 differentiates to 0.
3. x = 6, y = √(100 − 36) = 8, dx/dt = 2The values at this instant. y comes from the original equation.
4. dydt = −xy · dxdt = −68 · 2Solve step 2 for dy/dt and substitute.
5. dy/dt = −1.5 ft/sNegative because the height is decreasing: the top slides down at 1.5 ft/s.

Example 3: filling a cone

Water pours at 9 ft³/min into a tank shaped like an upside-down cone, 10 ft tall with a top radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?

1. r = h2Similar triangles: the water is a smaller cone with the same shape, and 5/10 = 1/2. This holds at every depth.
2. V = 13πr²h = 13πh²4h = πh³12Replace r using step 1, so only one changing quantity is left.
3. dVdt = πh²4 · dhdtDifferentiate in t.
4. 9 = π · 364 · dhdt = 9π dhdtPlug in dV/dt = 9 and h = 6.
5. dh/dt = 1π ≈ 0.318 ft/minThe level rises more slowly as the cone gets wider.

Example 4: a plane and a radar station

A plane flies at an altitude of 6 mi, at 500 mph, on a path that passes directly over a radar station. How fast is its distance from the station changing when that distance is 10 mi and the plane is still approaching?

1. s² = x² + 6²x is the horizontal distance to the point above the station, s the straight-line distance. The altitude stays 6.
2. 2s dsdt = 2x dxdtDifferentiate in t. The 6² is a constant.
3. s = 10, x = √(100 − 36) = 8, dx/dt = −500The plane is approaching, so x is shrinking and its rate is negative.
4. ds/dt = 8 · (−500)10 = −400 mphThe distance is decreasing at 400 mph.

Common mistakes, and the exact line they happen on

1. Plugging in before differentiating

The cone in Example 3, at the moment the water is 6 ft deep.

1. r = 3 when h = 6Right so far: r = h/2.
2. V = 13π · 3² · h = 3πh, so 9 = 3π dhdtThis is the mistake. r is 3 only at this instant. Writing it in before differentiating treats the radius as fixed, as if the tank were a cylinder. That gives dh/dt = 3/π ≈ 0.955 ft/min, three times the real answer. Differentiate V = πh³/12 first, then plug in: 1/π ≈ 0.318 ft/min.

2. Dropping the chain rule's rate

The ripple in Example 1.

1. dAdt = 2πr = 20πThis is the mistake. 2πr is dA/dr, the change in area per centimeter of radius. The question asks per second. The chain rule multiplies by dr/dt = 3, giving 60π cm²/s. If your answer doesn't use the rate you were given, something is missing.

3. A positive rate for a shrinking distance

The plane in Example 4.

1. s dsdt = x dxdt, s = 10, x = 8Right so far.
2. dx/dt = 500, so ds/dt = 400 mphThis is the mistake. 500 mph is the plane's speed, but x is the distance still to go, and it's getting smaller. Its rate is −500, and the distance to the station is decreasing at 400 mph, not increasing. For each variable, ask whether it's growing or shrinking before you plug in.

Practice

1. A sphere's radius grows at 0.5 cm/s. How fast is its volume growing when the radius is 4 cm?

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   V = (4/3)πr³, so dV/dt = 4πr² · dr/dt = 4π · 16 · 0.5 = 32π ≈ 100.5 cm³/s.

2. The side of a square grows at 0.2 cm/s. How fast is its area growing when the side is 8 cm?

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   A = s², so dA/dt = 2s · ds/dt = 2 · 8 · 0.2 = 3.2 cm²/s.

3. A 13 ft ladder's top slides down a wall at 1 ft/s. How fast is the bottom moving away from the wall when the top is 5 ft high?

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   x² + y² = 169, so x · dx/dt + y · dy/dt = 0. At y = 5, x = 12, and dy/dt = −1: 12 · dx/dt = 5, so the bottom moves at 5/12 ≈ 0.417 ft/s.

4. Water flows at 3 m³/min into a cylindrical tank of radius 2 m. How fast is the level rising?

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   V = π · 2² · h = 4πh, so 3 = 4π · dh/dt and dh/dt = 3/(4π) ≈ 0.239 m/min, at every depth: the radius really is constant here.

5. A person 6 ft tall walks away from a 15 ft lamppost at 5 ft/s. How fast is their shadow getting longer?

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   Similar triangles: 15/(x + s) = 6/s, so 9s = 6x and s = 2x/3. Then ds/dt = (2/3) · 5 = 10/3 ≈ 3.33 ft/s, wherever the person is.