∑Richard Hummel Math

Related Rates

Using implicit differentiation with respect to time to find how fast one quantity changes when another is changing.

Expanding ripple: area grows as a thin ring around the edge
A growing radius sweeps out new areanew ringnew area per secondrdr/dtoutward speedarea rate = circumference × edge speed
Given
At this instant, r = 5 m and dr/dt = 2 m/s.
Geometry
The new area is a thin ring, so dA/dt = (2Ï€r)(dr/dt).
Answer
dA/dt = 20Ï€ square meters per second.
Definition

Related rates problems involve two or more quantities that are both changing with time. Since they are related by some equation, their rates of change (derivatives with respect to time) are also related.

Strategy:

  1. Draw a diagram and label all quantities.
  2. Write an equation relating the quantities.
  3. Differentiate both sides with respect to time tt (using the chain rule).
  4. Substitute known rates and values.
  5. Solve for the unknown rate.

A rate like drdt\frac{dr}{dt} means "how fast rr is changing with respect to time."

Key properties
  • Every related-rates problem is implicit differentiation with respect to time, applied to a geometric or physical constraint
  • The relating equation must hold at every instant, not just the instant you're solving for — that's what justifies differentiating it
  • Units of a rate are always the units of the quantity divided by time
  • The chain rule is what connects a rate in one variable to a rate in another
Common mistakes
  • Substituting known values before differentiating: plugging in a number too early turns a variable into a constant, and its derivative silently becomes zero
  • Mixing up which rate is given and which is unknown: always identify clearly what dx/dtdx/dt values are known and which one is being solved for before differentiating
Expanding ripple in a pond

A stone is dropped into a pond. The radius of the ripple increases at 22 m/s. How fast is the area increasing when the radius is 55 m?

Area: A=Ï€r2A = \pi r^2. Differentiate with respect to tt: dAdt=2Ï€rdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}

Substitute r=5r = 5 and drdt=2\frac{dr}{dt} = 2: dAdt=2π(5)(2)=20π≈62.8 m2/s\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \approx 62.8 \text{ m}^2/\text{s}

Try it

A spherical balloon is being inflated so that its volume increases at 100100 cm³/s. How fast is the radius increasing when the radius is 55 cm?

Solution

V=43πr3  ⟹  dVdt=4πr2drdtV = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

Solve for drdt\frac{dr}{dt}: drdt=14πr2⋅dVdt=1004π(25)=100100π=1π≈0.318 cm/s\frac{dr}{dt} = \frac{1}{4\pi r^2} \cdot \frac{dV}{dt} = \frac{100}{4\pi(25)} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}

Related concepts

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