Recipe

Solve a related-rates problem

Use when: two or more quantities change over time, tied together by a shared geometric or physical equation, and you're given one rate to find another

Goal: the requested rate, at the specific instant described

Taught in Related rates -- the general method

1. 1

   Draw the setup and give every changing quantity a variable (not yet a number).

   Why: naming every quantity before touching any calculus is what keeps the next step from mixing up "the rate right now" with "the value right now."

2. 2

   Find a single equation connecting the quantities, using geometry or a given physical relationship.

   Example

   For a circular ripple with radius r and area A: A=\pi r^2.

   Why: this equation is the bridge. Without it, there's no way to connect a known rate to an unknown one.

3. 3

   Apply \dfrac{d}{dt} to the whole equation, treating every quantity as a function of t (so each one picks up its own rate via the chain rule).

   Example

   \dfrac{d}{dt}[A]=\dfrac{d}{dt}[\pi r^2] gives \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}.

   • the equation has more than two variables — a third relating equation (or a given constraint) is needed before differentiating, to reduce to two.
   • the equation already has exactly two changing quantities — continue directly to Step 4.

   Why: differentiating first, before any numbers go in, is what preserves the general relationship between the RATES, not just the quantities themselves.

4. 4

   Plug in the specific numbers given for this instant, including the known rate.

   Why: substituting a specific value before differentiating would treat that quantity as a constant, silently deleting its own rate of change from the equation.

5. 5

   Isolate the unknown rate algebraically.

   Example

   With r=5 and dr/dt=1.5: \dfrac{dA}{dt}=2\pi(5)(1.5)=15\pi cm²/s.

   Why: by this point every other piece of the differentiated equation is a known number, so isolating the one unknown is ordinary algebra.