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When one rate drives another

• Blow up a balloon: its radius grows, and because of that, its volume grows too.
• The two rates are linked — knowing how fast the radius changes tells you how fast the volume changes.
• Problems like this are related rates 相关变化率: several quantities changing together in time.
• The engine that links them is the chain rule, applied with respect to time 时间.

Start with an equation relating the variables

• First, find an equation that ties the quantities together — usually geometry or a formula.
• Balloon: $V=\tfrac43\pi r^3$ links volume and radius.
• Ladder sliding down a wall: $x^2+y^2=L^2$ (Pythagoras) links the two distances.
• This relating equation is true at every instant, so we can differentiate it in time.

A rising balloon: related rates link how fast radius, volume and height change together

Differentiate with respect to time

• Every variable secretly depends on $t$, so differentiating brings in a rate for each — via the chain rule.
• $\dfrac{d}{dt}\big[V=\tfrac43\pi r^3\big]$ gives $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
• Each term contributes a "$\tfrac{d(\text{that variable})}{dt}$" factor — exactly like implicit differentiation, but the hidden variable is $t$.
• Now the equation links the rates $\tfrac{dV}{dt}$ and $\tfrac{dr}{dt}$.

Sort the givens from the unknown

• Read the problem and label: which rates are given, and which single rate is wanted?
• "The radius grows at $2\,\tfrac{\text{cm}}{\text{s}}$" → $\tfrac{dr}{dt}=2$ is given.
• "How fast is the volume growing?" → $\tfrac{dV}{dt}$ is the unknown.
• Also note any fixed quantities (like the radius at the instant asked about).