Linked quantities ⇒ linked rates: If a quantity A depends on r, and r is changing in time, then A is changing too.
The chain rule links the rates: The chain rule connects how fast each one changes: dA/dt = (dA/dr) × (dr/dt). So to get one rate, differentiate the formula that links them, then multiply by the rate you already know.
IB-style question — a spreading ripple
A circular ripple spreads on a pond so that its radius grows at a constant 3 cm s⁻¹.
Find how fast its area is growing at the instant the radius is 10 cm.
Step by step
- Write the link between area and radius.
- Differentiate with respect to r.
- Chain rule: multiply by the known rate dr/dt = 3.
- Substitute the moment's value r = 10.
Final answer
The area is growing at 60π ≈ 188 cm² s⁻¹ when r = 10 cm. (Substitute r = 10 only at the END, after differentiating.)
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Differentiate, substitute the rate, solve: Often you know dV/dt and want dr/dt. Same recipe: differentiate the linking formula, write the chain rule, put in the known rate, then solve for the unknown rate.
IB-style question — an inflating balloon
A spherical balloon is inflated so that its volume increases at 100 cm³ s⁻¹.
Find the rate at which its radius is increasing when r = 5 cm. (Volume of a sphere: V = (4/3)πr³.)
Step by step
- Differentiate the volume formula with respect to r.
- Chain rule links the rates.
- Put in dV/dt = 100 and r = 5, then solve for dr/dt.
- Divide.
Final answer
The radius is increasing at 1/π ≈ 0.318 cm s⁻¹ when r = 5 cm.