One rate you know, one you want — the chain rule bridges them: Picture pumping air into a balloon. The radius grows, and so does the volume — they grow together. If you know how fast the volume changes (dV/dt) you can find how fast the radius changes (dr/dt), and vice versa.
The bridge is the chain rule:
dV/dt = (dV/dr) · (dr/dt).
dV/dr comes from the volume formula V = (4/3)πr³, and one of the time-rates is given in the question. The recipe:
1. Write the formula linking the quantities.
2. Differentiate it with respect to time t.
3. Substitute the known rate and the value of the variable to solve for the unknown rate.
IB-style question — inflating sphere
Air is pumped into a spherical balloon so its volume increases at 12 cm³ per second.
Find the rate at which the radius is increasing when the radius is 3 cm. (V = (4/3)πr³.)
Step by step
- Write the formula and differentiate it with respect to r.
- Link the rates with the chain rule, putting in the known dV/dt = 12.
- Substitute r = 3 and solve for dr/dt.
- Divide to isolate the unknown rate.
Final answer
dr/dt = 1/(3π) ≈ 0.106 cm per second.
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Find the right formula first — then differentiate w.r.t. time: Most related-rates problems are really 'spot the formula' problems:
• Sliding ladder → Pythagoras x² + y² = L² (L constant).
• Filling cone → V = (1/3)πr²h, often with r and h in a fixed ratio so you reduce to one variable.
• Growing triangle/area → the area formula for that shape.
Differentiate the chosen formula with respect to t. Every variable that changes leaves a 'd/dt' rate; constants leave nothing. Then substitute the moment in question.
IB-style question — sliding ladder
A 5 m ladder leans on a wall. Its base slides away from the wall at 0.4 m s⁻¹.
How fast is the top sliding down when the base is 3 m from the wall?
Step by step
- Let x = base distance, y = height up the wall. Pythagoras links them (ladder length 5 is constant).
- Differentiate both sides with respect to time t.
- When x = 3: y = √(25 − 9) = 4. Substitute x = 3, y = 4, dx/dt = 0.4.
- Solve for dy/dt.
Final answer
dy/dt = −0.3 m s⁻¹ — the top slides DOWN at 0.3 m per second (the minus sign shows height is decreasing).
IB-style question — growing equilateral triangle
An equilateral triangle's sides grow at 2 cm s⁻¹.
Find the rate the area increases when the side length is 10 cm. (Area = (√3/4)x².)
Step by step
- Write the area formula and differentiate with respect to t (chain rule on x²).
- Substitute x = 10 and dx/dt = 2.
Final answer
dA/dt = 10√3 ≈ 17.3 cm² per second.