A related-rate question links two quantities that both change with time. You know how fast one changes and need the rate of the other. The tool is the chain rule: dy/dt = dy/dx × dx/dt.
This lesson extends using a derivative as a rate of change. It belongs to the module on tangents, normals and rates.
What is the method, step by step?
- Name the quantities and write the rate you are given and the rate you want.
- Write the formula that connects the two quantities.
- Differentiate that formula with respect to the quantity you know the rate of.
- Multiply by the given rate using the chain rule.
- Substitute the value at the required moment and write the units.
For example, if you want dA/dt and are given dr/dt, use dA/dt = dA/dr × dr/dt.
Worked example
An oil stain is a circle. Its radius grows at 2 cm per second. Find the rate at which its area is increasing when the radius is 5 cm.
Step 1, name: A is area, r is radius. Given dr/dt = 2. Wanted dA/dt when r = 5.
Step 2, formula: A = πr².
Step 3, differentiate: dA/dr = 2πr.
Step 4, chain rule: dA/dt = dA/dr × dr/dt = 2πr × 2 = 4πr.
Step 5, substitute r = 5: dA/dt = 4π × 5 = 20π ≈ 62.8.
The area is increasing at 20π cm² per second (about 62.8 cm² per second).
Check: units are cm² per second, as expected for area per time. ✓
The mistake to watch for
A frequent slip is to stop after differentiating the formula and treat dA/dr as the answer.
Mistaken working: A = πr², so dA/dr = 2πr. At r = 5 the rate is 10π.
This is the rate of change of area with respect to radius, in cm² per cm. The question wanted the rate with respect to time.
The correction is to write the target “dA/dt” first and look at the denominators. If the denominator of your answer is r, there is still a factor of dr/dt missing.
Check yourself
Try these, then open each answer.
1. A cube has side x cm, increasing at 3 cm per second. Find the rate at which its volume is increasing when x = 4.
Show answer
V = x³, so dV/dx = 3x². dV/dt = 3x² × 3 = 9x². At x = 4: 9 × 16 = 144 cm³ per second.
2. A sphere has volume V = (4/3)πr³. Its volume increases at 12π cm³ per second. Find the rate of increase of the radius when r = 3.
Show answer
dV/dr = 4πr². At r = 3 this is 36π. Since dV/dt = dV/dr × dr/dt, we have 12π = 36π × dr/dt, so dr/dt = 1/3 cm per second.
3. A point moves on the curve y = x² + 1 so that x increases at 0.5 units per second. How fast is y changing when x = 3?
Show answer
dy/dx = 2x = 6 at x = 3. dy/dt = 6 × 0.5 = 3 units per second.
Where this leads next
Once the chain is automatic, finish the module with explaining a rate sign in context, then test yourself with the mixed practice set. The non-calculator working trainer helps with exact answers in terms of π, and the quadratic structure explorer is useful when a rate condition turns into a quadratic.
Students who know the chain rule can still stall on choosing the right formula. A teacher who sees your rough setup can spot that quickly in online one-to-one Additional Mathematics tuition.