Solving Related Rates Problems
A repeatable recipe for related rates
- Related-rates word problems feel hard until you follow a fixed procedure.
- Same five steps every time: picture, equation, differentiate, substitute, solve.
- The one rule people break: substitute numbers only after differentiating.
- Master the recipe and every related-rates problem becomes routine.
The five steps
- 1. Draw & label the changing quantities; note what's given and wanted.
- 2. Relate the variables with an equation (geometry/formula).
- 3. Differentiate both sides with respect to $t$ (chain rule → rate factors).
- 4. Substitute the known values at that instant.
- 5. Solve for the unknown rate; report with units and sign.
Volume grows fastest when big
y = a·x³ (V vs r)
Because $\tfrac{dV}{dt}=4\pi r^2\tfrac{dr}{dt}$, a fixed radius rate makes the volume rate soar as $r$ grows.
Put the related-rates steps in order.
Picture, relate, differentiate, substitute-and-solve.
Differentiate first, plug in last
- A number that is changing must stay a variable while you differentiate.
- If you set $r=5$ before differentiating, its rate $\tfrac{dr}{dt}$ vanishes — and the whole method collapses.
- Substitute the instant's values only at step 4, after the derivative is taken.
- The only things you may fix early are true constants (like a ladder's fixed length).
With $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$, $r=5$, and $\dfrac{dr}{dt}=2$, find $\dfrac{dV}{dt}$ as a multiple of $\pi$ (enter the number before $\pi$).
$4\pi(25)(2)=200\pi$.
When should you substitute the instant's numeric values?
Differentiate first; substitute numbers only afterward, or the rate terms vanish.
Setting $r=5$ before differentiating is a valid shortcut.
It kills $\tfrac{dr}{dt}$ and breaks the method.
Check units and reasonableness
- The answer is a rate — attach units of output per input (e.g. $\tfrac{\text{cm}^3}{\text{s}}$).
- Check the sign: growing quantity → positive; shrinking → negative.
- Ask if the size is reasonable for the situation.
- A tidy conclusion sentence ("the volume is increasing at ... per second") earns the interpretation marks.
A related-rates answer is a rate, so it must be reported with its ____.
E.g. cm³ per second.
A related-rates answer comes out $\dfrac{dV}{dt}=-30\,\tfrac{\text{cm}^3}{\text{s}}$. This means the volume is...
A negative rate means the volume is shrinking at $30$ cm³/s.
The number-one related-rates mistake: plugging in the instant's values before differentiating. Keep every changing quantity symbolic through the differentiation step; substitute only afterward. Substituting early kills the rate terms and gives a wrong (often zero) answer.
A spherical balloon's radius grows at $\tfrac{dr}{dt}=2\,\tfrac{\text{cm}}{\text{s}}$. How fast is the volume growing when $r=5$ cm?
- Relate: $V=\tfrac43\pi r^3$. Differentiate: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
- Now substitute $r=5$, $\tfrac{dr}{dt}=2$: $\dfrac{dV}{dt}=4\pi(25)(2)=200\pi$.
- The volume is increasing at $200\pi\approx628\ \tfrac{\text{cm}^3}{\text{s}}$.
Solve a related rates problem in five steps: draw/label, write a relating equation, differentiate with respect to $t$, substitute the instant's values only after differentiating, and solve for the unknown rate. Report it with correct units and sign, and check the result is reasonable.