Solving Related Rates Problems
IntroducedA 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.
An inflating balloon: dV/dt and dr/dt are linked through V = (4/3)πr³
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.
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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 $\pi$).
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$4\pi(25)(2)=200\pi$.
When should you substitute the instant's numeric values?
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Differentiate first; substitute numbers only afterward, or the rate terms vanish.
Setting $r=5$ before differentiating is a valid shortcut.
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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 ____.
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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...
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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.
Write from substituting after differentiating
- A related-rates equation still contains two variables.
- The values at one instant are known.
When do you substitute those values?
A variable that gets plugged in too soon cannot be differentiated.
In a related-rates problem, when do you insert the instant's numbers?
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Keep the variables until the rates appear.