[Figure: FiveStudy logo]
FiveStudy
Calculus AB · Unit 4 · 4.4
AP Calculus · Contextual Applications of Differentiation
Introduction to Related Rates
Differentiate the relation with respect to time, and every varying quantity brings its own rate.
key idea$\text{One equation relating quantities}\ \xrightarrow{\ \frac{d}{dt}\ }\ \text{one equation relating their RATES}$
Setting up the model
If two quantities are linked by an equation, their rates are linked too. Differentiating the relation with respect to time converts a geometric fact into a statement about how fast things change — and every term picks up a chain-rule factor.
Turning a relation into a rate equation
- Every varying quantity is a function of $t$, even when $t$ never appears in the equation.
- Differentiate implicitly with respect to $t$, not with respect to $x$. Each variable contributes its own $\dfrac{d\,\cdot}{dt}$.
- A constant stays a number. Only quantities that genuinely change get a rate.
- The result relates rates, so knowing all but one lets you solve for the last.
| Relation | Differentiated with respect to $t$ | Note |
|---|
| $A=\pi r^{2}$ | $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$ | one varying quantity on each side |
| $V=s^{3}$ | $\dfrac{dV}{dt}=3s^{2}\dfrac{ds}{dt}$ | chain rule on $s$ |
| $x^{2}+y^{2}=L^{2}$ | $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$ | $L$ constant ⇒ right side is $0$ |
| $A=\tfrac12 bh$ | $\dfrac{dA}{dt}=\tfrac12\left(\dfrac{db}{dt}h+b\dfrac{dh}{dt}\right)$ | product rule — both vary |
| $V=\tfrac13\pi r^{2}h$ | eliminate $r$ first | two varying quantities — reduce to one |
Row 4 is where students slip: if both factors change, the product rule applies. Row 5 is why the set-up sometimes needs geometry before any calculus.
Worked examples
EX 1
One relation, one chain rule
expanding circleA circular ripple's radius grows at $2$ cm/s. How fast is its area growing when $r=10$ cm?
1
Write the relation
$A=\pi r^{2}$ — true at every instant, not just at $r=10$.
2
Differentiate with respect to $t$
$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$.
3
List what you know
$\dfrac{dr}{dt}=2$ cm/s, and we want $\dfrac{dA}{dt}$ when $r=10$.
4
Substitute
$\dfrac{dA}{dt}=2\pi(10)(2)=40\pi$.
5
Answer with units
The area is increasing at $40\pi\approx126$ cm$^{2}$ per second.
EX 2
Why the substitution comes last
the classic errorA student sets $r=10$ in $A=\pi r^{2}$ first, getting $A=100\pi$, then differentiates. What happens?
1
Follow the student's step
$A=100\pi$ is a constant.
2
Differentiate it
$\dfrac{dA}{dt}=0$.
3
Name the error
Substituting fixed the radius forever, so the model says the circle never changes — the opposite of the question.
4
State the rule
$r=10$ describes one instant. The relation $A=\pi r^{2}$ must be differentiated while $r$ is still a variable.
5
The order
Relate → differentiate → then substitute. This single ordering prevents most related-rates failures.
EX 3
Both quantities changing
triangle areaA triangle's base grows at $3$ cm/s while its height shrinks at $1$ cm/s. Find $\dfrac{dA}{dt}$ when $b=12$ and $h=8$.
1
Write the relation
$A=\tfrac12 bh$.
2
Recognise a product
Both $b$ and $h$ vary, so differentiating needs the product rule.
3
Differentiate
$\dfrac{dA}{dt}=\tfrac12\left(\dfrac{db}{dt}\,h+b\,\dfrac{dh}{dt}\right)$.
4
Substitute
$=\tfrac12\left(3\cdot8+12\cdot(-1)\right)=\tfrac12(24-12)=6$.
5
Interpret
The area is increasing at $6$ cm$^{2}$/s — the base is winning. Note $\dfrac{dh}{dt}=-1$ carries its sign into the arithmetic.
EX 4
Reading the rates from the wording
translationTranslate each phrase into a rate statement: (a) “water drains at 5 L/min”, (b) “the ladder slips down the wall”, (c) “the balloon is inflated at a constant $30$ cm$^{3}$/s”.
1
(a) draining
Volume is decreasing: $\dfrac{dV}{dt}=-5$ L/min. ‘Drains’ supplies the minus sign.
2
(b) slipping down
Height is decreasing: $\dfrac{dy}{dt}\lt 0$. No number is given, so it stays symbolic.
3
(c) constant inflation
$\dfrac{dV}{dt}=+30$ cm$^{3}$/s, and constant means this holds at every instant, not just one.
4
Note what is not a rate
“when the radius is $10$ cm” is an instant condition, not a rate. It is used only at the substitution step.
5
The habit
Sort every given into rate or instant value before starting. Rates go into the differentiated equation; instant values go in last.
Units & interpretation
| Phrase in the question | Symbol | Sign |
|---|
| “increasing at”, “growing”, “filling” | $\dfrac{dQ}{dt}$ | positive |
| “decreasing”, “draining”, “shrinking”, “falling” | $\dfrac{dQ}{dt}$ | negative |
| “when the radius is 10 cm” | $r=10$ | not a rate — substitute last |
| “at a constant rate” | $\dfrac{dQ}{dt}=k$ | holds at every instant |
Half the marks on an introductory related-rates question are for translating the English correctly before any calculus happens.
Watch out
Substitute the instant value last
Setting $r=10$ before differentiating makes the radius constant and forces $\tfrac{dA}{dt}=0$. Differentiate first — Example 2.
Differentiate with respect to $t$, not $x$
Every varying quantity depends on time, so each contributes its own $\tfrac{d\,\cdot}{dt}$ factor even though $t$ never appears in the original relation.
Two varying factors need the product rule
$A=\tfrac12 bh$ with both $b$ and $h$ changing is a product, not a constant multiple.
Let the wording set the sign
‘Draining’, ‘falling’ and ‘shrinking’ all mean a negative rate. Losing that minus flips the final answer.