Solving Related Rates Problems

AP Calculus · Contextual Applications of Differentiation

Solving Related Rates Problems

The calculus is one chain rule. The marks are in the set-up, the substitution order and the units.

key idea$\text{Relate the quantities, THEN differentiate with respect to }t$

Setting up the model

[Figure: The ladder problem, drawn. x and y change while the ladder length stays fixed — that constant is what makes x2+y2=L2 differentiable into a relation between rates.]

The ladder problem, drawn. $x$ and $y$ change while the ladder length stays fixed — that constant is what makes $x^2+y^2=L^2$ differentiable into a relation between rates.

[Figure: The filling cone. The radius is not independent of the height: similar triangles fix r/h, and substituting that before differentiating is what leaves one rate to solve for.]

The filling cone. The radius is not independent of the height: similar triangles fix $r/h$, and substituting that before differentiating is what leaves one rate to solve for.

Related-rates problems are not hard because of the calculus — the differentiation is usually one chain rule. They are hard because of the set-up, and that is exactly where the marks are. Follow the same six moves every time.

The six moves

Draw and label. Put a letter on every changing quantity. Label constants with numbers, variables with letters — a length that changes must not be written as a number.
List what you know and what you want, as rates: e.g. given $\dfrac{dr}{dt}=3$, want $\dfrac{dV}{dt}$ when $r=5$.
Find an equation relating the variables — geometry (area, volume, Pythagoras), trigonometry, or a given formula. This equation must hold at all times, not just the instant asked about.
Differentiate both sides with respect to $t$, implicitly. Every variable picks up a $\dfrac{d}{dt}$ factor by the chain rule.
Substitute the instant's values last. Only after differentiating may you plug in the numbers that hold at that moment.
Solve, then answer in a sentence with units. State what is increasing or decreasing and how fast.

Situation	Relating equation	Differentiated
Sphere inflating	$V=\tfrac43\pi r^{3}$	$\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$
Ladder sliding down a wall	$x^{2}+y^{2}=L^{2}$	$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$
Conical tank filling	$V=\tfrac13\pi r^{2}h$	eliminate $r$ via similar triangles first
Shadow lengthening	similar triangles, $\dfrac{a}{b}=\dfrac{c}{d}$	differentiate the proportion
Angle of elevation	$\tan\theta=\dfrac{y}{x}$	$\sec^{2}\theta\dfrac{d\theta}{dt}=\dots$
Two vehicles separating	$s^{2}=x^{2}+y^{2}$	$2s\dfrac{ds}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

Almost every AP related-rates problem is one of these six. Recognising the geometry is most of the work.

Worked examples

EX 1

Inflating sphere

$V=\tfrac43\pi r^{3}$

Air is pumped into a spherical balloon at $100\ \text{cm}^{3}/\text{s}$. How fast is the radius increasing when the radius is $5$ cm?

Label

Let $r$ be the radius and $V$ the volume, both functions of time $t$.

Known and wanted

Given $\dfrac{dV}{dt}=100\ \text{cm}^{3}/\text{s}$. Want $\dfrac{dr}{dt}$ when $r=5$ cm.

Relate

$V=\tfrac43\pi r^{3}$ — true at every instant.

Differentiate with respect to $t$

$\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$.

Substitute the instant

$100=4\pi(5)^{2}\dfrac{dr}{dt}=100\pi\dfrac{dr}{dt}$.

Solve and answer

$\dfrac{dr}{dt}=\dfrac{1}{\pi}\approx 0.318$. The radius is increasing at about $0.318$ cm/s when the radius is $5$ cm.

EX 2

Sliding ladder — a negative rate that means something

$x^{2}+y^{2}=169$

A 13-ft ladder leans against a wall. The base slides away at $2$ ft/s. How fast is the top sliding down when the base is $5$ ft from the wall?

Label and note the constant

Let $x$ be the distance from wall to base and $y$ the height of the top. The ladder length is constant at $13$, so it is a number, not a variable.

Known and wanted

$\dfrac{dx}{dt}=2$ ft/s. Want $\dfrac{dy}{dt}$ when $x=5$.

Relate

Pythagoras: $x^{2}+y^{2}=169$.

Differentiate

$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$, so $x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0$.

Find the missing instant value

When $x=5$: $y=\sqrt{169-25}=12$. Note this had to wait until after differentiating.

Solve and interpret

$5(2)+12\dfrac{dy}{dt}=0\Rightarrow\dfrac{dy}{dt}=-\tfrac{5}{6}$. The top slides down at $\tfrac56\approx0.833$ ft/s — the minus sign is the physics, not an error.

EX 3

Conical tank — eliminate a variable first

similar triangles

Water pours into a cone (vertex down) at $2\ \text{m}^{3}/\text{min}$. The cone is $4$ m tall with top radius $2$ m. How fast is the depth rising when the water is $3$ m deep?

Label

Let $h$ be the depth and $r$ the surface radius of the water, both varying.

Relate

$V=\tfrac13\pi r^{2}h$ — but this has two varying quantities besides $V$.

Eliminate $r$

Similar triangles give $\dfrac{r}{h}=\dfrac{2}{4}$, so $r=\dfrac{h}{2}$. Then $V=\tfrac13\pi\left(\tfrac{h}{2}\right)^{2}h=\dfrac{\pi h^{3}}{12}$.

Differentiate

$\dfrac{dV}{dt}=\dfrac{\pi h^{2}}{4}\dfrac{dh}{dt}$.

Substitute

$2=\dfrac{\pi(3)^{2}}{4}\dfrac{dh}{dt}=\dfrac{9\pi}{4}\dfrac{dh}{dt}$.

Solve and answer

$\dfrac{dh}{dt}=\dfrac{8}{9\pi}\approx0.283$. The depth is rising at about $0.283$ m/min when the water is $3$ m deep.

EX 4

Two quantities both changing

$s^{2}=x^{2}+y^{2}$

A car heads east at $60$ km/h; a truck heads north at $80$ km/h from the same crossroads. How fast is the distance between them increasing $2$ hours later?

Label

Let $x$ be the car's distance east, $y$ the truck's north, $s$ the distance between them.

Known and wanted

$\dfrac{dx}{dt}=60$, $\dfrac{dy}{dt}=80$. Want $\dfrac{ds}{dt}$ at $t=2$.

Relate

$s^{2}=x^{2}+y^{2}$.

Differentiate

$2s\dfrac{ds}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$.

Values at the instant

At $t=2$: $x=120$, $y=160$, so $s=\sqrt{120^{2}+160^{2}}=200$.

Solve and answer

$200\dfrac{ds}{dt}=120(60)+160(80)=7200+12800=20000$, so $\dfrac{ds}{dt}=100$ km/h. They are separating at $100$ km/h two hours in.

Units & interpretation

An unlabelled number is an incomplete answer on the free-response section. Build the units from the quantities: a rate $\dfrac{dQ}{dt}$ always carries the units of $Q$ divided by time.

Rate	Units	A correct closing sentence
$\dfrac{dV}{dt}$ for a balloon	cm$^{3}$/s	“The volume is increasing at $100$ cm$^{3}$ per second.”
$\dfrac{dr}{dt}$ for the same balloon	cm/s	“The radius is increasing at about $0.318$ cm per second.”
$\dfrac{dy}{dt}$ for the ladder	ft/s	“The top is sliding down at $\tfrac56$ ft per second.”
$\dfrac{d\theta}{dt}$	radians/s	“The angle is increasing at $\dots$ radians per second.”

A negative rate is reported as a decrease, in words. Writing “$-\tfrac56$ ft/s” with no interpretation leaves a mark on the table.

Watch out

Substituting the instant too early

If you set $x=5$ before differentiating, $x$ becomes a constant and its derivative vanishes. Differentiate first; substitute last. This is the single biggest source of lost marks.

Treating a varying length as constant

In the cone problem, $r$ changes as the water rises. Only genuinely fixed dimensions — the ladder's $13$ ft, the cone's full height — may be written as numbers.

Forgetting the chain-rule factor

Differentiating $V=\tfrac43\pi r^{3}$ with respect to $t$ gives $4\pi r^{2}\dfrac{dr}{dt}$, not $4\pi r^{2}$. Every variable contributes its own $\tfrac{d}{dt}$.

Dropping the sign's meaning

A negative rate is information: the quantity is decreasing. Report it as such rather than taking an absolute value.

Course map. Topic 4.5 · LO CHA-3.E — solve problems involving related rates of change in applied contexts (EK CHA-3.E.1).