Related Rates of Change

In short: Related rates link how fast two connected quantities change, using the chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$ . Write one formula linking the two quantities, differentiate it, then substitute the values you know.

Starting point: what you already know

A rate of change is a derivative. $\frac{d r}{d t}$ means "how fast $r$ changes each second".
The chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$ . Think of the $d x$ parts "cancelling".
Formulas for shapes: circle area $π r^{2}$ , sphere volume $\frac{4}{3} π r^{3}$ , sphere surface area $4 π r^{2}$ , cone volume $\frac{1}{3} π r^{2} h$ , cube volume $x^{3}$ .
Similar triangles: matching sides are in the same ratio.
Signs: a quantity that is getting smaller has a negative rate.

The method

Turning the chain the other way

Often you know how fast the "big" quantity changes and want the small one. Flip the derivative.

Related rates

Idea

Linked quantities
Linked rates

Tool

Chain rule
dy/dt = dy/dx × dx/dt

Steps

Know and want
Linking formula
Differentiate
Substitute last

Classic problems

Ripples
Balloons
Filling a cone
Sliding ladder

Year 12The sliding ladder

A 5 m ladder leans on a wall. Its foot slides away at 0.5 m/s. How fast does the top slide down when the foot is 3 m from the wall?

Let the foot be $x$ m out and the top $y$ m up, so $x^{2} + y^{2} = 25$ and $y = 25 - x^{2}$ .

$\frac{d y}{d x} = \frac{- x}{25 - x ^{2}} = - \frac{3}{4}$ when $x = 3$ .

$\frac{d y}{d t} = - \frac{3}{4} \times 0.5 = - 0.375$ m/s. The top slides down at 0.375 m/s. The negative sign says $y$ is shrinking.

Changing angles. A balloon rises at 4 m/s, 50 m from an observer. With $tan θ = \frac{h}{50}$ , differentiate with respect to $t$ : $sec^{2} θ \frac{d θ}{d t} = \frac{1}{50} \frac{d h}{d t}$ . When $h = 50$ , $θ = \frac{π}{4}$ and $sec^{2} θ = 2$ , so $\frac{d θ}{d t} = \frac{4/50}{2} = 0.04$ rad/s.

Areas with an angle. For a triangle with two sides of length $a$ and the angle $θ$ between them, $A = \frac{1}{2} a^{2} sin θ$ , so $\frac{d A}{d t} = \frac{1}{2} a^{2} cos θ \frac{d θ}{d t}$ .

Shadows. Use similar triangles to link the shadow $s$ to the distance $x$ from the lamp. A 2 m person and a 5 m lamp give $\frac{s}{2} = \frac{x + s}{5}$ , so $s = \frac{2}{3} x$ and $\frac{d s}{d t} = \frac{2}{3} \frac{d x}{d t}$ , whatever the distance.

Distance from an observer. A plane at height 3 km flying level: $D^{2} = x^{2} + 9$ , so $D \frac{d D}{d t} = x \frac{d x}{d t}$ .

Comparing two moments. In a cone filling at a steady rate, $V = k h^{3}$ , so $\frac{d h}{d t} = \frac{1}{3 k h ^{2}} \frac{d V}{d t}$ : the depth rate is proportional to $\frac{1}{h ^{2}}$ . Doubling the depth divides the rate by 4.

Three rates in a chain. From a volume rate to a surface area rate for a sphere: $\frac{d S}{d t} = \frac{d S}{d r} \times \frac{d r}{d V} \times \frac{d V}{d t} = \frac{8 π r}{4 π r ^{2}} \frac{d V}{d t} = \frac{2}{r} \frac{d V}{d t}$ .

Quick summary

Linked quantities have linked rates.
Chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$ , and $\frac{d x}{d y} = 1 \div \frac{d y}{d x}$ .
Find a formula with only the two quantities you need. Use similar triangles or Pythagoras to remove a third.
Differentiate first, substitute last. Check the units and the sign.

Quiz

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Try 3 questions now

The radius of a circle grows at 2 cm/s. Which expression is $\frac{d A}{d t}$ ?
1. A $2 π r$
2. B $4 π r$
3. C $π r^{2}$
4. D $2 r$
Show answer
B: $4 π r$
$\frac{d A}{d r} \times \frac{d r}{d t} = 2 π r \times 2$ .
With the same circle, find $\frac{d A}{d t}$ when $r = 5$ .
1. A $10 π$ cm²/s
2. B $20 π$ cm²/s
3. C $25 π$ cm²/s
4. D $50 π$ cm²/s
Show answer
B: $20 π$ cm²/s
$4 π \times 5 = 20 π$ .
A cube's side grows at 0.1 cm/s. How fast does the volume grow when the side is 10 cm?
1. A 10 cm³/s
2. B 30 cm³/s
3. C 100 cm³/s
4. D 3 cm³/s
Show answer
B: 30 cm³/s
$\frac{d V}{d x} = 3 x^{2} = 300$ , times 0.1.