The application of differentiation, including the chain rule, to related rates of change problems involving two or more time-dependent quantities

What this dot point is asking

VCAA wants you to handle word problems in which two or more time-dependent quantities are related by a geometric or algebraic equation, and one rate is known while another is asked for. The dot point pulls together the chain rule (Unit 3) and modelling reasoning. Paper 2 typically includes one related-rates problem per year.

The four-step procedure

Every related-rates problem yields to the same four steps.

Step 1. Identify the variables and relate them

Read the problem. Name the time-dependent variables (volume, radius, height, area, distance). Write the equation that relates them, drawn from geometry (cone, sphere, cylinder, triangle, circle), physics, or algebra.

If the equation has more variables than the question requires, eliminate the extras using auxiliary relationships (similar triangles, ratios, fixed proportions).

Step 2. Differentiate both sides with respect to time

Apply $\frac{d}{dt}$ to the entire equation. Each variable contributes its time derivative via the chain rule.

For example, if $V = \frac{4}{3} \pi r^3$ and both $V$ and $r$ depend on time, then:

The factor $\frac{dr}{dt}$ comes from the chain rule because $r$ is a function of $t$.

Step 3. Substitute the given numerical values

After differentiating, substitute the values at the specific moment the question describes. Note that the differentiation must happen before the substitution; you cannot plug in a value for $r$ first and then differentiate (because then $V$ would be a constant).

Step 4. Solve for the unknown rate

The equation from step 3 is linear in the unknown rate. Solve algebraically. State units.

Standard contexts

Expanding circle (area from radius)

Equation: $A = \pi r^2$.

Differentiate: $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.

Use when a ripple or oil slick is expanding.

Inflating sphere (volume from radius, surface area from radius)

Volume: $V = \frac{4}{3} \pi r^3$, so $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

Surface area: $S = 4 \pi r^2$, so $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.

Use for balloons, bubbles.

Cylinder (volume from radius and height)

If both vary: $V = \pi r^2 h$, $\frac{dV}{dt} = 2 \pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}$ (product rule).

Often only one of $r$ or $h$ varies; the fixed variable drops.

Cone (similar-triangle reduction)

A conical tank with fixed total height $H$ and total radius $R$. As water fills to depth $h$ with surface radius $r$, the ratio $\frac{r}{h} = \frac{R}{H}$ holds. Eliminate $r$ to get $V = \frac{\pi R^2}{3 H^2} h^3$, a function of $h$ alone.

Use for water tanks, sand piles.

Sliding ladder

A ladder of length $L$ slides down a wall. Let $x$ be the horizontal distance from the wall to the foot of the ladder and $y$ the height of the top of the ladder.

Equation: $x^2 + y^2 = L^2$.

Differentiate: $2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0$, so $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$.

Use when an object slides along two perpendicular axes.

Trough or rectangular tank

A trough has cross-section of varying area as the water rises. Compute the cross-sectional area as a function of depth, then $V = A(h) \cdot$ (length), giving $\frac{dV}{dt} =$ (length) $\cdot A'(h) \cdot \frac{dh}{dt}$.

Worked example: receding shadow

A 1.8 m tall person walks away from a 5 m tall street lamp at 1.2 m/s. How fast is the tip of their shadow moving?

Step 1. Set up

Let $x$ be the distance from the lamp to the person and $s$ the distance from the lamp to the tip of the shadow. By similar triangles (lamp / shadow tip and person / shadow tip), $\frac{5}{s} = \frac{1.8}{s - x}$, giving $5(s - x) = 1.8 s$, i.e. $3.2 s = 5 x$, so $s = \frac{25}{16} x$.

Step 2. Differentiate

$\frac{ds}{dt} = \frac{25}{16} \frac{dx}{dt}$.

Step 3. Substitute

$\frac{dx}{dt} = 1.2$.

Step 4. Solve

$\frac{ds}{dt} = \frac{25}{16} \cdot 1.2 = 1.875$ m/s.

The tip of the shadow moves at 1.875 m/s, faster than the person.

Examples in context

Example 1. Spreading oil slick. An oil slick spreads as a circle whose area grows at $\frac{dA}{dt} = 8$ m$^2$/s. Since $A = \pi r^2$, differentiating gives $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$. When the radius is $r = 4$ m, $8 = 2\pi(4)\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{8}{8\pi} = \frac{1}{\pi} \approx 0.318$ m/s.

Example 2. Draining cylindrical tank. A cylindrical tank of fixed radius $2$ m drains so its volume falls at $\frac{dV}{dt} = -0.6$ m$^3$/min. With $V = \pi r^2 h = 4\pi h$ (radius fixed), differentiating gives $\frac{dV}{dt} = 4\pi\frac{dh}{dt}$. So $-0.6 = 4\pi\frac{dh}{dt}$ and $\frac{dh}{dt} = \frac{-0.6}{4\pi} \approx -0.0477$ m/min (the level falls).

Try this

Q1. A circle's radius grows at $\frac{dr}{dt} = 3$ cm/s. Find the rate of area increase when $r = 5$ cm. [3 marks]

• Cue. $\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(5)(3) = 30\pi \approx 94.2$ cm$^2$/s.

Q2. A spherical balloon's volume increases at $20$ cm$^3$/s. Find $\frac{dr}{dt}$ when $r = 5$ cm. [3 marks]

• Cue. $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$; $20 = 100\pi\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{1}{5\pi} \approx 0.0637$ cm/s.

Q3. A $5$ m ladder slides down a wall; its foot moves out at $0.5$ m/s. When the foot is $3$ m from the wall, find the rate the top descends. [4 marks]

• Cue. $x^2 + y^2 = 25$, at $x = 3$, $y = 4$; $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$; $\frac{dy}{dt} = -\frac{3(0.5)}{4} = -0.375$ m/s.