NSWMaths Extension 1Syllabus dot point

How do we link the rate of change of one quantity to the rate of change of another using the chain rule?

Solve related-rates problems by linking two changing quantities via an equation and differentiating with respect to time

A focused answer to the HSC Maths Extension 1 dot point on related rates. Linking two changing quantities through an equation, differentiating implicitly with respect to time, and substituting instantaneous values, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to model a physical situation in which two or more quantities are changing simultaneously, find an equation linking them, differentiate that equation with respect to time, substitute known instantaneous values, and solve for the unknown rate.

The answer

The general method

Draw a diagram and label every variable that changes.
Write down an equation that relates the variables. Geometry, similar triangles, area or volume formulas are common starting points.
Differentiate both sides of the equation with respect to $t$ . Every variable that changes with time picks up a $\frac{d}{dt}$ via the chain rule.
Substitute the given values (the known instantaneous values and the known rates).
Solve for the unknown rate.
State the answer with units and direction (sign).

The chain rule in disguise

If $V = f(r)$ and $r$ changes with time, then $\frac{dV}{dt} = f'(r) \frac{dr}{dt}$ . This is just the chain rule.

For multi-variable equations like $x^2 + y^2 = 25$ , implicit differentiation with respect to $t$ gives

2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0,

so $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$ .

Choosing the relation

The trickiest step is finding the right equation linking the variables. Useful approaches:

For a cone or sphere, use the volume formula.
For a right triangle (ladder against wall), use Pythagoras.
For shadow problems, use similar triangles.
For inflating balloons, use $V = \frac{4}{3} \pi r^3$ .

If the equation has too many variables, eliminate one using a constraint (for example, similar triangles in a cone give $r = k h$ ).

Differentiating implicitly

Every variable that changes with time gets a $\frac{d \cdot}{dt}$ attached. The product rule, chain rule and quotient rule apply as usual.

$\frac{d}{dt}(x^2) = 2 x \frac{dx}{dt}$ ,

$\frac{d}{dt}(\sin x) = \cos x \frac{dx}{dt}$ ,

$\frac{d}{dt}(x y) = \frac{dx}{dt} y + x \frac{dy}{dt}$ .

Direction matters

A negative rate means the variable is decreasing. State direction explicitly: "the water level is rising at $0.3$ m/s" or "the shadow is shortening at $1.2$ m/s".

Worked example

Inflating balloon

A spherical balloon is inflated so its volume increases at $50$ cm $^3$ /s. How fast is the radius increasing when the radius is $10$ cm?

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

Given $\frac{dV}{dt} = 50$ and $r = 10$ :

$50 = 4 \pi (100) \frac{dr}{dt}$ , so $\frac{dr}{dt} = \frac{50}{400 \pi} = \frac{1}{8 \pi}$ cm/s $\approx 0.040$ cm/s.

Ladder sliding

A $10$ m ladder leans against a vertical wall. The base slides away from the wall at $0.5$ m/s. How fast is the top sliding down when the base is $6$ m from the wall?

$x^2 + y^2 = 100$ . Implicit derivative: $2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0$ .

When $x = 6$ , $y = 8$ . Given $\frac{dx}{dt} = 0.5$ :

$2(6)(0.5) + 2(8) \frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = -\frac{6}{16} = -0.375$ m/s.

The top is sliding down at $0.375$ m/s.

Conical tank

Water flows into a cone (vertex down) at $4$ cm $^3$ /s. The cone has radius $6$ cm at the top and depth $12$ cm. How fast is the water level rising when the depth is $3$ cm?

Similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .

Volume of water: $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi h^3}{12}$ .

$\frac{dV}{dt} = \frac{3 \pi h^2}{12} \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$ .

Substitute $h = 3$ , $\frac{dV}{dt} = 4$ :

$4 = \frac{9 \pi}{4} \frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{16}{9 \pi}$ cm/s $\approx 0.566$ cm/s.

Shadow problem

A $1.8$ m person walks away from a $6$ m streetlight at $1.5$ m/s. How fast is the tip of the shadow moving?

Let $x$ be the distance from the person to the lamp post, and $s$ the length of the shadow.

Similar triangles: $\frac{6}{x + s} = \frac{1.8}{s}$ , so $6 s = 1.8(x + s)$ , so $s = \frac{1.8 x}{4.2} = \frac{3 x}{7}$ .

The tip is at distance $x + s = x + \frac{3 x}{7} = \frac{10 x}{7}$ .

$\frac{d}{dt}\!\left( \frac{10 x}{7} \right) = \frac{10}{7} \frac{dx}{dt} = \frac{10}{7} \cdot 1.5 = \frac{15}{7}$ m/s $\approx 2.14$ m/s.

Right triangle with one constant side

A boat is being pulled toward a dock by a rope wound at $2$ m/s. The dock is $4$ m above water. How fast is the boat approaching when the rope is $5$ m long?

Let $x$ = horizontal distance from boat to dock base, $L$ = rope length.

$x^2 + 16 = L^2$ . Differentiate: $2 x \frac{dx}{dt} = 2 L \frac{dL}{dt}$ , so $\frac{dx}{dt} = \frac{L}{x} \frac{dL}{dt}$ .

When $L = 5$ : $x = 3$ . Given $\frac{dL}{dt} = -2$ (rope shortening):

$\frac{dx}{dt} = \frac{5}{3} \cdot (-2) = -\frac{10}{3}$ m/s.

The boat approaches at $\frac{10}{3} \approx 3.33$ m/s.