Related rates of change: linking changing quantities via implicit differentiation

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 deep idea is that an everyday geometric fact (the volume of a cone, the length of a ladder, Pythagoras) holds at every instant of the motion, so it holds as an identity in time. Differentiating that identity with respect to $t$ turns a relationship between quantities into a relationship between their rates. Everything else is the chain rule and careful bookkeeping.

The answer

The general method

1. Draw a diagram and label every variable that changes. Mark which rate you are given and which you want.
2. Write down an equation that relates the variables. Geometry, similar triangles, area or volume formulas are common starting points.
3. If the equation has more than the two variables you care about, use a constraint to eliminate the extras before differentiating.
4. 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.
5. Substitute the given values (the known instantaneous values and the known rates) only now, after differentiating.
6. Solve for the unknown rate, then 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, read with $t$ as the underlying variable.

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

so $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$. Notice that nothing was solved for $y$ first: implicit differentiation handles the relationship directly, which is exactly why it is the right tool here.

Choosing the relation

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

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

If the equation has too many variables, eliminate one using a constraint. The classic case is a cone, where similar triangles force a fixed ratio between radius and height.

The cone trap: reduce to one variable first

The cone is where most students go wrong, so it is worth slowing down. The cone-volume formula $V = \frac{1}{3}\pi r^2 h$ has two changing lengths, $r$ and $h$. If you differentiate it as it stands you get a $\frac{dr}{dt}$ term you were never given and cannot find. The fix is to notice that the water surface and the cone wall always form similar triangles, so $r$ and $h$ stay in a fixed ratio. Substitute that ratio to get $V$ in terms of a single variable before differentiating. The stage-by-stage figures below build exactly this argument for a draining cone, the most examined related-rates context in Extension 1.

Stage 1, name the changing quantities. Water drains from an inverted cone. At any instant the water has surface radius $r$ and depth $h$, and both shrink as the cone empties. Mark them on the diagram before doing any algebra.

Stage 2, link $r$ and $h$ with similar triangles. The full cone here has top radius $6$ and height $12$. The triangle formed by the water surface is similar to the triangle of the whole cone, so $\dfrac{r}{h} = \dfrac{6}{12} = \dfrac{1}{2}$, giving $r = \dfrac{h}{2}$. This is the constraint that removes the extra variable.

Stage 3, write the volume in one variable. Substitute $r = \dfrac{h}{2}$ into $V = \dfrac{1}{3}\pi r^2 h$:

Now $V$ depends on $h$ alone, so the differentiation will only produce the rate $\dfrac{dh}{dt}$, which is the one the question controls.

Stage 4, differentiate with the chain rule. Differentiating $V = \dfrac{\pi h^3}{12}$ with respect to time gives

Substitute the instantaneous depth and the level's rate of fall to finish. Because the level is falling, $\dfrac{dh}{dt}$ is negative, so $\dfrac{dV}{dt}$ comes out negative too: the volume is decreasing, as it must be.

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 the direction explicitly: "the water level is rising at $0.3$ m/s" or "the shadow is shortening at $1.2$ m/s". A bare number with no sign interpretation often drops the final communication mark.

How exam questions ask about related rates

• "At what rate is the area / volume changing when ...?" The classic form. Write the area or volume formula, differentiate with respect to $t$, substitute the instant. Watch for a hidden constraint (a cone, a fixed perimeter) that needs eliminating first.
• "A ladder slides down a wall ..." or "a boat is pulled towards a dock ..." A Pythagoras problem. The fixed length (ladder, rope) is the constant; differentiate $x^2 + y^2 = L^2$.
• "How fast is the shadow lengthening / the tip of the shadow moving?" A similar-triangles problem. Be careful which length the question wants: the shadow itself, or the position of its tip.
• "Leave your answer in terms of $\pi$." A signal that the numbers are chosen to cancel cleanly; do not reach for a decimal.
• "Is the rate increasing or decreasing?" Asks for the sign and its meaning, not just a magnitude.