Related rates of change problems, the use of the chain rule to connect the rates of change of related variables, the setting up of a relating equation from the geometry or context, and the evaluation of an unknown rate at a given instant

What this dot point is asking

VCAA wants you to solve related-rates problems: two or more quantities change with time, and given the rate of one you must find the rate of another. The engine is the chain rule, which links the rates through an equation relating the quantities. You set up that equation from the context, differentiate with respect to time, and substitute the instant of interest.

The chain-rule link

If two quantities $V$ and $r$ are related and both change with time, their rates are connected by the chain rule:

This says the rate at which $V$ changes equals how fast $V$ changes with $r$, times how fast $r$ changes with time. Given any two of the three rates, you can find the third.

A reliable procedure

1. Identify the variables and rates. Note which rate is given and which is wanted, with units.
2. Write a relating equation. Use the geometry or the context to connect the variables, for example $V = \frac{4}{3}\pi r^3$ for a sphere.
3. Differentiate with respect to time. Apply the chain rule to every variable, since each depends on $t$.
4. Substitute the instant. Put in the known rate and the values of the variables at the moment asked about.
5. Solve and interpret. State the unknown rate with correct units and sign (positive for increasing, negative for decreasing).

The order matters: differentiate symbolically first, then substitute numbers, because substituting a numerical value for a variable before differentiating would freeze it and lose its rate.

Choosing the relating equation

The relating equation comes from the situation. Volume and radius of a sphere give $V = \frac{4}{3}\pi r^3$. A right-angled triangle (such as a ladder against a wall) gives Pythagoras $x^2 + y^2 = L^2$. A cone draining gives the cone volume formula, often with the radius expressed in terms of the height using similar triangles to reduce to one variable.

Examples in context

Example 1. For a square of side $s$ and area $A = s^2$, $\frac{dA}{dt} = 2s\frac{ds}{dt}$.

Example 2. A ladder of length $L$ sliding down a wall satisfies $x^2 + y^2 = L^2$, so $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.

Try this

Q1. Write the related-rates equation linking the area and radius of a circle. [2 marks]

• Cue. $A = \pi r^2$, so $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$.

Q2. A cube's edge grows at $2\ \text{cm/s}$. Find $\frac{dV}{dt}$ when the edge is $4\ \text{cm}$. [3 marks]

• Cue. $V = s^3$, $\frac{dV}{dt} = 3s^2\frac{ds}{dt} = 3(16)(2) = 96\ \text{cm}^3/\text{s}$.

Q3. Why must you differentiate before substituting the instant? [1 mark]

• Cue. Substituting first turns a variable into a constant, losing its rate of change.