A derivative with respect to time measures how fast a quantity changes. When two quantities are linked by a formula, their rates of change are linked by the chain rule. Key formulas \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=\dfrac{\mathrm{d}y}{\mathrm{d}x}\times\dfrac{\mathrm{d}x}{\mathrm{d}t}\qquad\text{and}\qquad\dfrac{\mathrm{d}x}{\mathrm{d}y}=\dfrac{1}{\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}\) Method Write down the rate you are given and the rate you want to find. Make the rate you want the subject of a chain-rule equation. Differentiate the formula linking the two quantities to fill in the missing derivative. Substitute the values at the instant asked for. Translating the wording The question saysWrite \(y\) is increasing at \(3\) units per second\(\dfrac{\mathrm{d}y}{\mathrm{d}t}=3\) \(A\) is decreasing at \(0.2\ \text{cm}^2\ \text{s}^{-1}\)\(\dfrac{\mathrm{d}A}{\mathrm{d}t}=-0.2\) Find the rate of change of \(V\)Make \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) the subject Exam check A quantity decreasing at \(0.2\) units per second has derivative \(-0.2\). Keep the minus sign in the working; at the end, a negative rate of change is stated as a positive rate of decrease. The reciprocal relation \(\mathrm dx/\mathrm dy=1/(\mathrm dy/\mathrm dx)\) requires a differentiable local inverse and \(\mathrm dy/\mathrm dx\ne0\) at the point. Do not divide by a zero derivative.

Rates of Change

Secondary 4

TGM Original Questions

A derivative with respect to time measures how fast a quantity changes. When two quantities are linked by a formula, their rates of change are linked by the chain rule.

Key formulas

\(\dfrac{\mathrm{d}y}{\mathrm{d}t}=\dfrac{\mathrm{d}y}{\mathrm{d}x}\times\dfrac{\mathrm{d}x}{\mathrm{d}t}\qquad\text{and}\qquad\dfrac{\mathrm{d}x}{\mathrm{d}y}=\dfrac{1}{\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}\)

Method

Write down the rate you are given and the rate you want to find.
Make the rate you want the subject of a chain-rule equation.
Differentiate the formula linking the two quantities to fill in the missing derivative.
Substitute the values at the instant asked for.

Translating the wording

The question says	Write
\(y\) is increasing at \(3\) units per second	\(\dfrac{\mathrm{d}y}{\mathrm{d}t}=3\)
\(A\) is decreasing at \(0.2\ \text{cm}^2\ \text{s}^{-1}\)	\(\dfrac{\mathrm{d}A}{\mathrm{d}t}=-0.2\)
Find the rate of change of \(V\)	Make \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) the subject

Exam check

A quantity decreasing at \(0.2\) units per second has derivative \(-0.2\). Keep the minus sign in the working; at the end, a negative rate of change is stated as a positive rate of decrease.

The reciprocal relation \(\mathrm dx/\mathrm dy=1/(\mathrm dy/\mathrm dx)\) requires a differentiable local inverse and \(\mathrm dy/\mathrm dx\ne0\) at the point. Do not divide by a zero derivative.

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