Connected Rates of Changes

Connected Rates of Changes

\(\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm dx}\dfrac{\mathrm dx}{\mathrm dt}\)

Differentiate the relation linking the quantities, then substitute the values at the required instant.

Increasing gives a positive rate; decreasing gives a negative rate.

Example

The radius, \(r\) cm, of a hemisphere is increasing at a constant rate of \(0.5\) cm/s. Find the rate of increase of the volume of the hemisphere when \(r=3\).

\(V=\dfrac23\pi r^3\qquad\dfrac{\mathrm{d}V}{\mathrm{d}r}=2\pi r^2\)

\(\dfrac{\mathrm{d}V}{\mathrm{d}t}\)\({}=\dfrac{\mathrm{d}V}{\mathrm{d}r}\dfrac{\mathrm{d}r}{\mathrm{d}t}\)\({}=2\pi r^2(0.5)\)\({}=\pi r^2\)

When \(r=3\),

\(\dfrac{\mathrm{d}V}{\mathrm{d}t}\)\({}=\pi(3)^2\)\({}=9\pi\ \mathrm{cm}^3\mathrm{/s}\)

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Answer:\(9\pi\text{ cm}^3\text{/s}\)

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