dy dt= dy dx dx 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= 23pi r3 dVdr=2pi r2 dVdt = dVdr drdt =2pi r2(0.5) =pi r2 When r=3, dVdt =pi(3)2 =9pi cm3/s

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

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

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

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 \(r=3\).

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

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

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

Connected Rates of Change

Connected Rates of Change

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