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A circular patch of oil spreads on water…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A circular patch of oil spreads on water. Its area is $A = \pi r^2$ square centimetres (cm $^2$ ) when the radius is $r$ centimetres (cm). Find the rate at which the area changes with the radius when $r = 6$ cm. Now suppose the radius grows as $r = 2t$ centimetres at time $t$ seconds (s). Write the area as a function of $t$ and find the rate at which the area grows at $t = 3$ s, in square centimetres per second.

Need a reminder? The key ideaShow

Product rule: $(uv)' = u'v + uv'$ . Quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ . The derivative $\dfrac{dy}{dx}$ is a rate of change, in units of $y$ per unit of $x$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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