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Suppose the volume, $V$, of a spherical tumour with a radius of $r = 2\,\textrm{cm}$ uniformly grows at a rate of $dV/dt=0.3\,\textrm{cm}^3/\textrm{day}$, where $t$ is the time in days. At what rate is the surface area of the tumour increasing. The volume of a sphere is given by $V=\dfrac43\pi r^3$ and the surface area is given by $A=4\pi r^2$.

Przemysław Scherwentke
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Amanda
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1 Answers1

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You can find $r$ as a function of $V$:

$$r=\sqrt[3]{\dfrac{3V}{4\pi}},$$

then

$$A=4\pi r^2= 4 \pi \left(\dfrac{3V}{4\pi}\right)^{2/3},$$

the last step is derive

$$\dfrac{dA}{dt}={2\pi} \left(\dfrac{3V}{4\pi}\right)^{-1/3}\dfrac{dV}{dt}.$$

Now you can put the numbers in the last equation.

David
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