Question
The volume of a spherical cancerous tumor is given by the function$$V(r)=\frac{4}{3} \pi r^{3}$$where $r$ is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor with respect to its radius whena. $r=\frac{2}{3} \mathrm{cm}$b. $r=\frac{5}{4} \mathrm{cm}$
Step 1
The derivative of a function gives us the rate of change of the function. The derivative of the volume function $V(r)=\frac{4}{3} \pi r^{3}$ with respect to $r$ is given by: $$\frac{dV}{dr} = \frac{d}{dr} \left(\frac{4}{3} \pi r^{3}\right) = 4\pi r^{2}$$ Show more…
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