# Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 16

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a.) The equation $\displaystyle V = \frac{4}{3} \pi r^3$ represents the volume of a growing spherical cell where the radius is measured in micrometers $( 1 \mu m = 10^{-6} m)$. Determine the average rate of change of $V$ with respect to $r$ when $r$ changes from.

$(i) 5 \text{ to } 8 \mu m \qquad (ii) 5 \text{ to } 6 \mu m \qquad (iii) 5 \text{ to } 5.1 \mu m$

b.) Find the instantaneous rate of change of $V$ with respect to $r$ when $r = 5 \mu m$.

c.) Show that the rate of change of the volume of a sphere with respect to its radius is equal to its surface area.

a.)$(i) \text{from } 5 \text{ to } 8 \mu m,$

$\displaystyle \text{Average rate} = \frac{V(8)-V(5)}{8-5} = 172 \pi = 540.3539 \frac{\text{volume}}{\mu m}$

$(ii) \text{from } 5 \text{ to } 6 \mu m,$

$\displaystyle \text{Average rate} = \frac{V(6)-V(5)}{6-5} = \frac{364}{3} \pi = 381.1799 \frac{\text{volume}}{\mu m}$

$(iii) \text{from } 5 \text{ to } 5.1 \mu m,$

$\displaystyle \text{Average rate} = \frac{V(5.1)-V(5)}{5.1-5} = \frac{7651}{75} \pi = 320.4843 \frac{\text{volume}}{\mu m}$

The instantaneous rate of change can be solved by taking the derivative of $V$ with respect to $r$. So,

\begin{equation} \begin{aligned} \frac{dV}{dr} &= \frac{4}{3} \pi \frac{d}{dr} (r^3)\\ \\ \frac{dV}{dr} &= \frac{4}{\cancel{3}} \pi (\cancel{3}r^2)\\ \\ \frac{dV}{dr} &= 4 \pi r^2 \end{aligned} \end{equation}

when $r = 5 \mu m$

\begin{equation} \begin{aligned} \frac{dV}{dr} &= 4 \pi (5)^2\\ \\ \frac{dV}{dr} &= 314.1593 \frac{\text{volume}}{\mu m} \end{aligned} \end{equation}

c.) Recall that the surface area of the sphere is $A(r) = 4 \pi r^2$, and the volume is $\displaystyle V = \frac{4}{3} \pi r^3$. The rate of change

$\displaystyle V'(r) = \frac{4}{3} \pi \frac{d}{dr}(r^3) = \frac{4}{3} \pi (3r^2) = 4 \pi r^2$

Therefore, the rate of change of volume of the sphgere is equal to its surface area.