SOLUTION 7: Draw a sphere of radius $r$, and assume that $r$ is a function of time $t$.

$\ \ \ \$ a.) The surface area of a sphere of radius $r$ is $$S=4 \pi r^2$$ GIVEN: $\ \ \ \displaystyle{ dr \over dt } = 3 \ ft/hr.$

FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $r=10 \ ft.$

Now differentiate the surface area equation with respect to time $t$, getting

$$D \{ S \} = D \{ 4 \pi r^2 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 4 \pi \cdot 2r \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 8 \pi r \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dr \over dt } = 3$ and $r =10. \Big)$

$$\displaystyle{ dS \over dt } = 8 \pi (10)(3) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 240 \pi \ ft^2/hr. \approx 753.98 \ ft^2/hr.$$

$\ \ \ \$ b.) The volume of a sphere of radius $r$ is $$V = \displaystyle{ 4 \over 3 } \pi r^3$$ GIVEN: $\ \ \ \displaystyle{ dr \over dt } = 3 \ ft/hr.$

FIND: $\ \ \ \displaystyle{ dV \over dt }$ when $r=10 \ ft.$

Now differentiate the volume equation with respect to time $t$, getting

$$D \{ V \} = D \{ \displaystyle{ 4 \over 3 } \pi r^3 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = \displaystyle{ 4 \over 3 } \pi \cdot 3 r^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = 4 \pi r^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dr \over dt } = 3$ and $r =10. \Big)$

$$\displaystyle{ dV \over dt } = \displaystyle{ 4 \over 3 } \pi \cdot 3 (10)^2 (3) \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = 1200 \pi \ ft^3/hr. \approx 3769.91 \ ft^3/hr.$$