SOLUTION 11: Draw a sphere of radius $r$, and assume that $r$ is a function of time $t$.

The volume of a sphere of radius $r$ is $$V = \displaystyle{ 4 \over 3 } \pi r^3$$

The surface area of a sphere of radius $r$ is

$$S=4 \pi r^2$$

GIVEN: $\ \ \ \displaystyle{ dV \over dt } = 64 \pi \ meters^3/hr.$

FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $r=2 \ m.$

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 $\displaystyle{ dV \over dt } = 64 \pi$ and $r =2. \Big)$

$$64 \pi = 4 \pi (2)^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$ $$64 \pi = 16 \pi \displaystyle{ dr \over dt } \ \ \ \longrightarrow$$ $$\displaystyle{ dr \over dt } = 4 \ meters/hr.$$

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 $\displaystyle{ dr \over dt } = 4$ and $r =2. \Big)$

$$\displaystyle{ dS \over dt } = 8 \pi (2)(4) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 64 \pi \ m^2/hr. \approx 201.06 \ meters^2/hr.$$

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