A spherical balloon is being inflated at the rate of 12 cubic feet per second. What is the radius of the balloon when its surface area is increasing at a rate of 8 feet square feet per second? Volume= (4/3)(pi)r^3 Area= 4(pi)r^2?

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A spherical balloon is being inflated at the rate of 12 cubic feet per second. What is the radius of the balloon when its surface area is increasing at a rate of 8 feet square feet per second? Volume= (4/3)(pi)r^3 Area= 4(pi)r^2?

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Answer 1 · 2015-04-08T00:45:04

Let
$S$ $S$ be the surface area (in square feet);
$V$ $V$ be the volume of the sphere (in cubic feet);
$t$ $t$ be the time (in seconds); and
$r$ $r$ be the radius (in feet).

$\frac{d S}{d t} = 8$ $(d S)/(dt) = 8$ (given)
$\to \frac{d t}{d S} = \frac{1}{8}$ $rarr (dt)/(d S) = 1/8$
$\frac{d V}{d t} = 12$ $(dV)/(dt) = 12$ (given)
Therefore
$\frac{d V}{d S} = \frac{d V}{d t} \cdot \frac{d t}{d S} = \frac{12}{8} = \frac{3}{2}$ $(dV)/(d S) = (d V)/(dt) * (dt)/(dS) = (12)/(8) = 3/2$

Using the Volume and Surface Area formulae (as given)
$\frac{d V}{d r} = 4 π r^{2}$ $(dV)/(dr) = 4 pi r^2$
and
$\frac{d S}{d r} = 8 π r$ $(dS)/(dr) = 8 pi r$
$\to \frac{d r}{d S} = \frac{1}{8 π r}$ $rarr (dr)/(dS) = 1/(8 pi r)$
Therefore
$\frac{d V}{d S} = \frac{d V}{d r} \cdot \frac{d r}{d S} = \frac{4 π r^{2}}{8 π r} = \frac{r}{2}$ $(dV)/(dS) = (dV)/(dr) * (dr)/(dS) = (4 pi r^2)/(8 pi r) = r/2$

Equating our two solutions for $\frac{d V}{d S}$ $(dV)/(dS)$
$\frac{r}{2} = \frac{3}{2}$ $r/2 = 3/2$

and the radius ( $r$ $r$ ) is 3 feet.

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A spherical balloon is being inflated at the rate of 12 cubic feet per second. What is the radius of the balloon when its surface area is increasing at a rate of 8 feet square feet per second? Volume= (4/3)(pi)r^3 Area= 4(pi)r^2?

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