Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3.3 ft3/min. how rapidly is the radius of the balloon increasing when the radius is 0.65 feet?

Answer 1

The radius is increasing at a rate of  0.31cm/sec.

We know the volume of the balloon is increasing at a rate of 3.3 $ft^{3} min$

This is expressed as:

$\frac{dV}{dT}$ = 3.3

The volume of a sphere is given by:

V = $\frac{4}{3}$π$r^{3}$

The question asks to find the rate of which the radius is increasing when the radius is 0.65 feet.

We need to find:

$\frac{dr}{dt}$

Since we don't have t in our equation, we will be differentiating implicitly. We can use the chain rule for this:

$\frac{dV}{dT}$ = $\frac{dV}{dR}$.$\frac{dr}{dt}$

We already know  

$\frac{dV}{dT}$ = 3.3

∴

3.3 =  $\frac{dV}{dR}$.$\frac{dr}{dt}$

To find $\frac{dV}{dR}$  we differentiate  V = $\frac{4}{3}$π$r^{3}$

∴

3.3 = 4$\frac{3.3}{5.30}$$r^{2}$.$\frac{dr}{dt}$

Dividing ,

$\frac{3.3}{5.30}$ = $\frac{dr}{dt}$ = 0.622 ft/min = 0.31 cm/sec

The radius is increasing at a rate of  0.31cm/sec.

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