Question

The volume of a balloon an be approximated by V = 4 3 π r 3 V=43πr3. If air is leaking from the balloon at a rate of 56 cubic centimeters per second, how fast is the radius of the balloon shrinking at the moment the radius is 5 centimeters? Make sure to choose the correct units for your answer.

Answer 1

Radius is increasing at the rate of 0.1783 cm/s

Step-by-step explanation:

Let r be the radius of the balloon and V be its volume at any time t.

Thus, V= (4/3)πr³

Now, let's differentiate both sides with respect to t and we obtain;

dV/dt = 4πr²•(dr/dt)

From the question, we are given that; dV/dt = 56 cm³/s

Thus,

56 = 4πr²•(dr/dt)

dr/dt = 56/(4πr²)

So,we want to find dr/dt at r=5,

Thus,

dr/dt = 56/(4π•5²)

dr/dt = 0.56/π = 0.1783 cm/s

Answer 2

The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s

Step-by-step explanation:

Here we have

dV/dt = 56 cm³/s

`(dV)/(dt) = (d)/(dt)((4)/(3)\pi r^3) = (4)/(3)\pi\cdot3r^2 (dr)/(dt)`

When the radius is 5 cm we have

`56 \hspace {0.09cm}cm^3/s= (4)/(3)\pi\cdot3\cdot 5^2 \cdot (dr)/(dt) = 314.16 * (dr)/(dt)`

Therefore,

`56 \hspace {0.09cm}cm^3/s= 314.16 \hspace {0.09cm}cm^2* (dr)/(dt)`

From which,

`(dr)/(dt) = 56 \hspace {0.09cm}cm^3/s /314.16 \hspace {0.09cm}cm^2`

`(dr)/(dt) = 0.1783 \hspace {0.09cm}cm/s`

The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.