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

Answer:

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

Explanation:

Here we have

dV/dt = 56 cm³/s

$\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi\cdot3r^2 \frac{dr}{dt}$

When the radius is 5 cm we have

$56 \hspace {0.09cm}cm^3/s= \frac{4}{3}\pi\cdot3\cdot 5^2 \cdot \frac{dr}{dt} = 314.16 \times \frac{dr}{dt}$

Therefore,

$56 \hspace {0.09cm}cm^3/s= 314.16 \hspace {0.09cm}cm^2\times \frac{dr}{dt}$

From which,

$\frac{dr}{dt} = 56 \hspace {0.09cm}cm^3/s \div314.16 \hspace {0.09cm}cm^2$

$\frac{dr}{dt} = 0.1783 \hspace {0.09cm}cm/s$

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

Answer 2

Answer:

Radius is increasing at the rate of 0.1783 cm/s

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