Question

Gas is escaping from a spherical balloon at the rate of 900cm3/sec. How fast is the surface area, radius of the balloon shrinking when the radius of the balloon is 30cm?

Answer 1

Answer:

$60cm^2/s$ and $\frac{1}{4\pi} cm/s$

Step-by-step explanation:

The volume of a spherical balloon is given by:

$V = \frac{4}{3} \pi r^3$

Differentiating with respect to time, we get:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ ...1

It is given that the rate of change of volume is $-900cm^3/s$ (since the gas inside is escaping) and the radius is 30cm, so:

$-900 = 4\pi (30)^2 \frac{dr}{dt}$

Or, $\frac{dr}{dt} = -\frac{1}{4\pi} cm/s$

Now, the surface area of the balloon is given by:

$S = 4\pi r^2$

Differentiating, again, with respect to time:

$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

We now know the values of r and $\frac{dr}{dt}$ both, so:

$\frac{dS}{dt} = -8\pi * 30 * \frac{1}{4\pi} cm^2/s$

So, $\frac{dS}{dt} = -60cm^2/s$

Hence, the rate of decrease of area of balloon is equal to $60cm^2/s$, and the rate of decrease of its radius, is equal to $\frac{1}{4\pi} cm/s$.