a spherical balloon is being filled with helium at the constant rate of 4 ft^3/min. Calculate the rate at which the surface of the balloon is increasing at the instant when its volume is 32pi/3 ft^3
a spherical balloon is being filled with helium at the constant rate of 4 ft^3/min. Calculate the rate at which the surface of the balloon is increasing at the instant when its volume is 32pi/3 ft^3
For a sphere..
$\displaystyle V = \frac{4}{3}\pi r^3$
$\displaystyle S = 4\pi r^2$
The plan is to find the radius at that exact point, then solve for the change of the radius at the point. Then we plug that into the equation for the rate of change of the surface and get our answer.
1. Get radius length
$\displaystyle V = \frac{32\pi}{3} = \frac{4}{3}\pi r^3$
$\displaystyle r = 2 ft$
2. Get derivative of Volume (with respect to time), solve for rate of change of radius.
$\displaystyle \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
$\displaystyle 4 = 4\pi 2^2 \frac{dr}{dt}$
$\displaystyle \frac{dr}{dt} = \frac{1}{4\pi} ft/min$
3. Get derivative of Surface Area, plug in rate of change of radius, get rate of change of area.
$\displaystyle \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$
$\displaystyle \frac{dS}{dt} = 8\pi (2) \frac{1}{4\pi}$
$\displaystyle \frac{dS}{dt} = 4 ft^2/min$
Cheers!