Assume that the radius ,r , of a sphere is expanding at a rate of 10in./min. The volume of a sphere is V=43πr^3. Determine the rate at which the volume is changing with respect to time when r=5in. The volume is changing at a rate of how many in^3/min.?
You mean V = (4/3) pi r^3
dV/dt = (4 pi r^2) dr/dt
(You really do not need calculus. It is the surface area times dr/dt)
12.566
To find the rate at which the volume is changing with respect to time, we can use the chain rule of differentiation.
Given: The rate of change of the radius with respect to time is dr/dt = 10 in./min.
The volume of a sphere is given by V = (4/3)πr^3.
First, let's differentiate the volume equation with respect to time (t):
dV/dt = dV/dr * dr/dt
To find dV/dr, we differentiate the volume equation with respect to the radius (r):
dV/dr = 3 * (4/3)πr^2 = 4πr^2
Now, substituting the given value of r = 5 in.:
dV/dr = 4π(5)^2 = 100π in^2
Next, substitute the given rate of change dr/dt = 10 in./min:
dV/dt = 100π * 10 = 1000π in^3/min
So, the volume is changing at a rate of 1000π in^3/min.