If the radius of a sphere is increasing at a constant rate of 4 cm/sec, then the volume is increasing at a rate of ???? cm^3/sec when the radius is 5 cm.
geez - impatient much?
I couldnt find the post, did the first one post?
To find the rate at which the volume is increasing, we need to use the formula for the volume of a sphere:
V = (4/3)πr^3
We are given that the radius is increasing at a constant rate of 4 cm/sec. This means the rate of change of the radius is 4 cm/sec, or dr/dt = 4 cm/sec.
Now, let's differentiate both sides of the volume formula with respect to time (t):
dV/dt = d/dt[(4/3)πr^3]
Using the chain rule, we can rewrite this equation as:
dV/dt = (4/3)π * 3r^2 * dr/dt
Substituting the given value dr/dt = 4 cm/sec and the given value r = 5 cm, we can calculate the rate at which the volume is increasing:
dV/dt = (4/3)π * 3(5^2) * 4 cm^3/sec
Simplifying further:
dV/dt = (4/3)π * 3 * 25 * 4 cm^3/sec
dV/dt = 200π cm^3/sec
Therefore, the volume is increasing at a rate of 200π cm^3/sec when the radius is 5 cm.