A sphere is increasing in volume at the rate of 3(pi) cm^3/sec. At what rate is the radius changing when the radius is 1/2 cm? The volume of a sphere is given by V=4/3(pi)r^2
A. pi cm/sec
B. 3 cm/sec
C. 2 cm/sec
D. 1 cm/sec
E. .5 cm/sec
dv/dt = 4pi r^2 dr/dt
plug and chug
To find the rate at which the radius is changing, we need to use the relationship between the volume of the sphere and its radius.
The volume of a sphere is given by V = (4/3)πr^3.
We are given that the volume is changing at a rate of 3π cm^3/sec. Therefore, dV/dt = 3π cm^3/sec.
Now, let's find the rate at which the radius is changing, dr/dt, when the radius is 1/2 cm.
We can use the chain rule to relate the rate of change of the volume to the rate of change of the radius. The chain rule states that for a function y = f(g(x)), the derivative dy/dx can be found using the formula dy/dx = f'(g(x)) * g'(x).
In this case, y = V and x = r. Applying the chain rule to V = (4/3)πr^3, we have:
dV/dr = dV/dx * dx/dr
The left hand side, dV/dr, represents the rate of change of the volume with respect to the radius, which we want to find.
The right hand side, dV/dx, represents the rate of change of the volume with respect to x (which is the radius), and dx/dr represents the rate of change of the radius with respect to x.
Now, let's differentiate the volume equation with respect to r:
dV/dr = d((4/3)πr^3)/dr
= (4/3)π * 3r^2
= 4πr^2
Therefore, dV/dr = 4πr^2.
Now, we can substitute the given values into the equation to find the rate at which the radius is changing when the radius is 1/2 cm.
dV/dr = 4πr^2
3π = 4π(1/2)^2 (since r = 1/2 cm)
3 = 4 * (1/4)
3 = 1
Thus, the rate at which the radius is changing when the radius is 1/2 cm is 1 cm/sec.
Therefore, the correct answer is D. 1 cm/sec.