V=4/3 pi r ^3 is the formula for the volume of a sphere. Find dV for a sphere if the radius of 10" is increased by .1"

Please help. I have no idea how to do this problem!

dV/dr = 4π r^2

dV = (4πr^2)(dr)
sub in r = 10 and dr = .1
dV = 4π(100)(.1) = 40π

To find the change in volume (dV) for a sphere when the radius is increased by 0.1 inches, we can use calculus. The formula for the volume of a sphere is V = (4/3)πr^3.

First, let's differentiate the formula with respect to r to find the derivative:

dV/dr = d/dx [(4/3)πr^3]
= (4/3)π * 3r^2
= 4πr^2

This tells us how the volume changes as the radius r changes.

To find dV, we can substitute the given radius into the derivative formula and calculate:

dV = 4π(10)^2 * 0.1
= 4π(100) * 0.1
= 40π

So, dV is equal to 40π cubic inches.

Therefore, the change in volume (dV) for a sphere when the radius of 10 inches is increased by 0.1 inches is 40π cubic inches.