Given v=4/3(pi)r^(3) and s=4(pi)r^(2) for a sphere, and the fact that the radius of the sphere is decreasing at 1/2 inches per minute, find the rate at which the volume is changing, in cubic inches per minute for a sphere at the time when the radius is 6 inches show work so i can understand)

We are given that dr/dt= -1/2 in/min (because the radius is decreasing, dr/dt must be negative).
Then we are asked to find dV/dt when r=6 in.
Now let's start solving the problem.
V= 4/3π*(r^3)
⇒dV/dt= 4/3π*3*(r^2)*(dr/dt)
⇒dV/dt= 4π*(r^2)*(dr/dt)
Plug r=6(in) and dr/dt= -1/2(in/min) in the above equation, we have:
dV/dt= 4π*6^(2)*(-1/2)= -72π (in^(3)/min)
Last but not least, don't forget the units because if you will take the AP Calculus exam, it will cost you a lot of points.
Hope that will help :))

Ithink its 126 hope its