A spherical water balloon is being filled with water at the rate of 125 cubic inches per minute. At what rate is the radius increasing when the radius is 10 inches? At what rate is the surface area increasing?
v = 4π/3 r^3
dv/dt = 4πr^2 dr/dt
Now just plug in your numbers.
and for rate of surface area increasing?
To find the rate at which the radius is increasing, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where V is the volume and r is the radius. The problem states that the volume is increasing at a rate of 125 cubic inches per minute. We can differentiate both sides of the equation with respect to time t to find the rate of change of volume with respect to time:
dV/dt = (d/dt) [(4/3) * π * r^3]
The derivative of r^3 with respect to t is 3r^2(dr/dt), since the derivative of r^3 is 3r^2 and we have to multiply it by the derivative of r with respect to t:
dV/dt = (4/3) * π * 3r^2(dr/dt)
Simplifying the equation:
125 = 4πr^2(dr/dt)
To find the rate at which the radius is increasing, we need to solve for dr/dt. Rearranging the equation:
(dr/dt) = 125 / (4πr^2)
Now we can substitute the given radius value, r = 10 inches, into the equation to find the rate of change of the radius:
(dr/dt) = 125 / (4π*10^2)
(dr/dt) = 125 / (400π)
(dr/dt) ≈ 0.099 inches per minute
Therefore, when the radius is 10 inches, the rate at which the radius is increasing is approximately 0.099 inches per minute.
To find the rate at which the surface area is increasing, we can use the formula for the surface area of a sphere:
A = 4πr^2
Again, we can differentiate both sides of the equation with respect to time t to find the rate of change of surface area with respect to time:
dA/dt = (d/dt) [4πr^2]
The derivative of r^2 with respect to t is 2r(dr/dt), since the derivative of r^2 is 2r and we have to multiply it by the derivative of r with respect to t:
dA/dt = 4π(2r)(dr/dt)
Substituting the radius value, r = 10 inches, and the rate of change of the radius, (dr/dt) = 0.099 inches per minute, into the equation:
dA/dt = 4π(2 * 10)(0.099)
dA/dt = 7.896π
dA/dt ≈ 24.8 square inches per minute
Therefore, when the radius is 10 inches, the rate at which the surface area is increasing is approximately 24.8 square inches per minute.