Plaster is being applied uniformly to a cannon-ball at the rate of 24 !
cubic
inches per minute. When the radius of the ball and plaster is 6 inches at what
rate is the surface area increasing?
To find the rate at which the surface area is increasing, we can use the formula for the surface area of a sphere:
Surface Area = 4πr^2
where r is the radius of the sphere.
Given that the radius is 6 inches, we can differentiate the formula with respect to time to find the rate of change of the surface area, dA/dt:
dA/dt = d/dt (4πr^2)
To find dA/dt, we need to differentiate the formula with respect to time. Before doing that, we need to express the radius r as a function of time t.
Since plaster is being applied uniformly to the cannonball at a rate of 24 cubic inches per minute, the volume of the plaster is increasing at a constant rate. The volume of a sphere is given by:
Volume = (4/3)πr^3
Differentiating the volume with respect to time, we have:
dV/dt = d/dt ((4/3)πr^3)
Given that dV/dt = 24 cubic inches per minute, we can solve for dr/dt, the rate at which the radius is changing:
24 = d/dt ((4/3)πr^3)
Now, we can differentiate the formula for the surface area with respect to time and substitute the value of dr/dt:
dA/dt = d/dt (4πr^2)
= 8πr * dr/dt
Finally, substituting the value of dr/dt with the calculated value of 24, and substituting the radius r as 6, we can find the rate at which the surface area is increasing:
dA/dt = 8πr * dr/dt
= 8π(6) * 24
= 1,152π cubic inches per minute
Therefore, the surface area is increasing at a rate of 1,152π cubic inches per minute.