A spherical iron ball 10 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at a rate of 15 StartFraction in cubed Over min EndFraction comma how fast is the thickness of the ice decreasing when it is 3 in. thick? How fast is the outer surface area of ice decreasing?

a) Rate at which the thickness of the ice is decreasing, da/dt = - 0.024 in/min

b) Rate at which the outer surface area is decreasing, dA/da = - 4.22 in²/min

Step-by-step explanation:

The volume of the sphere, V = 4/3 πr³

The thickness of the ice = a

inner diameter = 10 in

inner radius = 10/2 = 5 in

outer radius = 5 + a

Therefore, volume of ice = volume of the outer radius - volume of the inner radius

V = 4/3 π (4+a) ³ - 4/3 π4³

V = 4/3 π (4³ + 12a² + 48a + a³ - 4³)

V = 4/3 π (a³ + 12a² + 48a)

dV/dt = 4/3 π (3a²da/dt + 24ada/dt + 48da/dt)

dV/dt = 4π (a²da/dt + 8ada/dt + 16da/dt) ... (1)

When a = 3 in, dV/dt = - 15 m³/min

Substitute these values into equation (1)

-15 = 4π (9da/dt + 24 da/dt + 16da/dt)

-15/4π = 49da/dt

-1.194 = 49da/dt

da/dt = - 1.194/49

da/dt = - 0.024 in/min

b) How fast is the outer surface area of ice decreasing

The area of a sphere, A = 4πr²

The outer surface area of the sphere, A = 4π (4+a) ²

dA/da = 8π (4+a) da/dt

a = 3 in, da/dt = - 0.024 in/min

dA/da = 8π (4+3) * (-0.024)

dA/da = - 4.22 in²/min