Answer:
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
