calculus problem >_<?

Question

A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at a rate of 10 inches cubed / minute, how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?

Answer 1

r_ball = 0.5 * D_ball = 4 in.

i(t) = thickness of ice as a function of time

r(t) = total spherical radius as a function of time

= r_ball + i(t)

V(t) = total spherical volume as a function of time

A(t) = total spherical surface area as a function of time


V(ball) = (4/3)*pi*(r_ball)^3

V = (4/3)*pi*r^3

V(ice) = V - V(ball) = (4/3)*pi*(r^3 - r_ball^3)

(r^3 - r_ball^3) = i^3 + 3*r_ball*i^2 + 3*r_ball^2*i

dV(ice)/dt = (4/3)*pi*[3*i^2*(di/dt) + 6*r_ball*i*(di/dt)
+ 3*r_ball^2*(di/dt)]

dV(ice)/dt = (4*pi)*(di/dt)*(i^2 + 2*r_ball*i + r_ball^2)

dV(ice)/dt = -10 in^3/min --------> Find di/dt when i = 2 in.

-10 = 4*pi*(4 + 16 + 16)*(di/dt)

di/dt = -10/(144*pi) ~ - 0.022 inches/min


A(t) = 4*pi*r^2 = 4*pi*(r_ball^2 + 2*r_ball*i + i^2)

dA(t)/dt = 4*pi*[2*r_ball*(di/dt) + 2*i*(di/dt)]

= 8*pi*(di/dt)*(r_ball + i)


If (di/dt) = - 0.022 inches/min, then dA/dt when i = 2 inches is:

dA(t)/dt = 8*pi*(-0.022)*6 = - 48*(0.022)*pi ~ - 3.317 in^2/min