A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm ^3/ min . When the thickness of ice is 5 cm, the rate at which the thickness of ice decreases is...

Step-by-step solution

Let r be the radius of the iron ball and x be the thickness of the ice. Given r = 10 cm. The volume of the ice is V = 4 3 (r+x)^3 - 4 3 r^3 . Differentiating with respect to time t : dV dt = 4 (r+x)^2 dx dt Given that the ice melts at a rate of 50 cm ^3/ min , we have dV dt = -50 . When x = 5 cm, r+x = 10 + 5 = 15 cm. Substituting the values: -50 = 4 (15)^2 dx dt -50 = 900 dx dt dx dt = - 50 900 = - 1 18 cm / min The rate at which the thickness of ice decreases is 1 18 cm / min .