The spherical iron ball, 12 inches diameter, is coated with a layer of ice of uniform thickness. So we're going to use volume formulas and surface area formulas. For a sphere. Volume is 4 thirds π R cubed surface area is 4 Pi r ^2. If the ice melts at a rate of 14 inches cube per minute, rate means it's the derivative or how the volume is changing in terms of time it's melting. So we know DVDT is going to be a -14 inches cube per minute. How fast is the thickness of the ice decreasing when it is 5 inches thick? So how fast is the thickness? The thickness is going to be the radius changing. So the DRDT when it's 5 inches thick. Now if the ice is five inches thick, we have to think about adding it to the original radius of the spherical iron ball. So the original radius is 6 inches because the diameter was 12. So we're at a radius of 6 + 5 or 11. So taking our first derivative of volume, we get DVDT equaling 4 Pi R-squared DRDT. We stick in that -14 inches cube per minute 4 Pi, the radius being 11 ^2 DRDT. We do some plugging and chugging. We get -7 / 242π inches per minute is the change in the radius. But in course composite saying the thickness of the ice is decreasing. So the word decreasing says they've already taken into account that negative. So we're just going to put in 7 / 242 pie inches per minute. Then the follow up for the second part says how fast is the outer surface area of ice decreasing? So now we're going to take the derivative of the surface area in time. In terms of time, we get eight π RDRDT. Well, we just found DRDT in the previous portion and we know the R is still 11. So eight π * 11 * -7 / 242π. We reduce that. We get -2811 inches cubed per minute. Oh, inches squared per minute because it's an area. So then we need to read and realize it says the outer surface area of ice is decreasing. So it's already taken into account that negative. So we have 2811 inches squared per minute.