Sand falls from a conveyor belt at a rate of 15 meter cube per minute onto the top of conical pile. The height of the pile is always 3/8 of the base diameter. How fast are the height and the radius changing when the pile is 7 meters high? So volume equals 1/3 Pi R-squared H and we know that H = 3/8 diameter. Diameter is just two R so H = 3/4 R Now it's it specifies we want to figure out the how fast are the height and the radius changing when the pile is 7. So if we put seven and for our H we can figure out our R is just 28 thirds. So we're going to take that volume equation and figure out the DVDT. So if volume equals 1/3 Pi R-squared H, and we stick H as 3/4 R, we get volume equaling 1/4 Pi r ^3 DVDT would be 3/4 Pi R-squared DRDT. We're given that the pile is increasing at 15 meter cube per minute equal 3/4 Pi. The radius was 28 thirds squared and we want to find DRDT. So when we compute that we get 45 / 196 Pi meters per minute, but we're asked in centimeters per minute. So we're going to multiply by 100 centimeters to 1m and we get 1125 / 49 Pi centimeters per minute. Now we also are asked how the heights changing. If we go back to the H equaling 3/4 R equation, we could get the derivative of the height changing in terms of time equaling 3/4 of how the radius is changing in terms of time. So DHDT equal 3/4. The DRDT we found a moment ago was 45 / 196 Pi, so 135 / 784 Pi meters per minute. Changing it into centimeters, we'd multiply by 100 / 1. There are a lot of different ways to do this. This is one method.