As a snowball melts, its surface area and volume decrease. The surface area, S, in square centimetres, is modelled by the equation S = 4π r2, where r is the radius, in centimetres. The volume, V, in cubic centimetres, is modelled by the equation V = (4/3)π r3.

a) How do you determine the average rate of change of the surface area and of the volume as the radius decreases from 25 cm to 20 cm?

b) How do you determine the instantaneous rate of change of the surface area and the volume when the radius is 10 cm?

2. Average rate of change is just the slope of the secant line. Plug in 25cm and 20cm for both equations to get the y-values that go with them. Then just use y2-y1/x2-x1 to get the slope. That's your average change.

i.e. 4πr^2 yeilds the coordinates (25, 4π(25)^2) and (20, 4π(20)^2). So, your answer is [4π(25)^2 - 4π(20)^2]/(25-20)

For instantaneous change at 10cm, just derive both equations with respect to x and plug in 10cm for x.

i.e. for (4/3)πr3, our derivative is surface area 4πr^2. Plug in for 10cm, and you get your answer.