Question

"if a snowball melts so that its surface area decreases at a rate of 1 cm 2 min, find the rate at which the diameter decreases when the diameter is 10 cm." (stewart 249) stewart, james. single variable calculus, 8th edition. cengage learning, 20150101. vitalbook file.

Answer 1

We need to find the rate of the diameter, which we can denote as d(d)/dt.


`(dA)/(dt) = (dA)/(dx) \cdot (dr)/(dt)`

`(dA)/(dt) = -1`, since it is decreasing.


`-1 = (dA)/(dr) \cdot (dr)/(dt)`


`A = 4\pi \cdot r^(2)`

`(dA)/(dr) = 8\pi \cdot r`

At r = 5:

`(dA)/(dr) = 40 \pi`


`(dr)/(dt) = -(1)/(40 \pi)`

Since the diameter is twice the radius and this is simply the rate at which the radius is decreasing, then the diameter will be decreasing twice as fast:


`(d(d))/(dt) = -(1)/(20\pi)`

Thus, the diameter is decreasing at a rate of 1/(20pi) cm/min.