Question

A collapsible spherical tank is being relieved of air at the rate of 2 cubic inches per minute. At what rate is the radius of the tank changing when the surface area is 12 square inches?

Answer 1

radius is changing at the rate of `1/6` inches/minute

Explanation:

Let us set up the following variables:

`{(r, "radius (inches)"), (t, "time (minutes)"), (A, "Surface Area (inches)"^2), (V, "Volume (inches)"^3) :}`

Then "tank is being relieved of air at the rate of 2 cubic inches per minute" `=> (dV)/dt=2` and we want to find `(dr)/dt` when `A=12`

Because the tank is spherical then:
`A = 4pir^2`
`V=4/3pir^3 => (dV)/(dr) = 4pir^2 (=A)`

By the chain rule we have;
`(dV)/dt = (dV)/(dr) * (dr)/dt`
`:. 2 = A * (dr)/dt`
`:. (dr)/dt = 2/A`

So, when `A=12 => (dr)/dt = 2/12 = 1/6`