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?

1 Answer
Steve M
Nov 28, 2016

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

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