Given information about the fill and empty rate, how do I find the volume after t minutes?

Oil is leaking from a pipeline on the surface of a lake and forms an oil slick whose volume increases at a constant rate of 2000 cubic centimeters per minute. A recovery device arrives on the scene and begins removing oil. The rate at which oil is removed is #R(t)=sqrt(t)# cubic centimeters per minute, where t is the time in minutes since the device began working.
If 60,000 cubic centimeters of oil had already leaked by the time the recovery device began removing oil, write (but do not evaluate) an expression that gives the volume of oil after t minutes.

1 Answer
Mar 4, 2017

#V(t)=60000 + \int_0^t 2000 - sqrt(t)dt#

Explanation:

In this question, we know that oil is being both added to and removed from the lake.

Initial Amount -- Since 60,000 cubic centimeters of oil had already leaked at time 0, we know V(0) = 60000

Filling -- Because of the leak, 2000 cubic centimeters of oil are being added to the lake each minute, meaning this amount is being added to the volume of the oil slick. In other words, our fill rate #F(t) = 2000#

Emptying -- Because of the recovery device, oil is being removed from the lake. From the given information, they tell us that the rate it's being removed at is #R(t)=sqrt(t)#

Note that both the fill & emptying information are rates versus the initial information, which is an amount.

Our volume would be calculated by the amount being pumped in minus the amount being pumped out. Because we are given rates, we must take the antiderivative of the rate to get the amount.

Let's call our rate of change (fill rate - empty rate), #V'(t)#.
#\int_0^t V'(t) dt = [V(t)]_0^t = V(t) - V(0)#
So, #V(t) - V(0) = \int_0^t V'(t) dt#
#V(t) = V(0) + \int_0^t V'(t) dt = V(0) + \int_0^t F(x) - E(x) dt#
#=60000 + \int_0^t 2000 - sqrt(t)dt#

So, #V(t)=60000 + \int_0^t 2000 - sqrt(t)dt#