<p>Water is pumped into an underground tank at a constant rate of 8 gallons per minute.
Water leaks out of the tank at the rate of
√t + 1 (Square root of t+1) gallons per minute, for 0 ≤ t ≤ 120
minutes. At time t = 0, the tank contains 30 gallons of water.
How many gallons leak out of the tank from time t = 0 to t = 3 minutes?
How much water is in the tank at t = 3
Write an expression for A(t), the total number of gallons of water in the tank at time t.
At what time is teh amount of water the maximum?</p>
<p>I think you're in the wrong forum. This could be completely wrong, but this is my idea:</p>
<p>You're given formulae that are rates. When you want to go from rate to amount, you integrate (ie: to get from a velocity function to a distance function, you integrate the velocity). Therefore --</p>
<p>How many gallons leak out:
integral (√t + 1) from 0 to 3.</p>
<p>Amount of water in the tank at t=3:
formula you need is 30 + 8t - (√t + 1)
because you start with 30, add 8 gallons per minute, and remove √t + 1. Integrate that, and plug in 3 for t.</p>
<p>A(t):
I think it should be the integrated formula we used in the above step? I'm not sure why this question would come after the preceding one, though.</p>
<p>Maximum:
You just need the first derivative line test here. Use the non-integrated version of a(t) (ie: 30 + 8t - (√t + 1)), since that is technically the first derivative of a(t). Set it equal to zero and solve it, and then plug in values as you would on a first derivative line test.</p>
<p>Anyway, I did this really quick, and I'm prone to silly errors when it comes to calculus. I hope, at the very least, some of the concepts can make sense to you now and that you can use some of the hints to, you know, actually solve the problem correctly. My apologies if this is bad advice. :/</p>
<p>Good luck!</p>