Expert: Paul Klarreich - 3/21/2006

Question
Hi,
I hope I'm finding you in the best of health.
Please go to the following site for my question(example)
http://apcentral.collegeboard.com/repository/sg_calculus_ab_00.pdf
Then go to question#4(it says AB-4 on the top left of the page)
By part B, I wasn't sure how they got 14/3.
I know how to get the 30 because that's the initial and the 8*3 is the constant rate times the amount of time that went by.But why did they subtract 14/3?Thanks.

Answer
Hi, jeff,

You wrote:
Subject:  related rates(kind of)

>> Actually, it isn't.

Question:  Hi,
I hope I'm finding you in the best of health.
Please go to the following site for my question(example)
http://apcentral.collegeboard.com/repository/sg_calculus_ab_00.pdf

>> OK. Found it.

Then go to question#4(it says AB-4 on the top left of the page)

By part B, I wasn't sure how they got 14/3.
I know how to get the 30 because that's the initial and the 8*3 is the constant rate times the amount of time that went by.  But why did they subtract 14/3?

Thanks
---------------------------
If that's your only problem, you're doing fine.  Anyway, the amount of water in (any) tank at t=3 should be:

I. Water in the tank at the start (t=0)

PLUS

II. Water that is pumped IN during t=0 to t=3

MINUS

III. Water that leaks OUT during t=0 to t=3

Amount I is clearly 30.
Amount II is the 8 gpm times 3 minutes.
Amount III what leaked out during t=0 to t=3.  That amount was already computed in Part A of the problem.

Is that what you don't understand?  The solution to part A, which involves EITHER a definite integral, OR solving a differential equation.  (both ways are correct)