Question

Two full 48-gallon tanks begin draining at t = 0. Tank Alpha's volume is changing
at a constant rate of -16/5 gal/min. The rate at which Tank Bravo's volume is
changing is given by r(t)=-1/3t-1 gal/min.

a) How much water is in each tank after 5 minutes?
b) Which tub drains rst? How do you know?

Answer 1

Cannot parse -1/3t-1

Is that (-1/3)(t-1) or -1/(3t)-1 or -1/(3t-1)?

none of those choices makes sense at t=0.

Answer 2

negative one third t minus 1

Answer 3

ok. If you mean (-1/3 t) - 1 then we have for the two tanks:

Va = 48 - ∫16/5 dt = 48 - 16/5 t
Vb = 48 - ∫1/3 t + 1 dt = 48 - t - 1/6 t^2

So,
Va(5) = 48 - 16 = 32
Vb(5) = 48 - 5 - 25/6 = 38 5/6

Now just solve for when Va and Vb = 0 to find which drains first.

If I still didn't get it right, and I suspect I might not have, since "negative one third t minus 1" is -1 at t=0, then just make the fix and reevaluate the integrals.

Answer 4

Tyler, if yu do not know how to use parentheses try to type it with lines like
1
--------
(3 t - 1)

or something.