1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 1st order DE - word problem II

  1. Feb 27, 2007 #1


    User Avatar
    Gold Member

    A tank initially contains 60 gal of pure water. Brine (saltwater) containing 1lb of salt per gallon enters the tank at 2gal/min. and the (perfectly mixed) solution leaves the tank at 3 gal/min; thus the tank is empty after exactly 1 hr.
    a)Find the amount of salt in the tank after t minutes.
    b)What is the maximum amount of salt ever in the tank?

    Let Q be the amount of salt.
    I'll work with part a) first:

    [tex]\frac{dQ}{dt} = rate_{entering} - rate_{leaving}[/tex]

    Rate of entering: 1lb/gal * 2gal/min = 2lb/min

    Rate of leaving: [tex]\frac{Q}{(60-t)} \cdot \frac{3gal}{min} = \frac{3Q}{60-t}[/tex]

    This gives me the first order linear differential to be:
    [tex]\frac{dQ}{dt} = 2 - \frac{3Q}{60-t}[/tex]

    [tex]\frac{dQ}{dt} + \frac{3Q}{60-t} = 2[/tex]
    Well P(x) or P(t) in this case would be 3/(60-t), which when worked out would give me an integration factor of (60-t)^3.
    I would eventually get:
    [tex]Q\cdot (60-t)^3 = \int 2(60-t)^3[/tex]
    [tex]Q\cdot (60-t)^3 = \frac{(60-t)^4}{2} + C[/tex]

    [tex]Q = \frac{60-t}{2} + \frac{C}{(60-t)^3}[/tex]

    Before I go ahead and find a specific solution, is my general solution correct?
    Last edited: Feb 27, 2007
  2. jcsd
  3. Feb 27, 2007 #2


    User Avatar
    Homework Helper

    looks right to me
  4. Feb 27, 2007 #3


    User Avatar
    Science Advisor

    You've "lost a sign". The negative in "60-t" will introduce a negative into the integral. The integrating factor is (60-t)-3.

  5. Feb 27, 2007 #4


    User Avatar
    Gold Member

    Wow. OK, so using the integrating factor (60-t)^-3:

    [tex]Q\cdot (60-t)^{-3} = \int 2(60-t)^{-3}[/tex]

    [tex]Q\cdot (60-t)^{-3} = \frac{1}{(60-t)^2} + C[/tex]

    [tex]Q = -(60-t) + \frac{C}{(60-t)^{-3}}[/tex]

    [tex]Q = C\cdot(60-t)^{-3} - (60-t)[/tex]

    Using the initial values Q(0) = 0
    I Get get C = 1/3600, which gives me a specific solution of:

    [tex]Q = \frac{(60-t)^3}{3600} - (60-t)[/tex]

    Looks good?

    And for part b), is that like finding the maxima of the function Q(t)?
    Last edited: Feb 27, 2007
  6. Feb 27, 2007 #5


    User Avatar
    Science Advisor

    You've lost that same sign again!
    Multiplying both sides by (60-t)3 does NOT change
    to - (60-t)!

    Your formula is
    [tex]Q(t)= \frac{(60-t)^3}{3600}+ (60-t)[/tex]

    Yes, find when the "maximum amount of salt in the tank" is exactly like maximizing that function!
  7. Feb 28, 2007 #6

    i'm working on a similar problem. i'm trying to figure out why it is (60-t) & not (60+t)...is it because the tank will empty out and we are calculating time backwards?
  8. Feb 28, 2007 #7


    User Avatar
    Science Advisor

    Who is calculating time backwards? The tanks starts with 60 gallons of water in it and loses one gallon per minute: after 1 minute, 60- 1= 59 gallons, after 2 minutes 60- 2= 58 gallons, etc. until, after 60 minutes 60-60= 0 gallons and the tank is empty. The formula is 60- t because the tank is emptying- the water is going out. It it were 60+t, the tank would be getter "fuller"- water would be coming in.
  9. Feb 28, 2007 #8
    thanks. i didnt understand the equation fully. was thinking that in graph it would be Q vs t ...

    now i got it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook