1st order DE - word problem II

[ranger]

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?

 

[mjsd]

looks right to me

 

[HallsofIvy]

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.

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

 

[ranger]

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)?

 

[HallsofIvy]

You've lost that same sign again!
Multiplying both sides by (60-t)³ does NOT change
[tex]\frac{1}{(60-t)^2}[/tex]
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!

 

[snoothie]

hi,

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?

 

[HallsofIvy]

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.

 

[snoothie]

thanks. i didnt understand the equation fully. was thinking that in graph it would be Q vs t ...

now i got it.