# 1st order DE - word problem II

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:

$$\frac{dQ}{dt} = rate_{entering} - rate_{leaving}$$

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

Rate of leaving: $$\frac{Q}{(60-t)} \cdot \frac{3gal}{min} = \frac{3Q}{60-t}$$

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

$$\frac{dQ}{dt} + \frac{3Q}{60-t} = 2$$
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:
$$Q\cdot (60-t)^3 = \int 2(60-t)^3$$
$$Q\cdot (60-t)^3 = \frac{(60-t)^4}{2} + C$$

$$Q = \frac{60-t}{2} + \frac{C}{(60-t)^3}$$

Before I go ahead and find a specific solution, is my general solution correct?

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## Answers and Replies

mjsd
Homework Helper
looks right to me

HallsofIvy
Science Advisor
Homework Helper
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:

$$\frac{dQ}{dt} = rate_{entering} - rate_{leaving}$$

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

Rate of leaving: $$\frac{Q}{(60-t)} \cdot \frac{3gal}{min} = \frac{3Q}{60-t}$$

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

$$\frac{dQ}{dt} + \frac{3Q}{60-t} = 2$$
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:
$$Q\cdot (60-t)^3 = \int 2(60-t)^3$$
$$Q\cdot (60-t)^3 = \frac{(60-t)^4}{2} + C$$

$$Q = \frac{60-t}{2} + \frac{C}{(60-t)^3}$$

Before I go ahead and find a specific solution, is my general solution correct?

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

$$Q\cdot (60-t)^{-3} = \int 2(60-t)^{-3}$$

$$Q\cdot (60-t)^{-3} = \frac{1}{(60-t)^2} + C$$

$$Q = -(60-t) + \frac{C}{(60-t)^{-3}}$$

$$Q = C\cdot(60-t)^{-3} - (60-t)$$

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

$$Q = \frac{(60-t)^3}{3600} - (60-t)$$

Looks good?

And for part b), is that like finding the maxima of the function Q(t)?

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HallsofIvy
Science Advisor
Homework Helper
You've lost that same sign again!
Multiplying both sides by (60-t)3 does NOT change
$$\frac{1}{(60-t)^2}$$
to - (60-t)!

Your formula is
$$Q(t)= \frac{(60-t)^3}{3600}+ (60-t)$$

Yes, find when the "maximum amount of salt in the tank" is exactly like maximizing that function!

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
Science Advisor
Homework Helper
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.

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

now i got it.