differential equation word problem

Question

A tank contains 1000 L of brine (salt water) with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept throughly mixed and drains from the tank at the same rate. How much salt is in the tank after t minutes?

Answer 1

V = 1000 gal,

denote y(t) as the amount of salt in the tank at time t.

1. inflow of salt = 0 kg/min (because pure water enters)

denote y(t) as the amount of salt in the tank at time t.

outflow of salt = fraction of salt mass lost. To find this, we divide the rate of brine outflow by the total volume of the container, i.e. 10/1000 = 0.01. Thus, 1% of the brine will be lost per minute. multiply this by y to find the loss of salt per unit time.

1. outlfow of salt = -0.01y

y' is the change in mass of salt with respect to time, Hence, y'(t) = salt inflow rate - salt outflow rate.

1. y' = -0.01y

y' is dy/dt, so rearrange and separate

1. dy/y = -0.01dt

now integrate both sides, remembering that the tank starts with 15 kg salt, so y₀ = 15 kg.

1. lny - ln15 = ln(y/15) = -0.01t
2. y = 15e^-0.01t

Answer:

1. y(t) = 15e^-0.01t kg salt

Answer 2

Let s(t) be a function which tells us how much salt there is in the tank after t minutes. Clearly:

s(0)=15

To set up a differential equation which will allow us to solve for s(t) we need only recognize that the ratio of the amount of salt which leaves the tank per minute (let's call that amount "a") to the total amount of water which leaves the tank per minute (10 L) is equal to the ratio of the current amount of salt in the tank (s(t)) to the total amount of water in the tank (which is a constant 1000 L). Therefore, we can easily solve for "a" by setting up the following proportion:

a/10=s(t)/1000 -> a=s(t)/100

Therefore, the rate at which salt leaves the tank (in units of kg per minute) is just -a:

ds/dt=-a=-s/100

Since this is a separable differential equation we need to bring all the "s" terms to the left:

ds/s=-dt/100

Integrating both sides we obtain:

ln[s(t)]-ln[s(0)]=-t/100 -> ln[s(t)]=ln[s(0)]-t/100 -> s(t)=exp[ln[s(0)]-t/100] -> s(0)e^-t/100

Since s(0)=15 we obtain:

s(t)=15e^-t/100