A flask contains 6 liters of salt water. Suppose water containing M grams of sal

Question

A flask contains 6 liters of salt water. Suppose water containing M grams of salt per liter is pumped in at a rate of 3 liters per hour and the well-stirred mixture is pumped out at the same rate.

A. Write a differential equation for the amount, y= f(t), of salt in the flask at time, t.

B. In the long run, we would like the amount of salt in the flask to approach 12 grams. What should we select for the value of M?

C. Sketch and label the graphs of the constant solutions and the solution curves corresponding to y(0)=10 and

y(0)=13 of the differential equation obtained in (A) when the amount of salt in the flask approaches 12 grams.

I'm having trouble with this so please provide an explanation if you can, thanks!

Explanation / Answer

(a) We have:
{rate of change in the salt} = {rate salt pumped in} ? {rate salt pumped out} .
The rate that the salt is pumped in is 3M grams per hour, because every hour 3 liters pumped in,
and each liter contains M grams of salt. On the other hand, if at a time t there are y(t) grams of
salt in the ?ask, and half of the tank (3 out of 6 liters) is pumped out per hour, then half of the
amount of y(t) is pumped out per hour. Thus we get that
dy/dt = 3M ?1/2y(t).
(b) Constant solution occurs when 3M ?1/2y(t)
y = 0, i.e. when
y = 6M=12
M=2.
The solution corresponding to the y(0) = 8 initial condition is under the vertical line y = 6M, and
the slope of the tangent of this solution at t = 0 is
dy/dt(0) = 3M ?1/2
y(0) =3M ?1/2*10
3M-5=1 > 0
so this solution is increasing. The sketch of the graph is as follows:
(c) As t ? ? the solution y(t) approaches the constant solution, so the amount of salt in the ?ask
will approach 6M=12 grams in the long run

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