Archive for April, 2010

Mixture problem in Calculus

April 29, 2010 7 comments

I keep forgetting the way to solve this problem. So, when I in future have no clue I know where to look.

The problem
A 200-galon tank is half full of distilled water. At time t=0, a solution containing 0.5 pound/galon of salt enters the tank at the rate of 5 gal/minutes, and the well-stirred mixture is withdrawn at the rate of 3 gal/minutes.
a. Give a mathematical model representation of the above problem.
b. At what time, will the tank be full?
c. At the time the tank is full, how many pounds of salt will it contain?

The mathematical model
The model describing the above process is based on the following formula:

rate of change of the amount of salt  = (rate of salt arrived) – (rate of salt departed)

Suppose V(t) is the volume of the liquid in the tank at time t. The volume of liquid inside the tank will increase as the inflow (5 gal/minutes) is bigger than the outflow (3 galon/minutes).  Thus, we have

V(t)= 100 galon + (5 gal/min – 3 gal/min) \times t min,
V(t)=(100 + 2t) gal,

as we know that initially the tank is half full (100 gal).

Suppose y(t) is the amount of salt inside the tank at time t. The rate in of salt is as follows,

rate in (pound/min) = (0.5 pound/gal) \times (5 gal/min) = 2.5 pound/min,

while the rate out is
rate out (pound/min) = \frac{y(t)}{V(t)} \times (pound/gal) outflow (gal/min)
= \frac{y}{100+2t} \times 3 (pound/min)
= \frac{3y}{100+2t} (pound/min).

Therefore, the mathematica model is given by,

\frac{dy}{dt}=2.5 - \frac{3y}{100+2t}.          (1)

The solution
First we determine time t at which the tank will be full. We use the following equation,

V(t)=(100 + 2t),

We solve for t when the volume V(t) is 200 galon,  then we obtain t=50.

To solve the differential equation (1) we need the initial condition
as follow,


We are going to use Maple to solve the problem

> model := diff(y(t),t) = 2.5 - (3*y(t))/(100+2*t);
> init := y(0)=0;
> dsolve({model,init});

We obtain the solution that is given by

y(t) = 50+t-\frac{2500 \sqrt{50}}{(50+t)^{3/2}}

When t=50, the amount of salt inside the tank will be

y(50) = 82.32233047.