# Rate of change problem and solving equation

1. Jul 12, 2011

### CrimsonRed

1. The problem statement, all variables and given/known data
a) A large tank initially holds a mixture of 100 litres of water and 10 kg of salt. Fresh water is pumped in at a rate of 2L/min. The mixture is kept uniform by stirring and is pumped out at a rate of 2L/min. If x is the amount of salt present at any time t minutes, find the amount of salt present in the tank after 10 minutes.

b) 0 = 1 - te^(-t/2)

2. Relevant equations

3. The attempt at a solution
a. I would think this is a rate of change problem. I can imagine that the amount of salt will keep decreasing over time while the water volume stays at 100 litres. But I'm not able to form any equation to represent the problem.
Can I get any help to solve this problem?

b. I try to solve the equation but I don't have any luck. I think I need a numerical method to solve this equation. Can this equation be solved without using a numerical method? If numerical method the only way, which method should I use?

2. Jul 12, 2011

### Saraphim

Try describing the rate of change as an equation involving x(t), the amount of salt at any given time in minutes. Once you've done this, you should have a linear differential equation. Do you know how to solve this?

3. Jul 12, 2011

### CrimsonRed

thanks for your reply saraphim. I think that's my problem, I'm having difficulty to describe the question into an equation. I think once I' m able to form the equation I'm able to solve it, hopefully.

4. Jul 12, 2011

### Saraphim

Let x'(t) be the rate of change of salt in the tank and x(t) the amount of salt.

Then:

x'(t) = -2 L/minute * x(t)/(100 L)

since x(t)/(100 L) is the concentration at any given time.

5. Jul 12, 2011

### yoshtov

but isn't this an exponential decay equation? the derivative w.r.t. time shouldn't yield a linear equation. Here's my thought:

Start with the definition of exponential decay $$N(t)=N_{0}e^{-\lambda t}$$ The goal at hand is to find the value of lambda. We know that, in one minute, we will have eliminated 1/50 of the original fresh water supply, so it stands to reason that, in the same amount of time, we will have eliminated 1/50 of the original salt supply...I think. $$\frac{9.8}{10}=e^{-\lambda }$$ $$ln(0.98)=-\lambda =-0.020203$$ $$\lambda =0.020203$$$$N(t)=10e^{-0.020203t}$$$$N(10)=10e^{-0.020203\times 10}=8.17kg$$

As for the second part of your question, here's what I got:$$1=te^{\frac{-t}{2}}$$$$\frac{1}{t}=e^{\frac{-t}{2}}$$$$\frac{1}{t}=\frac{1}{\sqrt{e^{^{t}}}}$$$$t=\sqrt{e^{t}}$$$$t^{2}=e^{^{t}}$$ I could go no further, as my colleagues and I lack the insight to reduce it down any further.

Last edited: Jul 13, 2011
6. Jul 13, 2011

### CrimsonRed

a. I was just asuming it is a rate of change problem. I don't quite understand whether it is more exact to be considered as exponential decay equation.
By solving above equation from Saraphim,
x(t) = 10e$^{\frac{-1}{50}t}$
I can get the amount of salt after 10 minutes is 8.19kg.
By evaluating the answer, Can I assume that both ways (rate of change or exponential decay) resulting the same equation and answer?

b. As for my second question, I was also manage to reduce the equation, but it is supposed that we can get t as a numeric value.

7. Jul 13, 2011

### Saraphim

It is not a linear model (even though it is a linear differential equation) - if you plug in x(t)=10*exp(-t/50) into the equation for x'(t) this will be quite clear, and your results are, as far as I can see, correct.

Regarding b) I have no idea. Try posting a new question with that specifically.