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A saline solution contains $0.05\rm\,kg/liter $ salt. The saline solution flows into another tank with the speed of $3\rm\,liter/min$. What speed (in units $\rm kg/min$) is salt supplied to the tank?

My answer:

$$+3\rm\frac{liter}{min} \times 0.05 \frac{kg}{liter}=0.15 \frac{kg}{min},~~v=?$$

The salt is added to the tank with the speed of $0.15\rm\,kg/min$ - which is the correct answer.

But is there any way to describe this with a differential equation?

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You don't need differential equations to combine ratio and units information. Only multiplication was necessary. It certainly sounds like it could be the initial setup to a forthcoming related rates problem, requiring differential equations to solve, though. | Also, it looks like you want $$3\cdot 0.005=0.\color{Red}015,$$ unless the given value was $0.05$, not $0.005$ as in your first sentence. –  anon May 11 '12 at 8:14
    
Thank you for your comment. But I have to answer with a differential equation. Maybe something like y=Ce^(-ax). Is there a way to do this? –  user31113 May 11 '12 at 8:18
    
Yes you are right. I have mistyped it. is 0,05kg salt per liter. Sorry. I fixed it now. –  user31113 May 11 '12 at 8:19
1  
Well, you've computed the rate-of-change of the liters of salt in the tank (unless liquid also flows out of the tank), so setting the derivative of liters-of-salt-in-the-tank equal to this rate would be a valid differential equation. –  anon May 11 '12 at 8:22
    
I dont understand. Sorry. Nothing flows out of the tank. The tank is empty at first, then the saline solution (salt+water) is added to tank (3 liter per minute). –  user31113 May 11 '12 at 9:10

2 Answers 2

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An ordinary differential equation involves derivatives of a dependent variable (like $y$) with respect to an independent variable (like $x$). Here the two relevant variables are liters of salt in the tank and time. You know the rate of change of the first variable with respect to the second (indeed, it is a constant) so just set the derivative (of the first variable wrt the second, of course) equal to this known rate and you've written down a differential equation describing the situation.

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here the involved variables are linearly dependent...so while in calculation you are simply multiplying the rate,it is actually a multiplication by dy/dt...basically it is a very trivial kind of differential equation.you are forgeting that the rate you are talking about is basically a derivative...

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