Solving Exponential Growth Homework: Find Water to Eliminate 50% Salt

In summary, the amount of salt, S, in a tank with a constant volume, V, and continuous flow, depends on the amount of water, X, that has been pumped in. The rate of change of salt with respect to water is given by ds/dx = -S/V. To eliminate 50% of the salt, we can use the equation S(x)= (1/2)S(0), where S(0) is the initial amount of salt. By substituting V = 10,000 gallons, we can solve for the amount of water, X, needed to eliminate 50% of the salt. The general solution for S(x) is given by S
  • #1
lovemake1
149
1

Homework Statement


Water is pumped into a tank. Volume V, is kept constant by continuos flow. The amount of salt S, depends on the amount of water that ahs been pumped in, call it X.

ds/dx = -S/V

Find the amount of water needed to eliminate 50% of the salt. Take v AS 10,000 gallons


Homework Equations





The Attempt at a Solution



We know that volume V is kept constant by continuos flow. d/dv ?
im througly confused...
I understand that exponential growth has the traditional formula: Ce^kt
but how can we ues it with this question?

helps appreciated.
 
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  • #2
lovemake1 said:

Homework Statement


Water is pumped into a tank. Volume V, is kept constant by continuos flow. The amount of salt S, depends on the amount of water that ahs been pumped in, call it X.

ds/dx = -S/V

Find the amount of water needed to eliminate 50% of the salt. Take v AS 10,000 gallons


Homework Equations





The Attempt at a Solution



We know that volume V is kept constant by continuos flow. d/dv ?
im througly confused...
So am I! There is no differentiation with respect to V (please do not use both small and capital letters to mean the same thing). V is a constant- replace it with the 10000 you are given.

I understand that exponential growth has the traditional formula: Ce^kt
but how can we ues it with this question?

helps appreciated.
I take it this is a "differential equations" course. You would be expected to recognize that you can write the original equation as
[tex]\frac{dS}{S}= -\frac{dx}{10000}[/tex]
and integrate:
[tex]ln(S)= -\frac{x}{10000}+ C[/tex]
so that
[tex]S(x)= e^C e^{-\frac{x}{10000}}= C' e^{-\frac{x}{V}[/itex]
where [itex]C'= e^C[/itex]. That is the "[itex]Ce^{kt}[/itex]" you have, except that your variable is x, the amount of water that has flowed through, not time.

Now solve the equation S(x)= (1/2)S(0).
 
  • #3
Ah I see the light !
But the line after "so that",
Shouldn't it be e^lns = e^-x/v + c ?
just confused to what C is
 

Related to Solving Exponential Growth Homework: Find Water to Eliminate 50% Salt

1. How do you solve for exponential growth in this scenario?

To solve for exponential growth in this scenario, you will need to use the formula A = Pert, where A is the final amount, P is the initial amount, r is the growth rate, and t is the time. You will also need to know the half-life of the salt in water, which is the time it takes for 50% of the salt to be eliminated.

2. What is the first step in finding the amount of water needed to eliminate 50% of the salt?

The first step is to determine the initial amount of salt in the water. This can be done by measuring the concentration of salt in the water or by knowing the amount of salt added to the water initially.

3. How do you calculate the growth rate for this problem?

The growth rate can be calculated by dividing the natural logarithm of 2 (ln 2) by the half-life. This will give you the rate at which the salt concentration decreases over time.

4. Can this problem be solved without knowing the initial amount of salt or the half-life?

No, in order to solve for the amount of water needed to eliminate 50% of the salt, you will need to know both the initial amount of salt and the half-life. Otherwise, there will not be enough information to solve the equation.

5. Is there a way to double-check the solution to ensure it is correct?

Yes, you can double-check the solution by plugging in the calculated values into the original formula and making sure the resulting amount of salt is indeed half of the initial amount. You can also check your calculations and make sure they are accurate.

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