Modeling Using Differential Equations

  • Thread starter olicoh
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  • #1
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Homework Statement


Suppose the rate at which the volume in a tank decreases is proportional to the square root of the volume present. The tank initially contains 25 gallons, but has 20.25 gallons after 3 minutes. Solve for the general solution (do not solve for V).

The Attempt at a Solution


dV/dt = k√(V)

That's as far as I got. I know I have to "separate" the variables and whatnot, but there is no 't' to separate and differentiate from. I guess the equation would be: ∫1/√(V) dV = k∫ ___ dt
So my question is, since there is no 't' in the equation, what do I differentiate instead?
 

Answers and Replies

  • #2
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your setup is correct, k is just some constant and whenever you integrate a constant with respect to x, t, z, etc. your left with k multiplied by the variable you integrated with respect to
so in your case, for the RHS youll get kt+C
 
  • #3
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your setup is correct, k is just some constant and whenever you integrate a constant with respect to x, t, z, etc. your left with k multiplied by the variable you integrated with respect to
so in your case, for the RHS youll get kt+C
That's what I thought. I just wanted to double check. Thanks!
 
  • #4
24
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I have another question (actually 3 questions):

3) Use the initial condition to find the constant of integration, then write the particular solution (do not solve for V).

Attempt at solution: 2√(25)=k(0) + C
My answer: 2√(V)=kt+10


4) Use the second condition to find the constant of proportion.

Attempt at solution: 2√(20.25)=k(3) + 10 --> 4=3k+10 --> -6=3k
My answer: k = -2


5) Find the volume at t = 5 minutes. Round your answer to two decimal places.


My attempt at the solution: 2√(V)=(-2)(5) + 10 --> 2√(V)=0
My answer: V(5)=0

^^^
For number 5... I'm not sure my answer is right. Using common sense, I don't think it's possible for the tank to be at zero gallons at 5 minutes. I think I might have done something wrong at number 4.
 

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