Determing the equation of a streamline

[Bluestribute]

Homework Statement

A fluid has velocity components of u=[30/(2x+1)]m/s and v=(2ty)m/s,where x and y are in meters and t is in seconds. Determine the equation of the streamline that passes through point (2 m, 6 m) at time t = 2 s. Write the equation in the form y={y(x)}m, where x is in meters.

Homework Equations

dy/dx = v/u

The Attempt at a Solution

So when I separate my variables, I get udy=vdx. That means my equations are (30/(2x+1))dy = (2ty)dx. Is it really just a matter of integrating such "simple" equations? Or do I need to do some other method/algebra to get the (2x+1) on the dx side and the (2ty) on the dy side? Then all I have to do is integrate and find my constants, correct?

 

[rude man]

Homework Equations

dy/dx = v/u

The Attempt at a Solution

So when I separate my variables, I get udy=vdx. That means my equations are (30/(2x+1))dy = (2ty)dx. Is it really just a matter of integrating such "simple" equations? Or do I need to do some other method/algebra to get the (2x+1) on the dx side and the (2ty) on the dy side? Then all I have to do is integrate and find my constants, correct?

Separation of variables would be dy/v = dx/u. I don't see why you can't do as you wrote: separate variables, solve the ODE including the "initial conditions". If you want we can compare answers if you post yours first.

 

[Bluestribute]

So:

dy/v = dy/2ty
dx/u = (1/30)(2x+1)dx

I get x² + x + C1 and ln(y)/2t + C2

How do I solve for those constants to get y=f(x) form using the given boundaries?

 

[rude man]

So:

dy/v = dy/2ty
dx/u = (1/30)(2x+1)dx

I get x² + x + C1 and ln(y)/2t + C2

What happened to "30"? Then, put an equal sign somewhere & I'm a buyer.
Since it's a 1st order ODE you know there is only one boundary condition and one constant to be determined by the boundary condition. You can obviously combine your C1 and C2 into one constant. Have another shot at it, you're getting warmer.

 

[Bluestribute]

I divided the thirty into the dx equation.

So it's legal to set these equal to each other, put in the conditions, solve for the constant (which, now I understand why no one was putting in both C1 and C2), then rearrange to have y on one side and x on the other?

 

[rude man]

I divided the thirty into the dx equation.

But it's missing in "I get x2 + x + C1 and ln(y)/2t + C2 ". Put it back in!

So it's legal to set these equal to each other, put in the conditions, solve for the constant (which, now I understand why no one was putting in both C1 and C2), then rearrange to have y on one side and x on the other?

Your equation is dy/v = dx/u. This has all y on the left and all x on the right. Integrate both sides, then solve for y explicitly. Try to show your work or I can't tell if & where you slipped.

 

[Bluestribute]

Right right.

(1/30) x² + x + C = ln(y)/2t

I plug in my initial x,y,t points:

0.2 + C/30 = 0.44794

Solving for C yields C = 7.4382

So now (and I'm going to just drop my thirty in the denominator):

(x² + x + 7.4382)/30 = ln(y)/2t
(2tx² + 2tx + 14.8764t)/30 = ln(y)
e^{(2tx2/30 + 2tx/30 + 14.8764t/30) = y}

 

[rude man]

Right right.

(1/30) x² + x + C = ln(y)/2t

I plug in my initial x,y,t points:

0.2 + C/30 = 0.44794

Solving for C yields C = 7.4382

So now (and I'm going to just drop my thirty in the denominator):

(x² + x + 7.4382)/30 = ln(y)/2t
(2tx² + 2tx + 14.8764t)/30 = ln(y)
e^{(2tx2/30 + 2tx/30 + 14.8764t/30) = y}

When you solved for the constant of integration, did you let t=2?
Do that and you got it! My numbers agree if you let t=2 and compute the numerical value of e^{(14.8764t/30)}.

 

[Bluestribute]

I think I did, though I might have accidentally did 0 instead . . .

 

[rude man]

I think I did, though I might have accidentally did 0 instead . . .

Your last entry was e(2tx2/30 + 2tx/30 + 14.8764t/30) = y. Needed to get rid of "t".