Partial Derivatives of x^2-y^2+2mn+15=0

iwan89
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x^2 - y^2 +2mn +15 =0

x + 2xy - m^2 + n^2 -10 =0

The Question is:
Show that
del m/ del x = [m(1+2y) -2 x n ] / 2 (m^2 +n^2)

del m / del y = [x m+ n y] / (m^2 +n^2)

note that del= partial derivativesMy effort on solving this question is
Fx1=2x Fm1=2n
Fx2 =2y Fm2 =-2m

del m /del x = -Fx1/Fm1 + -Fx2/Fm2

the solution brings me nowhere near to answer..This is not a homework
 
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Your post is somewhat confusing.

Could you use the advanced features and write your equations using latex?

Assuming I understand the first part of your post what does the fx1... stuff represent.

And what is the m and n are they variables dependent on x and y?

You mentioned this was for your research so what is the research and what level of math are you comfortable with?
 
@jedishrfu partial derivatives. Sorry sir I am not really good in latex. I am very sorry and i was hoping that you can understand my question thank you.
 
i edited my question...
 
Okay but I still don't understand what Fx1/Fm1 is ?

Do you mean dx1/dm1 the derivative of x1 with respect to m1?

You really need to use latex so we can see the real equations.
 
@jedishrfu please ignore my solution.try to draft your own solution.. the real question is

x^2 - y^2 +2mn +15 =0

x + 2xy - m^2 + n^2 -10 =0

The Question is:
Show that
del m/ del x = [m(1+2y) -2 x n ] / 2 (m^2 +n^2)

del m / del y = [x m+ n y] / (m^2 +n^2)

note that del= partial derivatives

im doing fluid mechanics
 
Okay can you at least write your equations on paper and upload a photo of the paper?

The del notation confuses me as there is a del operator which when evaluated uses partial derivatives as in the gradient of some scalar function of x,y, z.
 
Looks like the current PF doesn't have an editor mode to latex and all The specialized math notation so I guess the photo route is the way to go.
 
10726809_928957633784479_546825814_n.jpg

this is the real question :) thanks!
 
  • #10
x2 - y2 + 2uv + 15 = 0
x + 2xy - u2 + v2 - 10 = 0
Using LaTeX, show that:
$$ \frac{\partial u}{\partial x} = \frac{u(1 + 2y) - 2xv}{2(u^2 + v^2)}$$
and
$$\frac{\partial u}{\partial y} = \frac{xu + vy}{u^2 + v^2}$$

You can right-click on the LaTeX stuff to show the Tex commands I used.
 
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  • #11
I haven't worked this all the way through, yet, but I would isolate the uv term in the first equation and take the partials of u with respect to x and then with respect to y. I would also isolate the u2 term in the second equation and take the partials of u w.r.t x and w.r.t. y.

Edit: I have worked this through for ##\frac{\partial u}{\partial y}## successfully, so the strategy I laid out above works. All it takes is a little bit of deft algebra to eliminate the partials of v with respect to the two independent variables.
 
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  • #12
@Mark44 can you help me prove it? thank you so much :)
 
  • #13
iwan89 said:
@Mark44 can you help me prove it? thank you so much :)
See post 11.
 
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  • #15
jedishrfu said:
Okay can you at least write your equations on paper and upload a photo of the paper?

The del notation confuses me as there is a del operator which when evaluated uses partial derivatives as in the gradient of some scalar function of x,y, z.

I think his question is now clear; he said exactly what his symbols represent. He said ##\text{del} = \partial##.

However, we cannot help him until he demonstrates that he has done some work on the problem; he needs to show us his work. Those are the PF rules.
 
  • #16
Ray Vickson said:
I think his question is now clear; he said exactly what his symbols represent. He said ##\text{del} = \partial##.

However, we cannot help him until he demonstrates that he has done some work on the problem; he needs to show us his work. Those are the PF rules.

Yes, I agree. I was asking for a more precise definition from the OP before helping.

If you notice though the confusion I had was the use of m,n instead of u,v and the meaning behind the Fx1... In his posts. He mentioned it was for fluid mechanics which routinely use the del operator and I wanted to make sure things were right before guiding him to a solution.
 
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