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math help?

Blogs > Oxygen
Post a Reply
1 2 Next All
Oxygen
Profile Blog Joined November 2003
Canada3581 Posts
Last Edited: 2010-05-21 22:26:46
May 21 2010 22:18 GMT
#1
trying to remember some math and I don't know why this isn't working

given

[image loading]


show that

[image loading]



we differentiate G with respect to x

[image loading]


now I would think the previous step is correct, but then when I differentiate Gx (first derivative of G with respect to x) a second time I get a pretty big expression and I can't reduce it all. I think I'm already making a mistake somewhere... if someone could just tell me if the following expression is correct, I can do the rest

[image loading]



thoughts?


*
Dont drink and derive. TSL: Made with Balls.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 22:28 GMT
#2
In your original line, G(x, y) = F(\frac{x}{x^{2} + y^{2}}, \frac{y}{x^{2} + y^{2}}),
is it implied that
u = \frac{x}{x^{2} + y^{2}}
and
v = \frac{y}{x^{2} + y^{2}}

or is it that x and y are functions of u and v? I'll work on this but I don't want to put any effort in until this is clarified.
Who would sup with the mighty, must walk the path of daggers.
Oxygen
Profile Blog Joined November 2003
Canada3581 Posts
May 21 2010 22:33 GMT
#3
that's the thing, I'm uncertain, that's how the question is worded. before you get to work on it let me do it with u = \frac{x} ...
sec
Dont drink and derive. TSL: Made with Balls.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 22:41 GMT
#4
Well, I went ahead and worked it out.
If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate,
\frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}}
the mixed partials will cancel and then you'll have
\frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0.
Who would sup with the mighty, must walk the path of daggers.
Rkie
Profile Blog Joined October 2009
United States1278 Posts
May 21 2010 22:48 GMT
#5
what does the backwards 6 mean?
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 22:52 GMT
#6
It means "a partial derivative with respect to"
so in the above case,
it says that G is a function of both x and y and to take two derivatives of G with respect to only x while keeping y constant and summing that with the second partial derivative of G with respect to only y while keeping x constant. This sum is supposed to be equal to 0 which it turns out is true.
Who would sup with the mighty, must walk the path of daggers.
DeathByMonkeys
Profile Blog Joined March 2008
United States742 Posts
May 21 2010 22:52 GMT
#7
On May 22 2010 07:48 Rkie wrote:
what does the backwards 6 mean?


Haha, it's the partial derivative.
Rotation
Profile Blog Joined July 2009
United States118 Posts
Last Edited: 2010-05-21 22:53:10
May 21 2010 22:52 GMT
#8
On May 22 2010 07:48 Rkie wrote:
what does the backwards 6 mean?


It's a partial derivative

EDIT: beaten to it.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 22:56 GMT
#9
Also, it's not a backwards 6... hehe.
It's actually the scripted form of the lower case Greek letter delta. It's similar to the use of the upper case Greek delta to mean "change in".
Who would sup with the mighty, must walk the path of daggers.
DeathByMonkeys
Profile Blog Joined March 2008
United States742 Posts
May 21 2010 22:59 GMT
#10
On May 22 2010 07:18 Oxygen wrote:
trying to remember some math and I don't know why this isn't working

given

[image loading]


show that

[image loading]



we differentiate G with respect to x

[image loading]


now I would think the previous step is correct, but then when I differentiate Gx (first derivative of G with respect to x) a second time I get a pretty big expression and I can't reduce it all. I think I'm already making a mistake somewhere... if someone could just tell me if the following expression is correct, I can do the rest

[image loading]



thoughts?


Seems to me like it might be some kind of Green's/Stokes' Thm kind of problem. Where when you take y/(x^2 + y^2) with respect to y and you take x/(x^2 + y^2) with respect to x you get the same thing and they cancel. I can't remember what thats called... shit now that's going to bug me.
Oxygen
Profile Blog Joined November 2003
Canada3581 Posts
Last Edited: 2010-05-21 23:09:31
May 21 2010 23:06 GMT
#11
On May 22 2010 07:41 kingjames01 wrote:
Well, I went ahead and worked it out.
If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate,
\frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}}
the mixed partials will cancel and then you'll have
\frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0.


That's the part I'm not understanding. Given

[image loading]


[image loading]


right?

so
[image loading]
. specifically, how do you differentiate
[image loading]
? Fu is a function of u and v, which are in turn functions of x and y, so do you have to apply the product rule? in which case, what does
[image loading]
yield?


Dont drink and derive. TSL: Made with Balls.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 23:08 GMT
#12
What you're probably remembering is that for any function is at least twice differentiable and continuous, then the mixed partials will always equal.

So, if f is a C^{2} function, ie. f, fx, fy, fxx, fxy, fyx, fyy, then fxy = fyx.
In general this can be extended to higher orders and is not restricted to C^{2}. You would have seen this in courses in Complex Analysis, Differential Equations, Undergrad Mechanics, Undergrad Calc...

Who would sup with the mighty, must walk the path of daggers.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 23:14 GMT
#13
On May 22 2010 08:06 Oxygen wrote:
Show nested quote +
On May 22 2010 07:41 kingjames01 wrote:
Well, I went ahead and worked it out.
If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate,
\frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}}
the mixed partials will cancel and then you'll have
\frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0.


That's the part I'm not understanding. Given

[image loading]


[image loading]


right?

so
[image loading]





Yes, that's right so far.

Continuing,
\frac{\partial^{2} G}{\partial x^{2}} = \frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right)
= \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x}
= \left(\frac{\partial^{2} F}{\partial u^{2}} \frac{\partial u}{\partial x} + \frac{\partial^{2} F}{\partial u \partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x}
= \frac{\partial^{2} F}{\partial u^{2}} \left(\frac{\partial u}{\partial x}\right)^{2} + \frac{\partial^{2} F}{\partial u \partial v} \frac{\partial v}{\partial x} \frac{\partial u}{\partial x}

Do the same thing for the second partial of G wrt y and sum them. Evaluate the partial of u wrt x, u wrt y, v wrt x, v wrt y and when you substitute you'll see some very big simplifications.

Finally, if you are going to continue this you should pm me. Homework threads aren't allowed here. I can latex the solution if you are stuck but this should get you going. Good luck!
Who would sup with the mighty, must walk the path of daggers.
Oxygen
Profile Blog Joined November 2003
Canada3581 Posts
Last Edited: 2010-05-21 23:19:59
May 21 2010 23:15 GMT
#14
took the liberty ... http://www.homeschoolmath.net/worksheets/equation_editor.php

[image loading]


[image loading]


[image loading]


[image loading]


seems like I should be fine from here. thanks a lot for your help.

btw homework threads are generally not allowed when posted in general forum; they're fine in blogs, as long it's not "do my work for me". the huge movement against "homework" threads started a few years back when lots of people used to do that.
Dont drink and derive. TSL: Made with Balls.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 23:18 GMT
#15
Nice! I didn't know that you could convert it online! =)
I'm on my Windows side waiting for the beta to come back online, so I was just typing it freehand. haha.
Who would sup with the mighty, must walk the path of daggers.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 23:29 GMT
#16
Not a problem! Glad I could help.
Ah, okay, I didn't know that blogging it would be fine. =)
Yeah, I remember when the crackdowns started happening. Those were still back in my lurker days.
Who would sup with the mighty, must walk the path of daggers.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
May 21 2010 23:34 GMT
#17
By the way, I forgot to put in the second term in this line:

\frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right)
= \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x}

My mistake. If you want to see why, just pretend that expression inside of the bracket is a function called H. Well, it depends on x and y or u and v. So if you want to split the derivative of H wrt x into a partial wrt u then you'll have to do it wrt v as well since x depends on both u and v.
Who would sup with the mighty, must walk the path of daggers.
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
May 21 2010 23:38 GMT
#18
lol I was so confused and thought I had forgotten how to do the chain rule until that post^
No I'm never serious.
Oxygen
Profile Blog Joined November 2003
Canada3581 Posts
May 21 2010 23:39 GMT
#19
ah that makes way more sense. also I imagine the last partial is dv/dx not du/dx?
Dont drink and derive. TSL: Made with Balls.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
Last Edited: 2010-05-21 23:44:08
May 21 2010 23:42 GMT
#20
Yes, you're right, my mistake again, I copied and pasted it. =)
It's hard to read latex code if you're not being careful.

EDIT:
For anyone who is following this discussion, the correct latex code is:

\frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right)
= \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial v}{\partial x}
Who would sup with the mighty, must walk the path of daggers.
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