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Oxygen
Canada3581 Posts
given show that we differentiate G with respect to x 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 thoughts?   | ||
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kingjames01
Canada1603 Posts
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. | ||
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Oxygen
Canada3581 Posts
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kingjames01
Canada1603 Posts
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. | ||
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Rkie
United States1278 Posts
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kingjames01
Canada1603 Posts
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. | ||
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DeathByMonkeys
United States742 Posts
On May 22 2010 07:48 Rkie wrote: what does the backwards 6 mean? Haha, it's the partial derivative. | ||
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Rotation
United States118 Posts
On May 22 2010 07:48 Rkie wrote: what does the backwards 6 mean? It's a partial derivative EDIT: beaten to it. | ||
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kingjames01
Canada1603 Posts
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". | ||
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DeathByMonkeys
United States742 Posts
On May 22 2010 07:18 Oxygen wrote: trying to remember some math and I don't know why this isn't working given show that we differentiate G with respect to x 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 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. | ||
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Oxygen
Canada3581 Posts
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 right? so | ||
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kingjames01
Canada1603 Posts
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... | ||
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kingjames01
Canada1603 Posts
On May 22 2010 08:06 Oxygen wrote: That's the part I'm not understanding. Given right? so 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! | ||
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Oxygen
Canada3581 Posts
 http://www.homeschoolmath.net/worksheets/equation_editor.phpseems 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. | ||
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kingjames01
Canada1603 Posts
I'm on my Windows side waiting for the beta to come back online, so I was just typing it freehand. haha. | ||
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kingjames01
Canada1603 Posts
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. | ||
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kingjames01
Canada1603 Posts
\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. | ||
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Nytefish
United Kingdom3810 Posts
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Oxygen
Canada3581 Posts
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kingjames01
Canada1603 Posts
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} | ||
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