| Blogs > Oxygen |
|
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?   | ||
|
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. | ||
|
Oxygen
Canada3581 Posts
sec | ||
|
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. | ||
|
Rkie
United States1278 Posts
| ||
|
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. | ||
|
DeathByMonkeys
United States742 Posts
On May 22 2010 07:48 Rkie wrote: what does the backwards 6 mean? Haha, it's the partial derivative. | ||
|
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. | ||
|
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". | ||
|
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. | ||
|
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 | ||
|
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... | ||
|
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! | ||
|
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. | ||
|
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. | ||
|
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. | ||
|
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. | ||
|
Nytefish
United Kingdom4282 Posts
| ||
|
Oxygen
Canada3581 Posts
| ||
|
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} | ||
| ||
|
|
Replay Cast
The PondCast
KCM Race Survival
WardiTV Winter Champion…
Classic vs Nicoract
herO vs YoungYakov
ByuN vs Gerald
Clem vs Krystianer
Replay Cast
Ultimate Battle
Light vs ZerO
WardiTV Winter Champion…
MaxPax vs Spirit
Rogue vs Bunny
Cure vs SHIN
Solar vs Zoun
Replay Cast
CranKy Ducklings
WardiTV Winter Champion…
[ Show More ] |
|
|
EPT
