anonymous
  • anonymous
Let f(x,y)=2xy/(x^2+y^2)^1/2 (x,y) not equal to (0,0) , f(0,0)=0 Show fyx(0,0) not equal to fxy(0,0) using method fxy= lim k->0 [ fx(0,k)-fx(0,0)]/k fyx=lim h->0 [ fy(h,0)-fy(0,0)]/h
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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apoorvk
  • apoorvk
umm, this may be stupid, but how can the function be defined for (0,0) when it already says (x,y) not equal to (0,0). [read second line of the problem] am i missing something?
anonymous
  • anonymous
Well nothing is stupid . Thats how you prove limits. Google it !
apoorvk
  • apoorvk
yeah i know limits, but then you must mean in the second line that lim f(0,0) = 0 x,y->0 right??

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anonymous
  • anonymous
yup !
anonymous
  • anonymous
What does fx(0,k) mean? x->0 and y -> k?
anonymous
  • anonymous
But f(0,0) isn't defined, is it?
anonymous
  • anonymous
Maybe @dumbcow knows how to do it, this stuff is a little advanced for me. I'm fairly certain this isn't a complicated problem, but I haven't taken multi-variable calculus yet.
apoorvk
  • apoorvk
okay so i seem to think fxy here is partial differential of f(x,y) wrt to x, and fyx is partial differential of f(x,y) wrt to y. see the structure. and same here, haven't taken advanced calculus, so don't think i ll be able help much other than guessing.
dumbcow
  • dumbcow
interesting, i find that fxy = fyx , im assumimg these are second partial derivatives \[\large f_{xy} = f_{yx} = \frac{6x^{2}y^{2}}{(x^{2}+y^{2})^{5/2}}\] http://www.wolframalpha.com/input/?i=d%2Fdx+d%2Fdy+2xy%2Fsqrt%28x^2%2By^2%29 http://www.wolframalpha.com/input/?i=d%2Fdy+d%2Fdx+2xy%2Fsqrt%28x^2%2By^2%29
dumbcow
  • dumbcow
since they are equavalent, the limit at (0,0) would also be the same
apoorvk
  • apoorvk
yeah i logically was also getting that conclusion, but ofcourse more based on logical guessing and less of calculations.
dumbcow
  • dumbcow
maybe the question was show they are equal ?
apoorvk
  • apoorvk
may be yes. high chances of that.
anonymous
  • anonymous
The question is that show they are not equal. I stated the method to be used , and that's the way I have to do it in my exam , and not just simple estimation.
anonymous
  • anonymous
Actually its a "step by step" sort of way , but I was not able to prove the conclusion , maybe some logical error Iam facing.
anonymous
  • anonymous
lets try this first off we need \(f_x(x,y)=\frac{2y^3}{(x^2+y^2)^{\frac{3}{2}}}\)
anonymous
  • anonymous
and so \(f_x(0,y)=\frac{2y^3}{(y^2)^{\frac{3}{2}}}=2\) if i am not mistaken
anonymous
  • anonymous
well hold on this is symmetric in x and y, so why these would not be the same is now confusing me. hmmm
anonymous
  • anonymous
Read the following link. Complete the steps, then mimic for your function http://www.unf.edu/~omilatov/MAC2313/differentiability.pdf
anonymous
  • anonymous
@ satellite , the method Iam talking about , you are not following that , maybe thats what misleading you !
anonymous
  • anonymous
@elias, I know the steps , its that Iam not able to prove the inequality , some logical error that too in the last steps are over shadowing the ans ! So see if you could solve it step by step
Zarkon
  • Zarkon
I would check to make sure that you wrote down/typed the correct function
anonymous
  • anonymous
yup the function is correct , i copied directly from my book .
anonymous
  • anonymous
and its sort of a proof , and not just straightforward estimation of fxy or fyx , maybe thats why many are getting confused !
anonymous
  • anonymous
Those two limits seem to be both infinite.
anonymous
  • anonymous
\[ f_x(x,y)=\frac{2 y^3}{\left(x^2+y^2\right)^{3/ 2}}\\ f_y(x,y)=\frac{2 x^3}{\left(x^2+y^2\right)^{ 3/2}} \] You can show from the definition that \[ f_x(0,0)= f_y(0,0)=0\] so \[\frac{ f_x(0,k) - 0}{k}=\frac{2}{\sqrt{k^2}} \] and \[\frac{ f_y(k,0) - 0}{k}=\frac{2}{\sqrt{k^2}} \]
anonymous
  • anonymous
So the limit when x tends to zero is infinite for both.
anonymous
  • anonymous
yea even I was getting the same ans ! but since the ques demands something different , I got stuck in between ! Then I thought since its coming to be infinity , so it must be implying that limits does not exist !
anonymous
  • anonymous
yes.
anonymous
  • anonymous
pardon ?
anonymous
  • anonymous
Do not worry about it.
anonymous
  • anonymous
From May its my papers , so have a reason to worry ! :-(
anonymous
  • anonymous
i think there is something wrong here because this function is symmetric in x and y, i.e. if you replace x by y and y by x you get the same thing. so why one should be different from the other is not at all clear to me
anonymous
  • anonymous
:-(((((( no idea ! :-(((((
anonymous
  • anonymous
Both limits diverge to \[ \infty \] In a way they are not different.
anonymous
  • anonymous
They are not different , but does not exist either !
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Iam gettting your point , infact I myself concluded the same ans , but ques says something else. I ll confirm it tomorrow and will let you know. Thanks a lot for your help ! :-)
apoorvk
  • apoorvk
phew... please don't forget to confirm what this was all about!
anonymous
  • anonymous
lol ! Yea this method is a bit "mugging up" thing , maybe you have never read that, so you were getting confused so as to what to do ! Anyways google it ,its a proper method ! And yea thanks ! :-)
apoorvk
  • apoorvk
definitely mugging up method i believe. atleast very theoretical or subjective. bet it's for a theory paper in your uni, right?
anonymous
  • anonymous
actually maths is a subsi in my course ! But yea it must be there in maths(h) ! IMP PROOOF ! :P
anonymous
  • anonymous
Please read http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives The above problem is a bit related to the link above

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