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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
 2 years ago
 2 years ago
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
 2 years ago
 2 years ago

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apoorvkBest ResponseYou've already chosen the best response.2
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?
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
Well nothing is stupid . Thats how you prove limits. Google it !
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
yeah i know limits, but then you must mean in the second line that lim f(0,0) = 0 x,y>0 right??
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
What does fx(0,k) mean? x>0 and y > k?
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
But f(0,0) isn't defined, is it?
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
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 multivariable calculus yet.
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
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.
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
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
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
since they are equavalent, the limit at (0,0) would also be the same
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
yeah i logically was also getting that conclusion, but ofcourse more based on logical guessing and less of calculations.
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
maybe the question was show they are equal ?
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
may be yes. high chances of that.
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
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.
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
Actually its a "step by step" sort of way , but I was not able to prove the conclusion , maybe some logical error Iam facing.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
lets try this first off we need \(f_x(x,y)=\frac{2y^3}{(x^2+y^2)^{\frac{3}{2}}}\)
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
and so \(f_x(0,y)=\frac{2y^3}{(y^2)^{\frac{3}{2}}}=2\) if i am not mistaken
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
well hold on this is symmetric in x and y, so why these would not be the same is now confusing me. hmmm
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Read the following link. Complete the steps, then mimic for your function http://www.unf.edu/~omilatov/MAC2313/differentiability.pdf
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
@ satellite , the method Iam talking about , you are not following that , maybe thats what misleading you !
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
@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
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.0
I would check to make sure that you wrote down/typed the correct function
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
yup the function is correct , i copied directly from my book .
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
and its sort of a proof , and not just straightforward estimation of fxy or fyx , maybe thats why many are getting confused !
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Those two limits seem to be both infinite.
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
\[ f_x(x,y)=\frac{2 y^3}{\left(x^2+y^2\right)^{3/ 2}}\\ f_y(x,y)=\frac{2 x^3}{\left(x^2+y^2\right)^{ 3/2}} \] 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 \[\frac{ f_y(k,0)  0}{k}=\frac{2}{\sqrt{k^2}} \]
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
So the limit when x tends to zero is infinite for both.
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
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 !
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Do not worry about it.
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
From May its my papers , so have a reason to worry ! :(
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
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
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
:(((((( no idea ! :(((((
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Both limits diverge to \[ \infty \] In a way they are not different.
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
They are not different , but does not exist either !
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
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 ! :)
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
phew... please don't forget to confirm what this was all about!
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
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 ! :)
 2 years ago

apoorvkBest ResponseYou've already chosen the best response.2
definitely mugging up method i believe. atleast very theoretical or subjective. bet it's for a theory paper in your uni, right?
 2 years ago

anjali_pantBest ResponseYou've already chosen the best response.0
actually maths is a subsi in my course ! But yea it must be there in maths(h) ! IMP PROOOF ! :P
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Please read http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives The above problem is a bit related to the link above
 2 years ago
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