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anonymous
 4 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
anonymous
 4 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

This Question is Closed

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe @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.

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lets try this first off we need \(f_x(x,y)=\frac{2y^3}{(x^2+y^2)^{\frac{3}{2}}}\)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well hold on this is symmetric in x and y, so why these would not be the same is now confusing me. hmmm

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

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

anonymous
 4 years ago
Best 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

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

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

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

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

anonymous
 4 years ago
Best 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 \[\frac{ f_y(k,0)  0}{k}=\frac{2}{\sqrt{k^2}} \]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea 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
 4 years ago
Best ResponseYou've already chosen the best response.0Do not worry about it.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0:(((((( no idea ! :(((((

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Iam 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
 4 years ago
Best ResponseYou've already chosen the best response.2phew... please don't forget to confirm what this was all about!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol ! 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
 4 years ago
Best ResponseYou've already chosen the best response.2definitely mugging up method i believe. atleast very theoretical or subjective. bet it's for a theory paper in your uni, right?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Please 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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