Register to reply

Check work on 2 variable function.

by tnutty
Tags: check, function, variable, work
Share this thread:
tnutty
#1
Sep26-09, 09:04 PM
P: 328
1. The problem statement, all variables and given/known data

Determine the set of points at which the function is continuous.

F(x,y) = R.

where R is a piecewise function of :

{
x^2*y^3 / (2x^2 + y^2) ; if(x,y) != (0,0)
1 ; if(x,y) = (0,0)
}

Obviously, the first function is not defined at point (0,0), but to find
the domain of the piecewise function, I first need to see if the
first function is at least continuous.

So here is my attempt at that :

Let A = x^2 * y^3 / (2x^2 + y^2 )

the |A| is =

x^2 * |y^3|
---------------
2x^2 + y^2

well x^2 <= 2x^2 + y^2, lets call that J

so A < J * |y^3| / (J) = |y^3| = sqrt(y^6), and we see that this function
is defined at point 0 , thus lim of A as (x,y) -->(0,0) = 0. ???

So if the above is true then the peicewise function should be
defined in region R^2???

I am not sure if this is correct. The book says that the answer is :

{ (x,y) | (x,y) != (0,0) }.

I think that means the function A is not defined at 0 thus the peicewise
function is not defined at point (0,0). What did I do wrong ?
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Billy Bob
#2
Sep27-09, 09:02 AM
P: 393
You showed [tex]\lim_{(x,y)\to (0,0)} F(x,y)=0[/tex] . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).
tnutty
#3
Sep27-09, 01:24 PM
P: 328
Quote Quote by Billy Bob View Post
You showed [tex]\lim_{(x,y)\to (0,0)} F(x,y)=0[/tex] . However, since F(0,0)=1 by the definition of F, we have that F is not continuous at (0,0).
I thought that given value defined in the piecewise, 1 in this case, was an arbitrary value.
So it did not matter, unless the limit of part A in the piecewise function did not exist?

Billy Bob
#4
Sep27-09, 03:02 PM
P: 393
Check work on 2 variable function.

If F(0,0) had been defined to be 0 instead of 1, then F would have been continuous everywhere.

If F(0,0) had been defined to be c, with c nonzero (c=1 is a special case), then as in your problem, F would not have been continuous at (0,0) but would be continuous everywhere else.


Register to reply

Related Discussions
Function of Function of Random variable Engineering, Comp Sci, & Technology Homework 1
Simple 2 variable integral just want to check my answer Calculus & Beyond Homework 2
2 variable function Advanced Physics Homework 0
Work input/output, efficient probelm, just need someone to check my work Introductory Physics Homework 1
Work question: please check my work Introductory Physics Homework 1