Hello,
I am having problems starting a homework question which reads:
F(x,y) is defined by the integral from (2,pi) to (x,y) as
(-2u(v^2)sin(u^2v)du + (cos((u^2)v) - u^2(v)(sin((u^2)v)))dv
Express F(x,y) as a function of x and y, eliminating the integral sign.
I usually give you guys something that I have done which is not far from your suggestions, but like I said at first, I don't know how to start.
I have done line integrals and closed interval integrals along with Green's Theorem, but I do not see how I can use these tools to solve the problem.
Thank you for any hints.
More likely a Physicist thing. Actually it would be:
F(x,y) = (Integral from (2,pi) to (x,y)) (-2u(v^2)sin(u^2v)idu + (cos((u^2)v) - u^2(v)(sin((u^2)v)jdv)
(or something) according to the problem statement. But, as ThePerfectHacker says, we need a path to integrate on. Unless this just happens to be path independent? Then again, we don't know which of "u" or "v" is the "x" or "y" component either, so even if the integral IS path independent, we still can't do the integral... (jcarlos, jump in here at any time! )
-Dan
As topsquark said the line integral is probably path independant (though I was lazy to check) thus, you use the fundamental theorem of line integrals. You do not need a path to integrate on. However, you need the points of the paramaterized curve (starting and ending) which you did not include.
Thank you everyone for your replies.
I posted just before I went to bed last night and I just got back to school; therefore, I could not get back to you as you were making comments to my question.
The question I have posted is exactly as worded in the text. I understand what topsquark is saying that the problem could be more physicist based if we include the direction vectors i and j. Regarding the path that the integral must be evaluated in, I am just as lost as you are and would consider the path to be independent. Like I said before, I have never seen this type of question before. I have done change of variables for an F(x,y) before by substitution with u and v, but this problem is the opposite with more twists in it. Perfect Hacker, could you please elaborate on your last reply at 9:21AM; no I did not include thet starting and end points of the parametrized curve. How can I parametrize the curve to get these points? Like I said before, the problem is written word by word from the text.
Thanks again
I assume these concepts are familar to you.
When you are given a path to integrate on it is in parametric form, meaning
R(t)=f(t)i+g(t)j---> Vector function notation
Alternativly,
x=f(t) and y=g(t) ---> Parametric form
This is called parametrizing your curve (expressing your curve as a parametric functions).
The starting point and endpoint is where you start to integrate (that value of t ) and where you end to integrate (that value of t).
For example, the path given by the point (0,0) and (2,4)
Along a parabolic arc y=x^2 is given by,
R(t)=ti+t^2j for 0<=t<=2
Alternaviely you can write,
x=t and y=t^2 for 0<=t<=2.
Now, if your vector field (F(x,y)) is independant of path then by the fundamental theorem of line integral you need to find the conservative of this vector field.
Meaning, find a function (called scalar potential) such that its gradient is the vector field. Once you have done that (assuming you vector field is independent) then all you do if evaluate it (scalar potential) at the endpoint and subtract it at the starting point, just like with a definite integral for area.
I think I can see what the original notation might mean now, though
I would still need the curve along which to integrate, also I would still
have to do some research before I could be sure of the detail.
==========================================
consider the curve defined by a single parameter lambda:
(u(lambda),v(lambda)),
then:
du=[d/dlambda] u(lambda) dlambda
and
dv=[d/dlambda] v(lambda) dlambda
==========================================
(curse this ongoing no LaTeX situation
RonL