Unit tanget outward normal vector and greens thrm

kekal6
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H(x,y)=<x^2/4,y^2/9,xy> the region E is 9x^2+4y^2<=36
also wat is given is the work is counterclockwise on R=<2cost,3sint> from -pi<=t<=pi
wat the questions are what is the unit tangent the outward normal vector with respect to the region E in terms of t. for the unit tangent i think its <(2cost)/sqrt(13),(3sint)/sqrt(13)>. i don't know if i did that right. i don't know where to start for the normal vector question though. also, i need to use greens thrm to find the amount of work needed to stop it if a particle was to move in the opposite direction.
 
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kekal6 said:
H(x,y)=<x^2/4,y^2/9,xy> the region E is 9x^2+4y^2<=36
also wat is given is the work is counterclockwise on R=<2cost,3sint> from -pi<=t<=pi
wat the questions are what is the unit tangent the outward normal vector with respect to the region E in terms of t. for the unit tangent i think its <(2cost)/sqrt(13),(3sint)/sqrt(13)>. i don't know if i did that right. i don't know where to start for the normal vector question though. also, i need to use greens thrm to find the amount of work needed to stop it if a particle was to move in the opposite direction.

Well, if r(t) = <2 cos(t), 3 sin(t)>, then the unit tangent vector is r'(t) / |r'(t)| and the unit normal vector is r''(t) / |r''(t)|. Use those to help you.
 
okay i got that part now the greens thrm one. my book only has it in terms of x and y but i have a feeling itd be easier using the R(t) one. the best it describes it is to do the integral of -pi to pi of F(r(t))dotr'(t) dt but i do not understand wat it fully means by the F(r(t)).
 
Remember that r(t) = <x(t), y(t)>. so F(r(t)) = <x(t)^2/4, y(t)^2/9, x(t)y(t)>. That's what it means. Then plug in what you have for x(t) and y(t). You know what they are, right?
 
ok i ended up getting 2sint+3cost the next little problem though is over that interval i get 0. should i just take the it from 0 to pi and multiply it by 2?
 
kekal6 said:
ok i ended up getting 2sint+3cost the next little problem though is over that interval i get 0. should i just take the it from 0 to pi and multiply it by 2?

Why is it such a problem to get 0? I'm thinking 0 might be the right answer here. Your function certainly isn't symmetric across z=0...
 
ok sounds good. just a short quick push on flux if u can answer. I am supposed to do the double int to the surface s of vector f dot n. can i do that with respect to the t or do i have to change it all the way back to x and y
 
kekal6 said:
ok sounds good. just a short quick push on flux if u can answer. I am supposed to do the double int to the surface s of vector f dot n. can i do that with respect to the t or do i have to change it all the way back to x and y

You should be able to do it with respect to t... but now you're getting into topics I'm not sure about.
 
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