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rado5
Nov9-09, 10:55 PM
1. The problem statement, all variables and given/known data

Find the unique vector which is perpendicular on x2+y2=100 at (x0,y0)=(6,8).

2. Relevant equations

\nablaF/ \left| \nablaF \left|

3. The attempt at a solution

I think the solution is \nablaF/ \left| \nablaF \left| but I don't know how to calculate \nablaF for x2+y2=100
I know \nablaF=(\partialF/\partialx,\partialF/\partialy,\partialF/\partialz)
Please tell me how I can find the answer. What is F here?

Dick
Nov9-09, 11:30 PM
F(x,y)=x^2+y^2-100. The level surface is F(x,y)=0. Now do the gradient. Or F(x,y)=x^2+y^2 and the level surface is F(x,y)=100, your choice.

rado5
Nov10-09, 11:26 AM
So in this case the solution is \nablaF/\left|\nablaF\left| = (0.6,0.8) Is it correct?
(\nablaF=(2x,2y)).

Dick
Nov10-09, 11:32 AM
So in this case the solution is \nablaF/\left|\nablaF\left| = (0.6,0.8) Is it correct?
(\nablaF=(2x,2y)).

If you mean to find the unit normal to the circle, sure, that's right.

rado5
Nov10-09, 11:49 AM
Thank you very much really. I actually translated the problem from persian to English, but I think it is now correct. Again thank you very much for your great generosity.