
#1
May1312, 07:21 PM

P: 43

1. The problem statement, all variables and given/known data
Consider the line perpendicular to the surface [itex]z=x^2+y^2[/itex] at the point where x = −1 and y = 2 Find a vector parametric equation for this line in terms of the parameter t. 3. The attempt at a solution I wasn't quite sure how to go about with this problem so I just went along with the following ideas. I first took the gradient of the function at that point: [itex]0=x^2+y^2z[/itex] [itex]∇F(x,y,z)= <2x,2y,1>[/itex] [itex]∇F(1,2,0)= <2,4,1>[/itex] Then I constructed the vector parametric equation of the line at that point: [itex]L(t) = P + t∇F[/itex] [itex]L(t) = <1,2,0> + t<2,4,1>[/itex] Afterwards, I submitted this equation, only finding that it was incorrect; can someone explain to me what went wrong here? 



#2
May1312, 07:26 PM

HW Helper
Thanks
PF Gold
P: 7,198

When ##x=1## and ##y=2##, ##z## isn't zero.




#3
May1312, 07:28 PM

P: 43

Wow haha that was a horrible miscalculation on my part. Thanks for pointing that out!



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