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anonymous
 5 years ago
Find an equation of the plane trangent to the hyperboloid \[2x^23y^2+z^2=7\] at the point (3,2,1).
anonymous
 5 years ago
Find an equation of the plane trangent to the hyperboloid \[2x^23y^2+z^2=7\] at the point (3,2,1).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try visiting this url it might be able to explain this concept better than i can: http://math.etsu.edu/multicalc/archives/Chap3/Chap36/index.htm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First you need to find the unit vector. You know how to find it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I think I figured it out. Grad F(x,y,x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would it be. \[6(x3) 6(y  2) +(z1) =0\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are right that's the equation for a plane passing through that point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sames right 6x6y+z=9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually that doesnt look right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got the same thing as lockdown but maybe i am wrong, i do know that you should most likely condense your answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I condensed it. I got 12x  12y + 2z  14 = 0 originally, but divided by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0meaning combine constants on the other side of the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here is a neat trick, Lockdown, rather than writing out the equation formula. Once you get your vector <6, 6, 1> Write it as an eq 6x6y+1= The other side can be determined in your head by the points (6*3=18) 12 + 1=7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know that trick is actually helpful to me too, thanks chaguanas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, the equation is taught in schools, mathematicians don't do it the eq way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, thats a good way of thinking about it, since they always want simplified anyways.
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