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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    try visiting this url it might be able to explain this concept better than i can: http://math.etsu.edu/multicalc/archives/Chap3/Chap3-6/index.htm

  2. anonymous
    • 5 years ago
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    did that help?

  3. anonymous
    • 5 years ago
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    First you need to find the unit vector. You know how to find it?

  4. anonymous
    • 5 years ago
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    yeah, I think I figured it out. Grad F(x,y,x)

  5. anonymous
    • 5 years ago
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    so would it be. \[6(x-3) -6(y - 2) +(z-1) =0\] ?

  6. anonymous
    • 5 years ago
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    You are right that's the equation for a plane passing through that point

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    Sames right 6x-6y+z=9

  9. anonymous
    • 5 years ago
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    actually that doesnt look right

  10. anonymous
    • 5 years ago
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    The 9 should be 7

  11. anonymous
    • 5 years ago
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    i got the same thing as lockdown but maybe i am wrong, i do know that you should most likely condense your answer

  12. anonymous
    • 5 years ago
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    yeah, I condensed it. I got 12x - 12y + 2z - 14 = 0 originally, but divided by 2

  13. anonymous
    • 5 years ago
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    Actually yeah

  14. anonymous
    • 5 years ago
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    meaning combine constants on the other side of the equation

  15. anonymous
    • 5 years ago
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    Here 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 6x-6y+1= The other side can be determined in your head by the points (6*3=18) -12 + 1=7

  16. anonymous
    • 5 years ago
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    you know that trick is actually helpful to me too, thanks chaguanas

  17. anonymous
    • 5 years ago
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    Yeah, the equation is taught in schools, mathematicians don't do it the eq way

  18. anonymous
    • 5 years ago
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    yeah, thats a good way of thinking about it, since they always want simplified anyways.

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