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anonymous

  • 5 years ago

Find all unit vectors perpendicular to the surface x^2 + 2yz^3 = 3 at the point (1, -1, -1)

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  1. anonymous
    • 5 years ago
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    find gradient and then neg. gradient should be all i believe, if I am not mistaken

  2. anonymous
    • 5 years ago
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    then once you find those, divide by length

  3. anonymous
    • 5 years ago
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    Yes

  4. anonymous
    • 5 years ago
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    \[f(x,y,z)=x^2+2yz^3-3=0 \rightarrow \nabla f=(2x,2z^3,0)\]will be normal to the surface at (x,y,z). Then,\[\pm \frac{\nabla f(x,y,z)}{|\nabla f(x,y,z)|}=\pm \frac{(2x,2z^3,0)}{\sqrt{4x^2+4z^6}}\]

  5. anonymous
    • 5 years ago
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    Substitute \[(x,y,z)=(1,-1,-1)\]

  6. anonymous
    • 5 years ago
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    Wait..

  7. anonymous
    • 5 years ago
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    forgot z

  8. anonymous
    • 5 years ago
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    why the 0 term at the end?

  9. anonymous
    • 5 years ago
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    yeah...because i just got up ;)

  10. anonymous
    • 5 years ago
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    oh, ok I see

  11. anonymous
    • 5 years ago
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    :)

  12. anonymous
    • 5 years ago
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    yeah, that is the gradient I got, with the last term being 6yz^2

  13. anonymous
    • 5 years ago
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    \[\nabla (x,y,z)=(2x,2z^3,6yz^2)\]

  14. anonymous
    • 5 years ago
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    Yeah

  15. anonymous
    • 5 years ago
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    how do you type in all those commands? that is cool, is there a manual around that I could look at some of those in?

  16. anonymous
    • 5 years ago
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    \[\pm \frac{\nabla (x,y,z)}{|\nabla (x,y,z)|}= \pm \frac{(2x,2z^3,6yz^2)}{\sqrt{4x^2+4z^6+36y^2z^4}}\]

  17. anonymous
    • 5 years ago
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    I wish there was a manual. I find everything by trial-and-error. Just use the equation editor on the bottom.

  18. anonymous
    • 5 years ago
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    You can take out a common factor of 2 between the numerator and denominator if you're desperate...

  19. anonymous
    • 5 years ago
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    But, cquinn, all you need to do now is sub (x,y,z)=(1,-1,-1).

  20. anonymous
    • 5 years ago
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    ahh, didn't recognize that at the bottom. will fiddle around with it some

  21. anonymous
    • 5 years ago
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    have fun!

  22. anonymous
    • 5 years ago
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    Thanks so much!! This helps a TON

  23. anonymous
    • 5 years ago
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    np

  24. anonymous
    • 5 years ago
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    fan us ;)

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spraguer (Moderator)
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