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

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

Mathematics
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find gradient and then neg. gradient should be all i believe, if I am not mistaken
then once you find those, divide by length
Yes

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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,\[\pm \frac{\nabla f(x,y,z)}{|\nabla f(x,y,z)|}=\pm \frac{(2x,2z^3,0)}{\sqrt{4x^2+4z^6}}\]
Substitute \[(x,y,z)=(1,-1,-1)\]
Wait..
forgot z
why the 0 term at the end?
yeah...because i just got up ;)
oh, ok I see
:)
yeah, that is the gradient I got, with the last term being 6yz^2
\[\nabla (x,y,z)=(2x,2z^3,6yz^2)\]
Yeah
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?
\[\pm \frac{\nabla (x,y,z)}{|\nabla (x,y,z)|}= \pm \frac{(2x,2z^3,6yz^2)}{\sqrt{4x^2+4z^6+36y^2z^4}}\]
I wish there was a manual. I find everything by trial-and-error. Just use the equation editor on the bottom.
You can take out a common factor of 2 between the numerator and denominator if you're desperate...
But, cquinn, all you need to do now is sub (x,y,z)=(1,-1,-1).
ahh, didn't recognize that at the bottom. will fiddle around with it some
have fun!
Thanks so much!! This helps a TON
np
fan us ;)

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