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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)
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find gradient and then neg. gradient should be all i believe, if I am not mistaken

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then once you find those, divide by length

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x,y,z)=x^2+2yz^33=0 \rightarrow \nabla f=(2x,2z^3,0)\]will be normal to the surface at (x,y,z). Then,\[\pm \frac{\nabla f(x,y,z)}{\nabla f(x,y,z)}=\pm \frac{(2x,2z^3,0)}{\sqrt{4x^2+4z^6}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Substitute \[(x,y,z)=(1,1,1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why the 0 term at the end?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah...because i just got up ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that is the gradient I got, with the last term being 6yz^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\nabla (x,y,z)=(2x,2z^3,6yz^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you type in all those commands? that is cool, is there a manual around that I could look at some of those in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pm \frac{\nabla (x,y,z)}{\nabla (x,y,z)}= \pm \frac{(2x,2z^3,6yz^2)}{\sqrt{4x^2+4z^6+36y^2z^4}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wish there was a manual. I find everything by trialanderror. Just use the equation editor on the bottom.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can take out a common factor of 2 between the numerator and denominator if you're desperate...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But, cquinn, all you need to do now is sub (x,y,z)=(1,1,1).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh, didn't recognize that at the bottom. will fiddle around with it some

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks so much!! This helps a TON
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