Question

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

Answer 1

find gradient and then neg. gradient should be all i believe, if I am not mistaken

Answer 2

then once you find those, divide by length

Answer 3

Yes

Answer 4

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

Answer 5

Substitute \[(x,y,z)=(1,-1,-1)\]

Answer 6

Wait..

Answer 7

forgot z

Answer 8

why the 0 term at the end?

Answer 9

yeah...because i just got up ;)

Answer 10

oh, ok I see

Answer 11

:)

Answer 12

yeah, that is the gradient I got, with the last term being 6yz^2

Answer 13

\[\nabla (x,y,z)=(2x,2z^3,6yz^2)\]

Answer 14

Yeah

Answer 15

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?

Answer 16

\[\pm \frac{\nabla (x,y,z)}{|\nabla (x,y,z)|}= \pm \frac{(2x,2z^3,6yz^2)}{\sqrt{4x^2+4z^6+36y^2z^4}}\]

Answer 17

I wish there was a manual. I find everything by trial-and-error. Just use the equation editor on the bottom.

Answer 18

You can take out a common factor of 2 between the numerator and denominator if you're desperate...

Answer 19

But, cquinn, all you need to do now is sub (x,y,z)=(1,-1,-1).

Answer 20

ahh, didn't recognize that at the bottom. will fiddle around with it some

Answer 21

have fun!

Answer 22

Thanks so much!! This helps a TON

Answer 23

np

Answer 24

fan us ;)