## anonymous 5 years ago Find all unit vectors perpendicular to the surface x^2 + 2yz^3 = 3 at the point (1, -1, -1)

1. anonymous

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

2. anonymous

then once you find those, divide by length

3. anonymous

Yes

4. anonymous

$f(x,y,z)=x^2+2yz^3-3=0 \rightarrow \nabla f=(2x,2z^3,0)$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

Substitute $(x,y,z)=(1,-1,-1)$

6. anonymous

Wait..

7. anonymous

forgot z

8. anonymous

why the 0 term at the end?

9. anonymous

yeah...because i just got up ;)

10. anonymous

oh, ok I see

11. anonymous

:)

12. anonymous

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

13. anonymous

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

14. anonymous

Yeah

15. anonymous

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

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

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

have fun!

22. anonymous

Thanks so much!! This helps a TON

23. anonymous

np

24. anonymous

fan us ;)