Question

Help!

Answer 1

Anyone knows how to figure this one out?

Answer 2

does it have anything to do with curvature??

Answer 3

work the tangent and the normal vectors

Answer 4

@IrishBoy123 working on 'em. i have the tangent already, doing the normal now

Answer 5

what is your tangent vector at t=0?

Answer 6

alright.. are they:
\[T = \left( \frac{ -\sin t }{ \sqrt{2} }, \frac{ \cos t }{ \sqrt{2} }, \frac{ 1 }{ \sqrt{2} } \right)\]
and
\[N = \left( \frac{ - \cos t }{ \sqrt{2} }, \frac{ -\sin t }{ \sqrt{2} }, 0 \right)\]
or am i messing up again??

Answer 7

in the notes, i have\[T = \frac{ r' }{ \left| r' \right| }\] and \[N = \frac{ T' }{ \left| T' \right| }\]
are they not really necessary?

Answer 8

they definitely are!!!

Answer 9

oh, you go that way. OK

Answer 10

@IrishBoy123 Oki!

Answer 11

and then i should just do their cross product?

Answer 12

y!!

Answer 13

alrighty! i'm on it!

Answer 14

i get \(<0,1,1> \times <-1,0,0> = <0, -1, 1>\)

Answer 15

yeah, i get the same (with the sqrt2) but the answer is supposed to be <0,1,1> so i'm trying to figure out there the problem is..

Answer 16

oooh okay! so (0,1,1) is good because is in the same plane

Answer 17

soz.

have you done it??!!

Answer 18

yes

the plane is \( y = z\)

do you agree?

Answer 19

yes

Answer 20

oh so B could also be the answer then?

Answer 21

oh sh1t. moment's reflection?

Answer 22

what do you think?

Answer 23

i think it could be? i'm not sure... what do you think?

Answer 24

i think you're right. both

Answer 25

what's with the person who made this paper tho? even the 'sudoku' piece from yesterday was shady. now two possible answers?? whaaa

Answer 26

lol!!

Answer 27

the sudoku was the hardest

this, the most interesting

Answer 28

haha yeah! i'll try the sudoku again tomorrow. anyway, thanks a lot for this one!!

hopefully my exam doesn't have multiple answers in its multiple choice XD

Answer 29

you need to do \(\partial \)'s in you partial differentiation

Answer 30

ie when \(f = f(x,y,z)\), then partial of f wrt x is: \(\dfrac{\partial f}{\partial x} = f_x = \partial_x f\)

Answer 31

for which one? the sudoku?

Answer 32

lol!!

for partial differentiation generally.

so, the soduko :-)) lol

Answer 33

oh ok

Answer 34

good night!!!!!

Answer 35

good night! and thanks again!