Question

Find the equations of the normal plane and osculating plane of the curve at:
x=2sin(3t), y=t, z=2cos(3t)
at (0, pi, -2)

Answer 1

I think they want you to find the unit tangent and unit Normal vectors and then use those to find the equation of the plane...

Answer 2

planes*

Answer 3

@Wislar you there?

Answer 4

would we need a binormal as well?

Answer 5

don't think so...

Answer 6

maybe for extra credit :)

Answer 7

let me check...

Answer 8

oscillating plane should be the one containing T and N so TXN gives the normal to the plane...

Answer 9

and TxN is B if i recall correctly ;)

Answer 10

http://www.encyclopediaofmath.org/index.php/Osculating_plane

this gives a method for the kissing plane

Answer 11

well.. sort of...

Answer 12

I'm here! I'm fine with the calculus part, I'm just not sure what the equation is.

Answer 13

direction of B

Answer 14

unit T X unit N gives the binormal

Answer 15

A little confused here/ What would I do once I found the binormal and normal vectors?

Answer 16

ive always kind of wondered why we would need unit vectors; shouldnt any 2 vectors when crossed give an orthogonal vector to both?

Answer 17

if you can find unit N and unit T, unit T is normal to the Normal plane and unit T X unit N is normal to the osculating plane

Answer 18

yes:)

Answer 19

given a vector<a,b,c>; and a point (x1,y1,z1)

a plane is constructed as\[a(x-x_1)+b(y-y_1)+c(z-z_1)=0\]

Answer 20

given a vector tha tis normal to the plane that is :)

Answer 21

ah

Answer 22

it helps to have alot of voices in the head that correct you ;)

Answer 23

a, b, and c are found with the normal vector right?

Answer 24

heh

Answer 25

yep

Answer 26

Awesome! Now what about the osculating plane?

Answer 27

a,b,c are the vectors that you want to find; and the "planes" for them are formed from the general construct of the plane equation