Question

Find the Normal and Osculating plane when t=1 for the position vector r(t) =

Answer 1

\[T(t)=\frac{R'(t)}{|R'(t)|}\\N(t)=\frac{T'(t)}{|T'(t)|}\\B(t)=T(t)\times N(t)\]
the osculating plane determined by the unit vectors T and N, the normal plane by the vectors N and B

Answer 2

Yes, I have already found T, N, and B. I can't seem to get them to work out though

Answer 3

They seem so convoluted that I'm pretty sure I did it wrong. I can't find the cross product when finding the planes they are so convoluted.

Answer 4

find the planes when t=1.

Answer 5

My answers are:

\[T= <\frac{ -\pi \sin(\pi t) }{ \sqrt{\pi^2+9} }, \frac{ \pi \cos(\pi t ) }{ \sqrt{\pi^2+9} }, \frac{ 3 }{ \sqrt{\pi^2+9} }>\]

Answer 6

Yes, I understand the question. Even after plugging in t=1 they are super convoluted, especially when taking the cross product.

Answer 7

Which makes me believe I did it wrong.

Answer 8

factor out the scalar \[\frac{1}{\sqrt{9+\pi^2}}\].
it's scalar anyway.

Answer 9

I don't need it when I do t=1?

Answer 10

\[T(1)=\frac{1}{\sqrt{\pi^2+9}}<0,-\pi,3>\]

Answer 11

ok, that's what I got

Answer 12

N(1)= <1,0,0>

Answer 13

So the question says "Find an equation for the osculating plane when t=1 (it contains the point with position vector r(1) and has normal vector \[T(1) X N(1)\] )

Answer 14

so I have to find T(1) X N(1)
I got \[<0, \frac{ 3 }{\sqrt{\pi^2+9} }, \frac{ \pi }{ \sqrt{\pi^2+9}}>\]

Answer 15

then I have have to do find the plane with that normal vector and that contains the point r(1)

Answer 16

So I ended up with \[3y+\pi z=3\pi\]

Answer 17

i have N(1)=pi<1,0,0>

Answer 18

nvm. pi can be "eliminated" in the equation of the plane

i also have the same equation for the osculating plane.

Answer 19

Thank you!