Hello, Ideasman!
Where is your difficulty?
If you have the formula for curvature, go for it!
This is a very unpleasant problem (especially without LaTeX)
. . I'll start it off for you.
Find the curvature at the following point and then sketch this curve.
. . r(t) .= .< cos(2t), 2·sin(2t), 4t >, . at t = 0, t = π/2
. . . . . . . . . . . . . . . . . . .| v x a |
Curvature Formula: . k .= .----------
. . . . . . . . . . . . . . . . . . . . |v|³
Given: .r(t) .= .< cos(2t), 2·sin(2t), 4t >
Then: .v(t) .= .< -2·sin(2t), 4·cos(2t), 4 >
Then: .a(t) .= .< -4·cos(2t), -8·sin(2t), 0 >
We will need:
Code:
| i j k |
| v x a | = | -2·sin(2t) 4·cos(2t) 4 |
| -4·cos(2t) -8·sin(2t) 0 |
We also need |v|³ ._________________________
. . . where |v| .= .√4·sin²(2t) + 16·cos²(2t) + 16