Find the curvature at the following point and then sketch this curve.

r(t) = <cos(2t), 2*sin(2t), 4t>, t = 0, t = Pi/2

Note: r is a vector

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- February 21st 2007, 02:40 AMIdeasmanCurvature (2nd)
Find the curvature at the following point and then sketch this curve.

r(t) = <cos(2t), 2*sin(2t), 4t>, t = 0, t = Pi/2

Note: r is a vector - February 21st 2007, 10:38 AMSoroban
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.

Quote:

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