- #1

- 37

- 0

function, say, r(s) and r(s) has constant curvature

(not equal to zero), then r(s) is a circle.

Thus, |T'(s)| = K but to prove it we would need

to show |T'(s)| = K => <-Kcos(s), -Ksin(s)>

and integrate component-wise two times,

right?

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- Thread starter hholzer
- Start date

- #1

- 37

- 0

function, say, r(s) and r(s) has constant curvature

(not equal to zero), then r(s) is a circle.

Thus, |T'(s)| = K but to prove it we would need

to show |T'(s)| = K => <-Kcos(s), -Ksin(s)>

and integrate component-wise two times,

right?

- #2

- 236

- 0

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