- #1
hholzer
- 37
- 0
If we parameterize the arc length of a vector valued
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?
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?