SeeDRankLou
02-27-2009, 04:08 AM
I have this problem, and I got the right answer, I'm just not sure I did this in the easiest manner that I could have. Here's what I did:
Find the curvature of r(t) in terms of t
r(t) = t cos(t) i + t sin(t) j
curvature (K) = | y'' x' - x'' y' | / ((x' ē + y' ē)^(3/2))
x(t) = t cos(t)
x'(t) = cos(t) - t sin(t)
x''(t) = -sin(t) - ( sin(t) + t cos(t) ) = -2 sin(t) - t cos(t)
y(t) = t sin(t)
y'(t) = sin(t) + t cos(t)
y''(t) = cos(t) + ( cos(t) - t sin(t) ) = 2 cos(t) - t sin(t)
(1) y'' x' = ( 2 cos(t) - t sin(t) ) ( cos(t) - t sin(t) ) = 2 cosē(t) - 3t cos(t) sin(t) + tē sinē(t)
(2) x'' y' = (-2 sin(t) - t cos(t) ) ( sin(t) + t cos(t) ) = -2 sinē(t) - 3t sin(t) cos(t) - tē cosē(t)
(1) - (2) = 2 ( cosē(t) + sinē(t) ) + tē ( sinē(t) + cosē(t) ) = 2 + tē
(3) x' ē = ( cos(t) - t sin(t) )ē = cosē(t) - 2t cos(t) sin(t) + tē sinē(t)
(4) y' ē = ( sin(t) + t cos(t) )ē = sinē(t) + 2t sin(t) cos(t) + tē cosē(t)
(3) + (4) = ( cosē(t) + sinē(t) ) + tē ( sinē(t) + cosē(t) ) = 1 + tē
K = | 2 + tē | / (( 1 + tē )^(3/2))
2 + tē > 0
K = ( 2 + tē ) / (( 1 + tē )^(3/2))
This seems like the long way of doing this. My question is if there was something simplier I could have done that I am overlooking. Like converting the vector function to an xy-function or something like that.
Find the curvature of r(t) in terms of t
r(t) = t cos(t) i + t sin(t) j
curvature (K) = | y'' x' - x'' y' | / ((x' ē + y' ē)^(3/2))
x(t) = t cos(t)
x'(t) = cos(t) - t sin(t)
x''(t) = -sin(t) - ( sin(t) + t cos(t) ) = -2 sin(t) - t cos(t)
y(t) = t sin(t)
y'(t) = sin(t) + t cos(t)
y''(t) = cos(t) + ( cos(t) - t sin(t) ) = 2 cos(t) - t sin(t)
(1) y'' x' = ( 2 cos(t) - t sin(t) ) ( cos(t) - t sin(t) ) = 2 cosē(t) - 3t cos(t) sin(t) + tē sinē(t)
(2) x'' y' = (-2 sin(t) - t cos(t) ) ( sin(t) + t cos(t) ) = -2 sinē(t) - 3t sin(t) cos(t) - tē cosē(t)
(1) - (2) = 2 ( cosē(t) + sinē(t) ) + tē ( sinē(t) + cosē(t) ) = 2 + tē
(3) x' ē = ( cos(t) - t sin(t) )ē = cosē(t) - 2t cos(t) sin(t) + tē sinē(t)
(4) y' ē = ( sin(t) + t cos(t) )ē = sinē(t) + 2t sin(t) cos(t) + tē cosē(t)
(3) + (4) = ( cosē(t) + sinē(t) ) + tē ( sinē(t) + cosē(t) ) = 1 + tē
K = | 2 + tē | / (( 1 + tē )^(3/2))
2 + tē > 0
K = ( 2 + tē ) / (( 1 + tē )^(3/2))
This seems like the long way of doing this. My question is if there was something simplier I could have done that I am overlooking. Like converting the vector function to an xy-function or something like that.