Solutions To Parametric Space Curve Problems

Space Curves (April 30 post):

(1) Find the curvature   k and the length of the curve from t = 0 to t =  p.

We have:    k  =   dT/ ds   And   dT/ ds =  dT/dt /  (ds/ dt)

dT/dt =   1/13 [(-24 sin  2t)i   -24  cos 2t) j

Where:  (dt/ds)   = 1/ 13 

Simplify:  dT/ ds   =  24 [- sin 2t i - cos 2t j] /  169

But:  dT/ ds =   k N    

Where:  N =  [- sin 2t i - cos 2t j]  

k   =  24/169

The length of the curve from t = 0 to t =  p can be obtained from: (dt/ds)   = 1/ 13 

So that:  ds =   13 dt  (for differential in arc length) 

Integrating:  s  =  ò ^p _o   13 dt  =13 p


2.  For a space curve defined by:  

x = exp (t) cos (t),   y = exp(t) sin (t), z = exp(t)

Find  the unit vector  T:

T = dR/ds = dt/ds [i dx/ dt  + j dy.dt + k dz/dt]

dx/dt =  e^t   [cos(t) - sin(t)}

dy/dt = e^t  [cos (t) + sin (t)}

dz/dt = e^t 

Whence:

T = (dt/ds)²   (e^t ) ²  [(cost - sin t)²  + (cos t + sin t)²  + 1]

= (dt/ds)²   (e^t ) ²  [1 + 2 sin t cost - 2 sin t cost + 1  + 1] =  3(dt/ds)²   (e^t ) ²

dt/ds  =   1/ Ö (3) e^t

^{Then: T =  1}/ Ö (3)   [(cos t - sin t)i + (cos t + sin t)j +  k]