## Friday, October 8, 2021

### Solutions To Problems For Parametric Curves In 3- Space

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 =   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 =  et   [cos(t) - sin(t)}

dy/dt = et  [cos (t) + sin (t)}

dz/dt = et

Whence:

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

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

dt/ds  =   1/ Ö (3) et

Then: T =  1Ö (3)   [(cos t - sin t)i + (cos t + sin t)j +  k