Vector Analysis Arc Length, Unit tangent, unit Normal and curvature

Question

Let C be the curve given by the equation, R(t)= sin t i + cos t j + log sec t k     (0≤t<π/2)

 

Find:

a.) the element of arc length, ds, along C, in terms of t.

b.) the unit tangent T.

c.) the unit normal N.

d.) the curvature k.

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Answer 1

 

Let C be the curve given by the equation, r(t)= sin t i + cos t j + log sec t k     (0≤t<π/2)

Let;

r(t) = < sin(t), cos(t), log(sec(t)) >

 

(*assuming logarithm is base 10*)

r'(t) = < cos(t), -sin(t), tan(t) >

|r'| = Sqroot( cos²(t) + (-sin)²(t) + tan²(t))

r''(t) = < -sint(t), -cos(t) sec(t)² >

 

*notation:

dr/dt = r'  (differentiation with respect to t)

r'' = 2nd derivative with respect to t

a.) Arc length, ds, along C, in terms of t.

 

Arc length, ds = |r'|dt   

ds = ∫ |r'| dt  over the interval  0≤t<π/2

b.) the unit tangent T.

 

Tangent vector = dr/dt = r' 

The derivative of r(t) gives a new vector function that is tangent to the defined curve.

 

Unit tangent vector, T = r'/|r'|

 

* from arc length you can see that |r'| = ds/dt

 T= r'/|r'| = (dr/dt)/(ds/dt) = dr/ds

c.) the unit normal N.

N = T'/|T'|

 

*The unit normal vector, N is orthogonal to the unit tangent vector, T.

d.) the curvature k.

The curvature measures how fast a curve is changing direction at a given point. There are several Eq.'s you can use to solve. I believe the one bellow will be the easiest.

 

k = |T'|/|r'|