can you find the unit tangent vector, unit normal and unit binormal , curvature and torsion for this as t= 1 ? ?

Question

x = 2t

y = t^2

z = lnt

and t = 1

Answer 1

Hi Dorsa,

Write the position vector as r(t) = (x(t), y(t), z(t)) = (2t, t^2, ln(t)). The velocity vector is then

v(t) = r'(t) = (x'(t), y'(t), z'(t)) and the acceleration vector is a(t) = v'(t) = r''(t) = (x''(t), y''(t), z''(t)). Computing second derivatives, we find v(t) = (2, 2t, 1/t) and a(t) = (0, 2, -1/t²). To compute the torsion, we will also use the derivative of the acceleration vector, which in physics is called the jerk:

j(t) = a'(t) = (x'''(t), y'''(t), z'''(t)) = (0,0, 2/t³).

At t = 1, the position, velocity, acceleration, and jerk vectors are given by

r(1) = (2,1, 0),

v(1) = (2, 2,1),

a(1) = (0, 2, -1),

j(1) = (0,0,2).

We now use these vectors to compute the unit tangent, normal, binormal, curvature, and torsion at t = 1.

The unit tangent vector t is given by t=v/||v||, where ||v|| is the length of v (the speed). The speed at t = 1 is

||v(1)||=(2² + 2² + 1²)^1/2 = 9^1/2 = 3, so the unit tangent vector at t = 1is

t(1) = v(1)/||v(1)|| = (1/3)(2, 2,1) = (2/3, 2/3, 1/3).

The unit normal vector n is given by the derivative of the unit tangent vector over its length: n = t'/||t'||. To compute this, we need to compute the unit tangent vector at time t, so we can take the derivative. The velocity at time t was v(t) = (2, 2t, 1/t) and its length at time t is

||v(t)|| = (2² + (2t)² + (1/t)²)^1/2 = (4+4t²+1/t²)^1/2 = 1/t (4t²+4t⁴+1)^1/2, where I have factored out a 1/t for future simplification. The unit tangent vector at time t is then

t(t) = v(t)/||v(t)|| = (2, 2t, 1/t)/(1/t (4t²+4t⁴+1)^1/2) = (2t, 2t², 1) / (4t²+4t⁴+1)^1/2

= ( 2t / (4t²+4t⁴+1)^1/2 , 2t² / (4t²+4t⁴+1)^1/2, 1 / (4t²+4t⁴+1)^1/2).

We now need to differentiate this vector. Unfortunately, this means using the quotient rule 3 times, one for each component (calculating unit normal vectors by hand is almost always a mess). Doing so gives, after some simplification,

t'(t) =( (2-4t²) / (1+2t²)², 4t / (1+2t²)², -4t / (1+2t²)²)

which at t = 1 is

t'(1) = (-2/9, 4/9, -4/9).

The length of this vector is ||t'(1)|| = ( (-2/9)² + (4/9)² + (-4/9)²)^1/2 = 2/3.

The unit normal vector at t = 1 is then

n(1) = t'(1) / ||t'(1)|| = (3/2) (-2/9, 4/9, -4/9) = (-1/3, 2/3, -2/3).

The unit binormal vector at t = 1 is given by the following cross product:

b(1) = t(1) x n(1) = (2/3, 2/3, 1/3) x (-1/3, 2/3, -2/3) = (-2/3, 1/3, 2/3).

The curvature at t = 1 can be computed by the formula

k(1) = || v(1) x a(1)|| / ||v(1)||^3 = || (2, 2,1) x (0, 2, -1) || / 3³ = ||(-4,2,4)|| / 27 = 6/27 = 2/9.

The torsion at t = 1 can be computed by the formula

T = (j(1) * (v(1) x a(1))) / || v(1) x a(1)||² , where the * is the dot product.

= ((0,0,2) * ((2, 2,1) x (0, 2, -1)) / || (2, 2,1) x (0, 2, -1) ||²

= ((0,0,2) *(-4,2,4) / ||(-4,2,4)||²

= 8 / 6²

= 8/36

= 2/9.

I hope this helps!