Find the torsion of the space curve.
-r(t) = (5t sin t + 5 cos t)i + (5t cos t) - 5 sin t)j - 6k
A) $\tau = 0$
B) $\tau = 1$
C) Undefined
D) $\tau = - 1$

Question

Find the torsion of the space curve.
-r(t) = (5t sin t + 5 cos t)i + (5t cos t) - 5 sin t)j - 6k 
A) $\tau = 0$
B) $\tau = 1$
C) Undefined
D) $\tau = - 1$

Quizplus · Accepted Answer

The torsion of a straight line is always 0 because torsion measures how a curve twists out of the plane of curvature, and a straight line does not twist. The given curve, $r(t) = (5t \sin t + 5 \cos t)i + (5t \cos t - 5 \sin t)j - 6k$, is not explicitly a straight line, but the constant term in the k-component (-6k) and the periodic functions in the i and j components suggest that the curve does not exhibit torsion. The absence of a component that would contribute to a change in the direction of the binormal vector (which torsion measures) implies that the torsion is 0.

Find the Torsion of the Space Curve τ=0\tau = 0τ=0 B) τ=1\tau = 1τ=1

Find the Torsion of the Space Curve $\tau = 0$ B) $\tau = 1$