Find the curvature of the space curve.
-r(t) = (3 t sin t + 3 cos t)i + 3j + (3t cos t - 3 sin t)k
A) $\kappa = - \frac { 1 } { 3 t }$
B) $\kappa = \frac { 1 } { 9 t ^ { 2 } }$
C) $\kappa = 3 t$
D) $\kappa = \frac { 1 } { 3 t }$

Question

Find the curvature of the space curve.
-r(t) = (3 t sin t + 3 cos t)i + 3j + (3t cos t - 3 sin t)k 
A) $\kappa = - \frac { 1 } { 3 t }$
B) $\kappa = \frac { 1 } { 9 t ^ { 2 } }$
C) $\kappa = 3 t$
D) $\kappa = \frac { 1 } { 3 t }$

Quizplus · Accepted Answer

The curvature $\kappa$ of a space curve defined by a vector function $r(t)$ is given by $\kappa = \frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}$. Calculating the derivatives $r'(t)$ and $r''(t)$, and then their cross product and norms, simplifies to $\kappa = \frac { 1 } { 3 t }$ for this curve.

Find the Curvature of the Space Curve κ=−13t\kappa = - \frac { 1 } { 3 t }κ=−3t1​

Find the Curvature of the Space Curve $\kappa = - \frac { 1 } { 3 t }$