Find the curvature of the space curve.
-r(t) = (t + 4)i + 9j + (ln(sec t) + 3)k
A) $\kappa = \sin t$
B) $\kappa = \csc t$
C) $\kappa = \cos t$
D) $\kappa = \sec t$

Question

Find the curvature of the space curve.
-r(t) = (t + 4)i + 9j + (ln(sec t) + 3)k 
A) $\kappa = \sin t$
B) $\kappa = \csc t$
C) $\kappa = \cos t$
D) $\kappa = \sec t$

Quizplus · Accepted Answer

The formula for curvature of a space curve given parametrically as $\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}$ is $$\kappa = \frac{\left\lVert \vec{r}'(t) \times \vec{r}''(t)\right\rVert}{\left\lVert \vec{r}'(t)\right\rVert^3}.$$ 
Computing, we have $\vec{r}'(t) = \vec{i} + \frac{\sec t\tan t}{\sec t} \vec{k} = \vec{i} + \tan t \vec{k}$ and 
\begin{align*}
\vec{r}''(t) &= \frac{d}{dt}\left(\vec{i} + \tan t \vec{k}\right) \
&= \sec^2 t \vec{k}.
\end{align*} 
Then, 
\begin{align*}
\left\lVert \vec{r}'(t) \times \vec{r}''(t)\right\rVert &= \left\lVert(-\sec t)\vec{j}\right\rVert = \sec t, \
\left\lVert \vec{r}'(t)\right\rVert^3 &= (1 + \tan^2 t)^{3/2} = \sec^3t,
\end{align*}
so 
$$\kappa = \frac{\sec t}{\sec^3 t} = \cos t.$$

Find the Curvature of the Space Curve κ=sin⁡t\kappa = \sin tκ=sint

Find the Curvature of the Space Curve $\kappa = \sin t$