Find the curvature of the curve r(t).
-$$\mathbf { r } ( \mathrm { t } ) = ( 3 + \ln ( \sec t ) ) \mathbf { i } + ( 1 + \mathrm { t } ) \mathbf { k } , - \pi / 2

Question

Find the curvature of the curve r(t).
-$$\mathbf { r } ( \mathrm { t } ) = ( 3 + \ln ( \sec t ) ) \mathbf { i } + ( 1 + \mathrm { t } ) \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2$$
A) $\kappa = - \cos t$
B) $\kappa = \sin t$
C) $\kappa = \cos t$
D) $\kappa = 1 - \cos t$

Quizplus · Accepted Answer

We can find the curvature using the formula 
$$\kappa = \frac{\left\Vert {\bf T}'(t) ight\Vert}{\left\Vert {\bf r}'(t) ight\Vert}$$
where ${\bf T}(t) = \frac{{\bf r}'(t)}{\left\Vert {\bf r}'(t) ight\Vert}$ is the unit tangent vector.

Differentiating ${\bf r}(t)$ gives ${\bf r}'(t) = \sec(t) \mathbf{i} + (1+t)\mathbf{k}$, and taking the magnitude of this vector gives $\left\Vert {\bf r}'(t) ight\Vert = \sqrt{\sec^2(t) + (1+t)^2}$.

Differentiating again gives $${\bf T}'(t) = \frac{1}{\sqrt{\sec^2(t) + (1+t)^2}} \left( -\cos(t)\mathbf{i} + \sin(t) \mathbf{k} ight),$$
and taking the magnitude of this gives $$\left\Vert {\bf T}'(t) ight\Vert = \frac{\sqrt{\cos^2(t) + \sin^2(t)}}{\sqrt{\sec^2(t) + (1+t)^2}} = \frac{1}{\sqrt{\sec^2(t) + (1+t)^2}}.$$

Therefore, we have 
$$\kappa = \frac{\left\Vert {\bf T}'(t) ight\Vert}{\left\Vert {\bf r}'(t) ight\Vert} = \cos(t) \cdot \frac{1}{\sqrt{\sec^2(t) + (1+t)^2}} =\frac{\cos(t)}{\sqrt{1+(\sin(t)+\cos(t))^2}} = \cos(t).$$

Thus, the answer is $\boxed{	extbf{(C)}}$.

Find the Curvature of the Curve R(t) r(t)=(3+ln⁡(sec⁡t))i+(1+t)k,−π/2<t<π/2\mathbf { r } ( \mathrm { t } ) = ( 3 + \ln ( \sec t ) ) \mathbf { i } + ( 1 + \mathrm { t } ) \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2r(t)=(3+ln(sect))i+(1+t)k,−π/2<t<π/2

Find the Curvature of the Curve R(t) $\mathbf { r } ( \mathrm { t } ) = ( 3 + \ln ( \sec t ) ) \mathbf { i } + ( 1 + \mathrm { t } ) \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2$