Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 6 } \right)$ is

A) $\frac { 1 } { 2 } \mathbf { i } + \frac { \sqrt { 3 } } { 2 } \mathbf { k }$
B) $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$
C) $\frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$
D) $\frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$
E) $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$

Question

Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 6 } \right)$ is

A) $\frac { 1 } { 2 } \mathbf { i } + \frac { \sqrt { 3 } } { 2 } \mathbf { k }$
B) $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$
C) $\frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$
D) $\frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$
E) $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$

Quizplus · Accepted Answer

The unit normal vector $\mathbf{n}(t)$ is perpendicular to the tangent vector $\mathbf{T}(t)$, which is the derivative of $\mathbf{r}(t)$ normalized. To find $\mathbf{n}(t)$, we first find the derivative of $\mathbf{r}(t)$, then normalize it to get $\mathbf{T}(t)$, and then differentiate $\mathbf{T}(t)$ and normalize again to get $\mathbf{N}(t)$, which is the principal normal vector. However, since the question directly asks for the unit normal vector at $t = \frac{\pi}{6}$ without specifying the process, we can simplify the process by understanding that the unit normal vector is perpendicular to the direction of motion and lies in the plane of curvature.Given $\mathbf{r}(t) = -\sin(2t)\mathbf{i} + 2t\mathbf{j} - \cos(2t)\mathbf{k}$, we can see that the components of the velocity vector $\mathbf{v}(t) = \mathbf{r}'(t)$ would involve $-2\cos(2t)\mathbf{i} + 2\mathbf{j} + 2\sin(2t)\mathbf{k}$. However, for the unit normal vector, we are interested in the direction that is perpendicular to this velocity vector and lies in the plane of the curve's motion.At $t = \frac{\pi}{6}$, evaluating the sine and cosine functions gives us specific values: $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$. The correct choice must reflect a vector that is perpendicular to the direction of motion at this point and normalized.Choice C, $\frac{\sqrt{3}}{2}\mathbf{i} + \frac{1}{2}\mathbf{k}$, correctly represents a unit vector that is perpendicular to the direction of motion at $t = \frac{\pi}{6}$, considering the components of the velocity and acceleration vectors at this point. It is normalized (since the square root of the sum of the squares of the coefficients equals 1) and correctly oriented based on the given function $\mathbf{r}(t)$.

Let r(t)=−sin⁡(2t)i+2tj−cos⁡(2t)k\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }r(t)=−sin(2t)i+2tj−cos(2t)k

Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$