Find the unit vectors that are parallel to the tangent line to the curve $y = 2 x ^ { 2 }$ at the point $( 2,8 )$.
A) $\frac { \pm ( 2 \mathbf { i } - 8 \mathbf { j } ) } { 65 }$
B) $\frac { \pm ( 2 \mathbf { i } + 8 \mathbf { j } ) } { 65 }$
C) $\frac { \pm ( i + 8 j ) } { \sqrt { 65 } }$
D) $\frac { \pm ( \mathbf { i } + \mathbf { j } ) } { \sqrt { 65 } }$
E) $\frac { \pm ( \mathbf { i } - 8 \mathbf { j } ) } { 65 }$

Question

Quizplus · Accepted Answer

We first find the derivative of the function using the power rule: 
$$\frac{dy}{dx} = 4x$$
Substituting $x=2$, we get the slope of the tangent line: 
$$m=\frac{dy}{dx}\bigg|_{x=2} = 8$$
We now need to find the unit vector parallel to this slope. Since the slope is positive, the tangent line goes in the positive $x$ and positive $y$ direction.

Recall that the unit vector parallel to a given vector $\mathbf{v} = \begin{pmatrix}a\b\end{pmatrix}$ is given by $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$, where $\|\mathbf{v}\|$ is the length (magnitude) of $\mathbf{v}$.

Thus, the tangent line has direction vector $\begin{pmatrix}1\8\end{pmatrix}$, and the unit vectors parallel to this are:
$$\pm \frac{1}{\sqrt{1^2+8^2}}\begin{pmatrix}1\8\end{pmatrix} = \pm \frac{1}{\sqrt{65}}\begin{pmatrix}1\8\end{pmatrix}$$
which matches the form of answer choice (C).

Find the Unit Vectors That Are Parallel to the Tangent y=2x2y = 2 x ^ { 2 }y=2x2

Find the Unit Vectors That Are Parallel to the Tangent $y = 2 x ^ { 2 }$