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) $\frac { \pm ( \mathbf { i } + 8 \mathbf { j } ) } { \sqrt { 65 } }$
D) $\frac { \pm ( \mathbf { i } + \mathbf { j } ) } { \sqrt { 65 } }$
E) $\frac { \pm ( i - 8 j ) } { 65 }$

Question

Quizplus · Accepted Answer

To find unit vectors parallel to the tangent line to the curve y = 2x^2 at the point (2,8), we need to find the slope of the tangent line at that point. Taking the derivative of y with respect to x, we get:
$\frac{dy}{dx} = 4x$
At the point (2,8), the slope of the tangent line is:
$\frac{dy}{dx}\Bigg|_{(2,8)} = 4(2) = 8$
So the tangent line has a slope of 8 at the point (2,8). To find unit vectors parallel to this tangent line, we can use the vector $\langle 1,8\rangle$, which has the same direction as the tangent line. To make this a unit vector, we divide by its magnitude:
$\frac{\langle 1,8\rangle}{|\langle 1,8\rangle|} = \frac{\langle 1,8\rangle}{\sqrt{1^2+8^2}} = \pm \frac{1}{\sqrt{65}}\langle 1,8\rangle$
The correct answer is therefore (C) $\frac{\pm(\mathbf{i}+8\mathbf{j})}{\sqrt{65}}$.

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 $y = 2 x ^ { 2 }$