# How do you find a unit vector a) parallel to and b) normal to the graph of f(x)=-(x^2)+5 at given point (3,9)?

Jan 6, 2017

Parallel unit vector: $\left(\vec{\frac{1}{\sqrt{37}} , \frac{6}{\sqrt{37}}}\right)$

Perpendicular unit vector: $\left(\vec{\frac{6}{\sqrt{37}} , - \frac{1}{\sqrt{37}}}\right)$

#### Explanation:

Step 1: Determine the general equation for the slope of the tangent
The slope of a line is given by the derivative of the function.
Given $f \left(x\right) = {x}^{2} + 5$
the slope is $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 2 x$ (using the exponent rule for exponents)

Step 2: Determine the specific slope of the tangent at the given point
At $\left(3 , 9\right)$, $x = 3$
So the slope is $\frac{\mathrm{df} \left(3\right)}{\mathrm{dx}} = 2 \cdot 3 = 6$

Step 3: Determine the unit vector with slope of the tangent
Consider a unit vector with slope of $6$ and a base at the origin:
$\textcolor{w h i t e}{\text{XXX}} \frac{y}{x} = 6$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = 6 x$

and since it is a unit vector:
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{x}^{2} + {y}^{2}} = 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \sqrt{{x}^{2} + {\left(6 x\right)}^{2}} = 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \sqrt{37} x = 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \frac{1}{\sqrt{37}}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = \frac{6}{\sqrt{37}}$

Unit vector parallel to the tangent: $\left(\vec{x , y}\right) = \left(\vec{\frac{1}{37} , \frac{6}{37}}\right)$

Step 4: Determine the unit vector perpendicular to the tangent
Remember that if a line has a slope of $m$ then all lines perpendicular to it will have a slope of $- \frac{1}{m}$.
So we are looking for a unit vector with slope $\left(- \frac{1}{6}\right)$
Using the same logic as in Step 3, we have
$\textcolor{w h i t e}{\text{XXX}} \left(\vec{x , y}\right) = \left(\vec{\frac{1}{\sqrt{37}} , - \frac{6}{\sqrt{37}}}\right)$