# How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: f(x) = sqrt(25+x^2) and point: (3,4)?

Jul 26, 2018

Unit vector parallel to the graph at (3,4) is is
n^->=4/5i +3/5j^
Unit vector normal to the graph at (3,4) is is
${n}^{\to} = \frac{3}{5} i - \frac{4}{5} j$

#### Explanation:

$f \left(x\right) = \sqrt{25 + {x}^{2}}$
slope of the tangent indicates a vector parallel to the graph at a point
Differentiating
Let
$y = f \left(x\right)$
$y = \sqrt{25 + {x}^{2}}$
Squaring both sides
${y}^{2} = 25 + {x}^{2}$
${y}^{2} - {x}^{2} = {5}^{2}$
Now, Applyig chain rule and differentiating
$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x = 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$
At
(x,y)-=(3,4);
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{4}$
Slope of a parallel line is
$\sqrt{{3}^{2} + {4}^{2}} = 5$
Unit vector parallel to the graph at (3,4) is is
n^->=4/5i +3/5j^
Normal is perpendicular to the parallel.
Thus, slope of the normal is
${m}_{1} {m}_{2} = - 1$
Slope of anormal line is
$- \frac{4}{3}$
Unit vector normal to the graph at (3,4) is is
${n}^{\to} = \frac{3}{5} i - \frac{4}{5} j$