# What is the equation of the line that is normal to f(x)= xsqrt( 3x+2)  at  x=1 4 ?

##### 1 Answer
May 28, 2016

$y - 28 \sqrt{11} = - \frac{2 \sqrt{11}}{65} \left(x - 14\right)$

#### Explanation:

First, find the point of tangency.

$f \left(14\right) = 14 \sqrt{42 + 2} = 14 \sqrt{44} = 28 \sqrt{11}$

The function and normal line will pass through the point $\left(14 , 28 \sqrt{11}\right)$.

First, find the slope of the tangent line by differentiating the function.

$f \left(x\right) = x {\left(3 x + 2\right)}^{\frac{1}{2}}$

To differentiate, we will have to use the product rule.

$f ' \left(x\right) = {\left(3 x + 2\right)}^{\frac{1}{2}} \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} {\left(3 x + 2\right)}^{\frac{1}{2}}$

The two derivatives present are:

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

Via the chain rule:

$\frac{d}{\mathrm{dx}} {\left(3 x + 2\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(3 x + 2\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(3 x + 2\right)$

$= \frac{3}{2} {\left(3 x + 2\right)}^{- \frac{1}{2}}$

Plugging these both back in to find $f ' \left(x\right)$:

$f ' \left(x\right) = {\left(3 x + 2\right)}^{\frac{1}{2}} \cdot 1 + x \cdot \frac{3}{2} {\left(3 x + 2\right)}^{- \frac{1}{2}}$

$f ' \left(x\right) = \sqrt{3 x + 2} + \frac{3 x}{2 \sqrt{3 x + 2}}$

Getting a common denominator:

$f ' \left(x\right) = \frac{2 \left(3 x + 2\right)}{2 \sqrt{3 x + 2}} + \frac{3 x}{2 \sqrt{3 x + 2}}$

$f ' \left(x\right) = \frac{9 x + 4}{2 \sqrt{3 x + 2}}$

The slope of the tangent line is

$f ' \left(14\right) = \frac{126 + 4}{2 \sqrt{42 + 2}} = \frac{130}{2 \sqrt{44}} = \frac{65}{2 \sqrt{11}}$

The slope of the normal line is the opposite reciprocal of the slope of the tangent line, since the two lines are perpendicular.

The opposite reciprocal of $\frac{65}{2 \sqrt{11}}$ is $- \frac{2 \sqrt{11}}{65}$.

The equation of a line with slope $- \frac{2 \sqrt{11}}{65}$ passing through $\left(14 , 28 \sqrt{11}\right)$ is:

$y - 28 \sqrt{11} = - \frac{2 \sqrt{11}}{65} \left(x - 14\right)$

Graphed are the function and the normal line:

graph{(y-28sqrt11+(2sqrt11)/65(x-14))(y-xsqrt(3x+2))=0 [-72.4, 227.8, -13.4, 136.7]}