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

Jun 5, 2016

color(blue)(y=-sqrt(5)/3 x + (4sqrt(5))/3color(white)(2/2)

#### Explanation:

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

Using calculus

Let $u = 3 x + 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 3$....(1)

By substitution

$y = {u}^{\frac{1}{2}} \implies \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} {u}^{- \frac{1}{2}}$ ....(2)

By comparing equations (1) and (2)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\cancel{\mathrm{du}}} \times \frac{\cancel{\mathrm{du}}}{\mathrm{dx}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times 3$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \sqrt{3 x + 2}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{3 x + 2}}{\sqrt{3 x + 2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \sqrt{3 x + 2}}{3 x + 2}$
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$\textcolor{b l u e}{\text{Determine gradient of normal}}$
Condition for $x = 1$

Gradient of given curve $\to \frac{3 \sqrt{5}}{5}$

$\text{So gradient of normal is } \to - \frac{5}{3 \sqrt{5}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{5}}{\sqrt{5}}$ giving:

$\text{ } - \frac{\cancel{5} \sqrt{5}}{3 \times \cancel{5}}$

color(brown)("gradient "->-(sqrt(5))/3
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$\textcolor{b l u e}{\text{Determine point through which the normal passes}}$

For $x = 1$

$y = \sqrt{3 x + 2} \text{ "vec("becomes")" } y = \sqrt{5}$

$\implies \text{Point } \to \left(x , y\right) \to \left(1 , \sqrt{5}\right)$

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$\textcolor{b l u e}{\text{Determine equation of normal}}$

$y = - \frac{1}{m} x + c \text{ "vec("becomes")" } \sqrt{5} = \left(- \frac{\sqrt{5}}{3} \times 1\right) + c$

add$\text{ } \frac{\sqrt{5}}{3}$ to both sides

$\sqrt{5} + \frac{\sqrt{5}}{3} = 0 + c$

$\implies c = \frac{4 \sqrt{5}}{3}$

color(blue)(y=-sqrt(5)/3 x + (4sqrt(5))/3color(white)(2/2)