# How do you find the equation of the tangent and normal line to the curve y=x^3 at x=1?

Dec 12, 2017

The equation of the tangent $y = 3 x - 2$
Then the equation of the normal is $y = - \frac{1}{3} x + \frac{4}{3}$

#### Explanation:

Given -

$y = {x}^{3}$

At x=1; y= 1^3=1

$\left(1 , 1\right)$
It is at this point there is a tangent and a normal.

The slope of the tangent is equal to the slope of the given curve at $x = 1$

The slope of the curve at any given point is, its first derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2}$

Slope of the curve at $x = 1$

$m = 3 \times {1}^{1} = 3$

Then the slope of the tangent ${m}_{1} = 3$

The equation of the tangent

$y - {y}_{1} = {m}_{1} \left(x - {x}_{1}\right)$
$y - 1 = 3 \left(x - 1\right)$
$y - 1 = 3 x - 3$
$y = 3 x - 3 + 1$
$y = 3 x - 2$

If the two lines cut vertically then ${m}_{1} \times {m}_{2} = - 1$

${m}_{2} = \frac{- 1}{{m}_{1}} = \frac{- 1}{3} = - \frac{1}{3}$

Then the equation of the normal is -

$y - {y}_{1} = {m}_{1} \left(x - {x}_{1}\right)$

$y - 1 = - \frac{1}{3} \left(x - 1\right)$
$y - 1 = - \frac{1}{3} x + \frac{1}{3}$
$y = - \frac{1}{3} x + \frac{1}{3} + 1$
$y = - \frac{1}{3} x + \frac{4}{3}$