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

##### 1 Answer

Dec 12, 2017

The equation of the tangent

Then the equation of the normal is

#### Explanation:

Given -

#y=x^3#

At

#(1,1)#

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

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

#dy/dx=3x^2#

Slope of the curve at

#m=3xx1^1=3#

Then the slope of the tangent

The equation of the tangent

#y-y_1=m_1(x-x_1)#

#y-1=3(x-1)#

#y-1=3x-3#

#y=3x-3+1#

#y=3x-2#

If the two lines cut vertically then

#m_2=(-1)/(m_1)=(-1)/3=-1/3#

Then the equation of the normal is -

#y-y_1=m_1(x-x_1)#

#y-1=-1/3(x-1)#

#y-1=-1/3x+1/3#

#y=-1/3x+1/3+1#

#y=-1/3x+4/3#