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

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Answer 1 · 2017-12-12T02:31:31

The equation of the tangent #y=3x-2#
Then the equation of the normal is #y=-1/3x+4/3#

Given -

#y=x^3#

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

#(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 #x=1#

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

#dy/dx=3x^2#

Slope of the curve at #x=1#

#m=3xx1^1=3#

Then the slope of the tangent #m_1=3#

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_1 xx m_2=-1#

#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#

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