How do you find the equation of the tangent and normal line to the curve y=x^3y=x3 at x=1?
1 Answer
The equation of the tangent
Then the equation of the normal is
Explanation:
Given -
y=x^3y=x3
At
(1,1)(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
The slope of the curve at any given point is, its first derivative.
dy/dx=3x^2dydx=3x2
Slope of the curve at
m=3xx1^1=3m=3×11=3
Then the slope of the tangent
The equation of the tangent
y-y_1=m_1(x-x_1)y−y1=m1(x−x1)
y-1=3(x-1)y−1=3(x−1)
y-1=3x-3y−1=3x−3
y=3x-3+1y=3x−3+1
y=3x-2y=3x−2
If the two lines cut vertically then
m_2=(-1)/(m_1)=(-1)/3=-1/3m2=−1m1=−13=−13
Then the equation of the normal is -
y-y_1=m_1(x-x_1)y−y1=m1(x−x1)
y-1=-1/3(x-1)y−1=−13(x−1)
y-1=-1/3x+1/3y−1=−13x+13
y=-1/3x+1/3+1y=−13x+13+1
y=-1/3x+4/3y=−13x+43
