Find the equations of the tangent and normal to the given curves at the indicated points:
y = x3 at (1, 1)
It is given that equation of curve is y = x3
On differentiating with respect to x, we get
= 3x2
Therefore, the slope of the tangent at (1, 1) is 3.
Then, the equation of the tangent is
y – 1 = 3(x – 1)
⇒ y = 3x - 2
Then, slope of normal at (1,1)
=
Now, equation of the normal at (1,1)
y – 1 = (x – 1)
⇒ 3y -3 =- x + 1
⇒ x + 3y - 4 = 0