Find the equation of the tangent and the normal to the given curve at

1.	Find the equation of the tangent and the normal to the given curve at the indicated point: y = x3 at P(1, 1).
Answer» Consider y = x³ as the equation of curve By differentiating both sides w.r.t. x dy/dx = 3x² By substituting the values (dy/dx)_{(1, 1)} = 3(1)² = 3 Here the required equation of tangent at point (1, 1) y – y₁ = m(x – x₁) We can write it as (y – y₁)/(x – x₁) = m By substituting the values (y – 1)/(x – 1) = 3 By cross multiplication y – 1 = 3x – 3 So we get 3x – y – 2 = 0 Here the required equation of normal at point (1, 1) y – y₁ = – 1/m(x – x₁) We can write it as (y – y₁)/(x – x₁) = – 1/m By substituting the values (y – 1)/(x – 1) = – 1/3 By cross multiplication 3y – 3 = – x + 1 So we get x + 3y – 3 – 1 = 0 x + 3y – 4 = 0

Find the equation of the tangent and the normal to the given curve at the indicated point: y = x3 at P(1, 1).

Answer»

Consider y = x³ as the equation of curve

By differentiating both sides w.r.t. x

dy/dx = 3x²

By substituting the values

(dy/dx)_{(1, 1)} = 3(1)² = 3

Here the required equation of tangent at point (1, 1)

y – y₁ = m(x – x₁)

We can write it as

(y – y₁)/(x – x₁) = m

By substituting the values

(y – 1)/(x – 1) = 3

By cross multiplication

y – 1 = 3x – 3

So we get

3x – y – 2 = 0

Here the required equation of normal at point (1, 1)

y – y₁ = – 1/m(x – x₁)

We can write it as

(y – y₁)/(x – x₁) = – 1/m

By substituting the values

(y – 1)/(x – 1) = – 1/3

By cross multiplication

3y – 3 = – x + 1

So we get

x + 3y – 3 – 1 = 0

x + 3y – 4 = 0