Question

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

Answer 1

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