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 = x4 – 6x3 + 13x2 – 10x + 5 at the point where x = 1
Answer» It is given that y = x⁴ – 6x³ + 13x² – 10x + 5 By substituting the value of x = 1 y = (1)⁴ – 6(1)³ + 13(1)² – 10(1) + 5 On further calculation y = 1 – 6 + 13 – 10 + 5 = 3 Point of contact is (1, 3) Consider y = x⁴ – 6x³ + 13x² – 10x + 5 as the equation of curve By differentiating w.r.t. x dy/dx = 4x³ – 18x² + 26x – 10 So we get (dy/dx)_{(1, 3)} = 4(1)³ – 18(1)² + 26(1) – 10 = 2 Here the required equation of tangent is y – y₁ = m(x – x₁) By substituting the values y – 3 = 2x – 2 On further calculation 2x – y – 2 + 3 = 0 So we get 2x – y + 1 = 0 Here the required equation of normal i y – y₁ = -1/m(x – x₁) By substituting the values y – 3 = -1/2(x – 1) On further calculation 2y – 6 = – x + 1 So we get 2y – 6 + x – 1 = 0 x + 2y – 7 = 0

Find the equation of the tangent and the normal to the given curve at the indicated point: y = x4 – 6x3 + 13x2 – 10x + 5 at the point where x = 1

Answer»

It is given that

y = x⁴ – 6x³ + 13x² – 10x + 5

By substituting the value of x = 1

y = (1)⁴ – 6(1)³ + 13(1)² – 10(1) + 5

On further calculation

y = 1 – 6 + 13 – 10 + 5 = 3

Point of contact is (1, 3)

Consider y = x⁴ – 6x³ + 13x² – 10x + 5 as the equation of curve

By differentiating w.r.t. x

dy/dx = 4x³ – 18x² + 26x – 10

So we get

(dy/dx)_{(1, 3)} = 4(1)³ – 18(1)² + 26(1) – 10 = 2

Here the required equation of tangent is

y – y₁ = m(x – x₁)

By substituting the values

y – 3 = 2x – 2

On further calculation

2x – y – 2 + 3 = 0

So we get

2x – y + 1 = 0

Here the required equation of normal i

y – y₁ = -1/m(x – x₁)

By substituting the values

y – 3 = -1/2(x – 1)

On further calculation

2y – 6 = – x + 1

So we get

2y – 6 + x – 1 = 0

x + 2y – 7 = 0