Tangent and Normals

Differentiation has several applications in mathematics as well as in our daily lives. Some of them include:

• Rate of Change of a Quantity
• Increasing and Decreasing Functions
• Tangent and Normal to a Curve
• Minimum and Maximum Values
• Newton’s Method
• Linear Approximations

In this article, you will learn about one of the most commonly used applications of differentiation called tangents and normals.

As we know, tangent is a line that touches the curve at exactly one point, whereas normal is the line perpendicular to the tangent of that curve.

Let us derive the equation of the tangent line and the normal line to a curve at a given point using differentiation.

Equation of Tangent to a Curve

The point slope form of a equation of a line is:

y – y₁ = m(x – x₁)

Here,

m = Slope of the line

(x₁, y₁) = Point on the line

This is the equation of a straight line passing through a given point (x₁, y₁) having finite slope m.

Suppose y = f(x) be the equation of a curve and the slope of the tangent to the curve y = f(x) at the point (x₁, y₁) can be represented as

Thus, the equation of the tangent to the curve y = f(x) at (x₁, y₁) is:

y – y₁ = f'(x₁)(x – x₁)

Or

The equation of tangent to the curve y = f(x) at x = a is given by:

y – f(a) – f’(a)(x – a)

Here, f’(a) is the slope or gradient of the tangent to the curve y = f(x).

Equation of Normal to a Curve

We know that when two lines with slopes m₁ and m₂ are perpendicular to each other, then m₁m₂ = -1

That means, the product of slope of perpendicular lines is -1.

Normal is perpendicular to the tangent as shown in the above figure. So, the the slope of the normal to the curve y = f(x) at (x₁, y₁) is given by:

-1/Slope of the tangent

= -1/f'(x₁)

Also, consider f'(x₁) ≠ 0.

Therefore, the equation of the normal to the curve y = f(x) at (x₁, y₁) is given by:

y – y₁ = [-1/f'(x₁)] (x – x₁)

(y – y₁)f'(x₁) = -(x – x₁)

(y – y₁)f'(x₁) + (x – x₁) = 0

Tangents and Normals Examples

Example 1:

Find the equation of tangent and normal to the curve f(x) = x³ – 12x² + 6x + 1 at x = 2.

Solution:

Let the given equation be:

y = f(x) = x³ – 12x² + 6x + 1

Also, x = 2, i.e. a = 2

Substituting x = 2 in f(x),

f(2) = (2)³ – 12(x)² + 6(2) + 1

= 8 – 12(4) + 12 + 1

= 8 – 48 + 13

= -27

The point on the curve at which the tangent line passes through is (2, -27).

Slope of the tangent = f’(a) = f’(2)

f’(a) = f’(2)

= 3(2)² – 24(2) + 6

= 3(4) – 48 + 6

= 12 – 48 + 6

= 30

Thus, the equation of tangent to the curve at x = 2 is:

y – f(2) = f’(2)(x – 2)

y – (-27) = 30(x – 2)

y + 27 = 30x – 60

30x – 60 – y – 27 = 0

30x – y – 87 = 0

Since tangents and normals are perpendicular to each other, the slope of the normal = -1/slope of the tangent

Slope of normal = -1/f’(2) = -1/30

Equation of the normal is:

y – f(a) = [-1/f’(a)] (x – a)

y – (-27) = (-1/30)(x – 2)

30(y + 27) = -(x – 2)

30y + 810 = -x + 2

x + 30y + 810 – 2 = 0

x + 30y + 808 = 0

This is the required equation of normal to the curve f(x) at x = 2.

Example 2:

Find the equation of all lines having slope 2 and being tangent to the curve

Solution:

Given,

Slope of the tangent to the given curve at any point (x, y) is given by:

As per the give, slope is 2.

i.e. 2/(x – 3)² = 2

⇒ (x – 3)² = 2/2

⇒ (x – 3)² = 1

⇒ x – 3 = ± 1

When x – 3 = 1, x = 4

When x – 3 = -1, x = 2

Substituting x = 2 in the given equation,

y + [2/(2 – 3)] = 0

y – 2 = 0

y = 2

Substituting x = 4 in the given equation,

y + [2/(4 – 3)] = 0

y + 2 = 0

y = -2

Thus, there are two tangents to the given curve with slope 2 and that will be passing through the points (2, 2) and (4, – 2).

Now, the equation of a tangent to the curve with slope 2 and passing through (2, 2) is:

y – 2 = 2(x – 2)

y – 2 = 2x – 4

2x – 4 – y + 2 = 0

2x – y – 2 = 0

Now, the equation of a tangent to the curve with slope 2 and passing through (4, -2) is:

y – (-2) = 2(x – 4)

y + 2 = 2x – 8

2x – 8 – y – 2 = 0

2x – y – 10 = 0

Hence, the equations

Some of the important points about tangents and normals are listed below:

• If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then:

Slope of the tangent = dy/dx = tan θ

• If slope of the tangent line is zero, then tan θ = 0 and thus, θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x₁, y₁) is given by y = y₁.
• If θ →π/2, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at (x₁, y₁) is given by x = x₁.