Normal Using Differentiation





In this page normal using differentiation we are going to see the how to find equation of normal using the concept differentiation.

Procedure of finding equation of normal using differentiation :

Step 1 : Differentiate the given equation of the curve once. If we differentiate the given equation we will get slope of the curve that is slope of tangent drawn to the curve.

Step 2 : We have to apply the given points in the general slope to get slope of the particular tangent at the particular point.

Step 3 : From slope of tangent we have to find the slope of normal (-1/m).

Step 4 : Now we have to apply the point and the slope in the formula

(y - y₁) = (-1/m) (x - x₁)

Now we are going to see some example problems to understand this concept.

Example 1:

Find the equation of the tangent to the curve y = x³ at the point (1,1)

Solution:

y = x³

Step 1 : Differentiate the equation of the curve once to find the slope of the curve that is slope of the tangent drawn to the curve.

      dy/dx = 3 x²

slope of the curve at the point (1 , 1)

  dy/dx = 3 (1)²

slope of the curve that is lope of tangent at the point (1,1) dy/dx = 3

slope of normal = -1/3

Equation of the normal :

(y - y₁) = (-1/m) (x - x₁)

(y - 1) = (-1/3) (x - 1)

 3 (y - 1) = -1 (x - 1)

3 y - 3 = - x + 1

x + 3 y - 3 - 1 = 0

 x + 3 y - 4 = 0


Example 2:

Find the equation of normal to the curve y = x² - x - 2 at the point (1,-2)

Solution:

y = x² - x - 2

Step 1 : Differentiate the equation of the curve once to find the slope of the curve that is slope of the tangent drawn to the curve.

      dy/dx = 2 x - 1

slope of the curve that is slope of the tangent at the point (1 , -2)

  dy/dx = 2 (1) - 1

           = 2 - 1

           = 1

slope of the tangent (dy/dx) = 1

slope of normal = -1/1

                     = -1

Equation of the normal:

(y - y₁) = (-1/m) (x - x₁)

(y - (-2)) = -1 (x - 1)

 y + 2 = - x + 1

 x + y + 2 - 1 = 0

 x + y + 1 = 0

normal using differentiation