Lecture Notes: Tangent and Normal Lines

Key Concepts

Tangent Line

• Definition: The tangent to a curve at a specific point is the instantaneous gradient at that spot.
• Characteristics:
  • It's a straight line that touches the curve at exactly one point.
  • Represents the slope of the curve at that point.

Normal Line

• Definition: The line perpendicular to the tangent at the point of tangency.
• Characteristics:
  • It has a gradient that is the negative reciprocal of the tangent's gradient.
  • Forms a 90-degree angle with the tangent line.

Gradient Formula

• Gradient Formula: Rise over Run
• Using Trigonometry:
  • ( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{rise}}{\text{run}} )
  • ( \tan(\theta) ) gives the gradient.
  • Example: Given an angle, ( \theta ), ( \tan(\theta) = 2 ) implies ( \theta = \tan^{-1}(2) ).

Method to Find Tangent and Normal

Finding the Gradient of a Tangent

• Formula: The gradient of the tangent at a point ( (x_1, y_1) ) can be found by differentiating the function and substituting ( x = x_1 ).

Finding the Gradient of a Normal

• Formula: If the gradient of the tangent ( m_t ) is known, the gradient of the normal ( m_n ) is given by:
  • ( m_n = -\frac{1}{m_t} )

Equation of a Line Using Point-Gradient Formula

• Formula: ( y - y_1 = m(x - x_1) )
  • ( m ): Gradient
  • ( (x_1, y_1) ): Point through which the line passes

Example Problems

Example 1: Finding the Equation of the Normal

• Task: Find the equation of the normal to ( y = x^3 + 3x^2 - 12x + 1 ) at ( x = 2 ).
• Steps:
  1. Differentiate the function to get ( f'(x) = 3x^2 + 6x - 12 ).
  2. Substitute ( x = 2 ) to find the gradient of the tangent: ( f'(2) = 24 ).
  3. Gradient of normal ( m_n = -\frac{1}{24} ).
  4. Find ( y_1 ) by substituting ( x = 2 ) into the original function.
  5. Use the point-gradient formula to express the equation of the normal.

Example 2: Finding the Point of Contact of the Normal

• Task: Given ( f(x) = 3x^2 - 4x ) and the gradient of the normal is (-\frac{1}{4}), find the point of contact.
• Steps:
  1. Differentiate ( f(x) ) to get ( f'(x) = 6x - 4 ).
  2. Since ( m_t = 4 ), solve for ( x ) where ( 6x - 4 = 4 ).
  3. Calculate ( x ) and use it to find ( y ) by substituting back into ( f(x) ).
  4. Result is the point of contact: ( (\frac{4}{3}, 0) ).

Final Thoughts

• Ensure understanding of perpendicular relationships between tangents and normals.
• Practice differentiating functions and applying these concepts to various problems.