Home / IB DP Maths 2026, 2027 & 2028 / Application and Interpretation HL / IB Mathematics AI SL Tangents and normal MAI Study Notes

IB Mathematics AI SL Tangents and normal MAI Study Notes - New Syllabus

IB Mathematics AI SL Concepts of population, sample MAI Study Notes

LEARNING OBJECTIVE

  • Tangents and normals at a given point, and their equations

Key Concepts: 

  • Tangent
  • Normal

MAI HL and SL Notes – All topics

 TANGENTS: DEFINITION AND GEOMETRIC INTERPRETATION

Definition: 

  • A tangent to a curve at a given point is a straight line that touches the curve at that point and has the same gradient (slope) as the curve.

Geometric Interpretation:

It represents the instantaneous direction of the curve at that point. It does not intersect the curve nearby the point of contact (locally).

 Tangent Line Equation

For a function \( y = f(x) \) and a point \( P(a, f(a)) \), the tangent line has:

  • Slope: \( f'(a) \), the derivative at \( x = a \)

  • Equation: \( y – f(a) = f'(a)(x – a) \)

 Analytical Approach

  1. Differentiate the function to find \( f'(x) \)
  2. Evaluate the derivative at \( x = a \) to get the slope \( m \)
  3. Use point-slope form: \( y – f(a) = m(x – a) \)

 Technology Approach (GDC)

  • Graph the function on a calculator
  • Use the ‘tangent’ or ‘dy/dx’ function to find slope at a point
  • Trace to point and graph tangent directly if supported

Example 1: Tangent Line

Find the equation of the tangent to the curve \( f(x) = 2x^3 – 3x^2 + x \) at the point where \( x = 1 \).

▶️Answer/Explanation
  • Given: \( f(x) = 2x^3 – 3x^2 + x \)
  • Differentiate: \( f'(x) = 6x^2 – 6x + 1 \)
  • Evaluate at \( x = 1 \): \( f'(1) = 6(1)^2 – 6(1) + 1 = 6 – 6 + 1 = 1 \)
  • Point on the curve: \( (1, f(1)) = (1, 2(1)^3 – 3(1)^2 + 1) = (1, 2 – 3 + 1) = (1, 0) \)
  • Use point-slope form: \( y – 0 = 1(x – 1) \)
  • Final equation: \( y = x – 1 \)

Example 2: Tangent (Analytical)

Find the equation of the tangent to the curve \( f(x) = \sqrt{x} \) at \( x = 4 \).

▶️Answer/Explanation
  • \( f(x) = x^{1/2} \Rightarrow f'(x) = \frac{1}{2\sqrt{x}} \)
  • \( f'(4) = \frac{1}{4} \)
  • Point: \( (4, \sqrt{4}) = (4, 2) \)
  • Tangent equation: \( y – 2 = \frac{1}{4}(x – 4) \)
  • Final answer: \( y = \frac{1}{4}x + 1 \)

 NORMAL: DEFINITION AND GEOMETRIC INTERPRETATION

Definition:

  •  A normal to a curve at a point is a line perpendicular to the tangent at that point and also passes through the same point on the curve.

  •     $\text{(Normal → At Point Q , Dashed Line  )}$

Geometric Interpretation:

  •  It is the instantaneous direction perpendicular to the curve at the given point.

 Normal Line Equation

For a curve \( y = f(x) \) at point \( (a, f(a)) \):

  • Slope of the normal: \( -\frac{1}{f'(a)} \), the negative reciprocal of the tangent’s slope

  • Equation: \( y – f(a) = -\frac{1}{f'(a)}(x – a) \)

 Analytical Approach

  1. Differentiate to find \( f'(x) \)

  2. Evaluate \( f'(a) \) → find normal slope as \( -\frac{1}{f'(a)} \)

  3. Use point-slope formula to write normal line equation

 Technology Approach (GDC)

  • Use ‘tangent’ to find slope, then compute negative reciprocal
  • Use graphing software or calculator to plot both tangent and normal at a point

Example 1: Normal Line

Find the equation of the normal to the curve \( f(x) = x^2 + 1 \) at the point where \( x = 2 \).

▶️Answer/Explanation
  • \( f(x) = x^2 + 1 \)
  • \( f'(x) = 2x \Rightarrow f'(2) = 4 \) → Tangent slope
  • Normal slope = \( -\frac{1}{4} \)
  • Point on curve: \( (2, f(2)) = (2, 5) \)
  • Normal equation: \( y – 5 = -\frac{1}{4}(x – 2) \)
  • Final answer: \( y = -\frac{1}{4}x + \frac{9}{2} \)

Example 2: Normal Line (Analytical Approach)

Find the equation of the normal to the curve \( f(x) = x^2 – 4x + 5 \) at the point where \( x = 3 \).

▶️Answer/Explanation
  • Given: \( f(x) = x^2 – 4x + 5 \)
  • Differentiate: \( f'(x) = 2x – 4 \)
  • Evaluate at \( x = 3 \): \( f'(3) = 2(3) – 4 = 2 \)
  • Normal slope = \( -\frac{1}{2} \)
  • Point on the curve: \( (3, f(3)) = (3, 9 – 12 + 5) = (3, 2) \)
  • Normal equation: \( y – 2 = -\frac{1}{2}(x – 3) \)
  • Final equation: \( y = -\frac{1}{2}x + \frac{7}{2} \)
Scroll to Top