Tangents and Normals to Curves

Introduction

• Visual intuition is essential for comprehending tangents and normals.
• Tangents and normals offer insights into instantaneous rates of change and perpendicular relationships in calculus.
• These concepts are applied in fields such as physics and engineering to describe motion and direction.

Understanding Tangents

Definitions and Properties

• Tangent Line:
  • Definition: A line that just touches a curve at a particular point, matching its slope without intersecting.
  • Key Property: Represents the immediate direction or slope of the curve at that point.

Calculating Tangent Lines

• Formula: $y - y_1 = m(x - x_1)$, where $m$ represents the gradient at the point $(x_1, y_1)$.
• Steps:
  • Differentiate: Determine the derivative $f'(x)$.
  • Evaluate: Insert $x = x_1$ to identify slope $m$.
  • Substitute the values into the formula to derive the equation of the tangent.

Relationship to Secants

• Tangent: Touches the curve at a single point.
• Secant: Intersects the curve at two or more points.

Tangent	Secant
Touches the curve at a single point	Intersects the curve at two or more points
Displays instantaneous slope	Shows average slope between points

Introduction to Normals

Definitions and Properties

• Normal Line:
  • Definition: A line that is perpendicular to a tangent at the point of tangency.
  • Key Property: Intersects the tangent at the contact point on the curve.

Calculating Normal Lines

• Formula: $y - y_1 = -\frac{1}{m}(x - x_1)$, where $m$ is the slope of the tangent.

Finding Gradients

Differentiation Techniques

• Differentiation: A method to determine gradients, crucial in calculus and practical applications like physics for motion and economics for cost functions.

Understanding Derivatives as Slope Indicators

• Derivatives indicate the slope of tangent lines.
• Consider the function $f(x) = x^2$:
  • $f'(x) = 2x$ reveals the slope at various points.
  • For $x = 1$, $f'(1) = 2$.

Examples and Solutions

Example: Curve $y = x^2$ at $(1, 1)$

• Tangent:
  • Differentiate: $f'(x) = 2x$.
  • At $x = 1$, slope = 2.
  • Equation of Tangent: $y - 1 = 2(x - 1)$ or simplified: $y = 2x - 1$.
• Normal:
  • Slope = $-\frac{1}{2}$.
  • Equation of Normal: $y - 1 = -\frac{1}{2}(x - 1)$ or simplified: $y = -\frac{1}{2}x + \frac{3}{2}$.

Practice Questions with Solutions

1. Determine the tangent and normal lines for $y = x^3 + 2x$ at point $(2, 12)$.

   • Solution:
     • $f'(x) = 3x^2 + 2$
     • At $x = 2$: $f'(2) = 3(2)^2 + 2 = 12 + 2 = 14$
     • Tangent: $y - 12 = 14(x - 2)$ or $y = 14x - 16$
     • Normal: $y - 12 = -\frac{1}{14}(x - 2)$ or $y = -\frac{1}{14}x + \frac{168+2}{14} = -\frac{1}{14}x + \frac{170}{14}$

2. Determine the tangent and normal lines for $y = \sin x$ at point $(\frac{\pi}{4}, \frac{\sqrt{2}}{2})$.

   • Solution:
     • $f'(x) = \cos x$
     • At $x = \frac{\pi}{4}$: $f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
     • Tangent: $y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(x - \frac{\pi}{4})$ or $y = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}(1-\frac{\pi}{4})$
     • Normal: $y - \frac{\sqrt{2}}{2} = -\frac{2}{\sqrt{2}}(x - \frac{\pi}{4})$ or $y = -\sqrt{2}x + \frac{\sqrt{2}}{2} + \frac{\pi\sqrt{2}}{4}$

3. Determine the tangent and normal lines for $y = e^x - x$ at point $(0, 1)$.

   • Solution:
     • $f'(x) = e^x - 1$
     • At $x = 0$: $f'(0) = e^0 - 1 = 1 - 1 = 0$
     • Tangent: $y - 1 = 0(x - 0)$ or $y = 1$ (horizontal line)
     • Normal: Since the tangent has slope 0, the normal has undefined slope (vertical line): $x = 0$

Perpendicularity and Common Solutions

• Perpendicular Slopes:

  • If the gradient of the tangent is 'm', the gradient of the normal is '-1/m'.

• Solution Strategies:

  • Correct application of the reciprocal and negative sign is crucial for normals.

Conclusion

• Understanding tangents and normals enhances proficiency in mathematical theory and practical application.
• Regular practice using both traditional and digital tools is important for effective exam preparation.

Quick Quiz: "If the tangent slope is 3, what is the normal's slope?" Answer: $-\frac{1}{3}$ (since the slopes of perpendicular lines multiply to give -1)