Applications of Derivatives: Tangent Lines, Normal Lines, and Curve Analysis

Slopes, Tangent Lines, and Normal Lines

• Derivative as Slope: For a curve $y = f(x)$, the derivative $\frac{dy}{dx}$ or $y'$ represents the slope $m$ of the tangent line at any point $P(x, y)$.
• Angle of Inclination ($\tau$): The smallest angle a line makes with the x-axis, where the slope is given by $m = \tan(\tau)$.
  • Vertical Tangent ($\tau = 90^\times$ or $90^\text{ degrees}$): $\tan(90^\text{ degrees})$ is undefined, indicating a vertical slope or asymptote.
  • Horizontal Tangent ($\tau = 0^\text{ degrees}$): $\tan(0^\text{ degrees}) = 0$, indicating a slope of zero.
  • Tangent at $45^\text{ degrees}$: $\tan(45^\text{ degrees}) = 1$, indicating a slope of $1$
• Point-Slope Form: The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Normal Line: Perpendicular to the tangent line at the point of tangency, with a slope given by the negative reciprocal $m_N = -\frac{1}{m_T}$.

Curve Analysis of $y = \frac{x^3}{3} - \frac{x^2}{2} + 3$

• Derivative Equation: Slope formula is $y' = x^2 - x$
• Angle of Inclination at $x = 2$:
  • Slope $m = 2^2 - 2 = 2$
  • Angle $\tau = \text{arctan}(2) = 63.43^\text{ degrees}$
• Points with Zero Slope ($y' = 0$):
  • Solving $x^2 - x = 0$ gives $x = 0$ and $x = 1$
  • Corresponding points are $(0, 3)$ and \begin{pmatrix} 1 & \frac{17}{6} \begin{pmatrix}
• Points with Angle of Inclination $\tau = 45^\text{ degrees}$:
  • Slope $m = \tan(45^\text{ degrees}) = 1$
  • Solving $x^2 - x - 1 = 0$ yields $x = 1.618$ and $x = -0.618$
  • Corresponding points are $(1.618, 3.103)$ and $(-0.618, 2.703)$

Tangent and Normal Lines to Parabola $y = x^2$

• Target Point: $(1, 1)$
• Derivative and Slope: $y' = 2x$, giving tangent slope $m_T = 2(1) = 2$
• Tangent Line Equation: $2x - y - 1 = 0$
• Normal Line Slope: $m_N = -\frac{1}{2}$
• Normal Line Equation: $x + 2y - 3 = 0$

Tangent and Normal Lines to Ellipse $4x^2 + 9y^2 = 25$

• Target Point: $(-2, -1)$
• Implicit Differentiation: $8x + 18y \times y' = 0 \rightarrow y' = -\frac{4x}{9y}$
• Tangent Slope at $(-2, -1)$: $m_T = -\frac{4(-2)}{9(-1)} = -\frac{8}{9}$
• Tangent Line Equation: $8x + 9y + 25 = 0$
• Normal Line Slope: $m_N = \frac{9}{8}$
• Normal Line Equation: $9x - 8y + 8 = 0$

Questions & Discussion

• Angle Calculation: For a slope of $2$, the angle $\text{arctan}(2)$ equals $63.43^\text{ degrees}$ or $1.11\text{ radians}$.
• Abbreviation: The notation PTS in solutions refers to points.
• Calculator Inputs: For quadratic solver equations $a x^2 + b x + c = 0$, inputs $a$, $b$, and $c$ represent numerical coefficients of $x^2$, $x$, and the constant term.