Worked examples

Worked Example 1: Tangent and Normal Equations to a Rational Function

Problem Statement: Find the equations of both the tangent line and the normal line to the curve y = (3x - 1) / (x + 1) at the point where x = 1.

Detailed Solution & Derivation: Step 1: Find the y-coordinate at x = 1: y1 = (3(1) - 1) / (1 + 1) = 2 / 2 = 1. Tangency point is P(1, 1). Step 2: Differentiate y using the Quotient Rule: dy/dx = [(x + 1)·d/dx(3x - 1) - (3x - 1)·d/dx(x + 1)] / (x + 1)² = [(x + 1)(3) - (3x - 1)(1)] / (x + 1)² = [3x + 3 - 3x + 1] / (x + 1)² = 4 / (x + 1)². Step 3: Calculate tangent slope m at x = 1: m = 4 / (1 + 1)² = 4 / 4 = 1. Step 4: Equation of Tangent Line: y - y1 = m(x - x1) -> y - 1 = 1(x - 1) -> y = x (or x - y = 0). Step 5: Normal slope m_normal = -1 / m = -1 / 1 = -1. Step 6: Equation of Normal Line: y - 1 = -1(x - 1) -> y - 1 = -x + 1 -> x + y - 2 = 0.

Worked Example 2: Kinematic Motion Analysis via Derivatives

Problem Statement: A projectile moves along a straight line such that its distance s (in meters) from a fixed point after t seconds is given by: s(t) = t³ - 6t² + 9t + 4. Determine: (1) The velocity at any time t; (2) The time instances when the particle is at rest; (3) The total distance covered during the first 4 seconds.

Detailed Solution & Derivation: Step 1: Velocity v(t) = s'(t) = 3t² - 12t + 9. Step 2: Particle is at rest when v(t) = 0: 3(t² - 4t + 3) = 0 -> 3(t - 1)(t - 3) = 0 -> t = 1 second and t = 3 seconds. Step 3: Because the particle reverses direction at t = 1 and t = 3, total distance must be computed in three sub-intervals: [0, 1], [1, 3], and [3, 4]. s(0) = 4 m; s(1) = 1 - 6 + 9 + 4 = 8 m; s(3) = 27 - 54 + 27 + 4 = 4 m; s(4) = 64 - 96 + 36 + 4 = 8 m. Distance d1 = |s(1) - s(0)| = |8 - 4| = 4 m. Distance d2 = |s(3) - s(1)| = |4 - 8| = 4 m. Distance d3 = |s(4) - s(3)| = |8 - 4| = 4 m. Total distance covered = 4 + 4 + 4 = 12 meters.