Worked examples

07 Worked examples

Determining Tangent and Normal Lines to an Algebraic CurveProblem Statement & Given Data: Find the equations of the tangent line and the normal line to the curve y = (2x - 1) / (x + 2) at the point where x = 1.Step-by-Step Mathematical & Chemical Solution: [1] Step 1: Determine the y-coordinate of the point of contact: Substitute x = 1 into the curve equation: [2] y1 = (2(1) - 1) / (1 + 2) = (2 - 1) / 3 = 1/3. [3] Point of tangency P is (1, 1/3). [4] Step 2: Differentiate y using the Quotient Rule: dy/dx = [(denominator)(derivative of numerator) - (numerator)(derivative of denominator)] / (denominator)². [5] dy/dx = [(x + 2)(2) - (2x - 1)(1)] / (x + 2)² = [2x + 4 - 2x + 1] / (x + 2)² = 5 / (x + 2)². [6] Step 3: Evaluate the slope m at x = 1: m = 5 / (1 + 2)² = 5 / 3² = 5/9. [7] Step 4: Equation of the Tangent Line: y - y1 = m (x - x1) ==> y - 1/3 = 5/9 (x - 1). [8] Multiply both sides by 9: 9y - 3 = 5(x - 1) ==> 9y - 3 = 5x - 5. [9] Rearrange in standard form: 5x - 9y - 2 = 0. [10] Step 5: Equation of the Normal Line: Slope of normal m_norm = -1 / (5/9) = -9/5. [11] y - 1/3 = -9/5 (x - 1) ==> Multiply by 15: 15y - 5 = -27(x - 1) ==> 15y - 5 = -27x + 27. [12] Rearrange in standard form: 27x + 15y - 32 = 0.💡 High-Yield Exam Shortcut & Pitfall Alert: Exam Tip: Always double-check your tangent and normal slopes by verifying that m_tangent × m_normal = (5/9) × (-9/5) = -1!