Applications of Derivatives
Tangent and minimum ordinate
Grade 12
Question:
<p>A curve passes through \((2, 0)\) and the slope of tangent at any point \((x, y)\) is \(x^2 - 2x\) \(\forall\, x \in R\). The point of minimum ordinate on the curve where \(x > 0\) is \((a, b)\), then find the value of \((a + 6b)\).</p>
Step-by-Step Solution
Key Concept: The slope function dy/dx = x² - 2x defines the curve through integration. The minimum ordinate (minimum y-value) for x > 0 occurs where dy/dx = 0, giving x = 2 as the critical point.
<p><strong>Step 1: Find the curve equation</strong></p><p>Given: dy/dx = x² - 2x and curve passes through (2, 0)</p><p>Integrate: y = ∫(x² - 2x)dx = x³/3 - x² + C</p><p><strong>Step 2: Use initial condition to find C</strong></p><p>At point (2, 0): 0 = (2)³/3 - (2)² + C</p><p>0 = 8/3 - 4 + C</p><p>0 = 8/3 - 12/3 + C</p><p>C = 4/3</p><p>So: y = x³/3 - x² + 4/3</p><p><strong>Step 3: Find critical points for x > 0</strong></p><p>dy/dx = 0 ⟹ x² - 2x = 0 ⟹ x(x - 2) = 0</p><p>For x > 0: x = 2 is the only critical point</p><p><strong>Step 4: Verify it's a minimum</strong></p><p>d²y/dx² = 2x - 2</p><p>At x = 2: d²y/dx² = 4 - 2 = 2 > 0 ✓ (minimum)</p><p><strong>Step 5: Find the y-coordinate at x = 2</strong></p><p>b = (2)³/3 - (2)² + 4/3 = 8/3 - 4 + 4/3 = 12/3 - 4 = 0</p><p>Point of minimum ordinate: (a, b) = (2, 0)</p><p><strong>Step 6: Calculate a + 6b</strong></p><p>a + 6b = 2 + 6(0) = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2