<p>Given the curve \(y = \sqrt{x}\) and the point \(\left(\dfrac{3}{2}, 0\right)\). Find the shortest distance from the point to the curve.</p>
Step-by-Step Solution
Key Concept: The shortest distance occurs along the normal to the curve. Set up the distance function D² = (x - 3/2)² + x, minimize using calculus, or equivalently find where the tangent at point (x, √x) is perpendicular to the line joining (x, √x) and (3/2, 0).
<p><strong>Step 1:</strong> Let P(x, √x) be a point on y = √x. Distance squared from P to (3/2, 0):</p><p>D² = (x - 3/2)² + (√x)² = (x - 3/2)² + x</p><p><strong>Step 2:</strong> Expand: D² = x² - 3x + 9/4 + x = x² - 2x + 9/4</p><p><strong>Step 3:</strong> Minimize by taking derivative and setting to zero:</p><p>d(D²)/dx = 2x - 2 = 0 ⟹ x = 1</p><p><strong>Step 4:</strong> At x = 1: Point on curve is (1, 1)</p><p>D² = (1 - 3/2)² + 1² = 1/4 + 1 = 5/4</p><p>D = √(5/4) = √5/2 ≈ 1.1180</p><p><strong>Verification:</strong> Slope of tangent at (1,1): dy/dx = 1/(2√x)|ₓ₌₁ = 1/2. Slope of line from (1,1) to (3/2, 0) is -2. Product = -1 ✓ (perpendicular)</p><p>∴ Answer: 1.1180</p>
Correct Answer: 1.1180