Differential Equations
Geometric Applications
Grade None
Question:
<p>The differential equation of the family of curves for which the length of the normal is equal to a constant k, is given by</p>
<p>(a) \(y\left(\frac{dy}{dx}\right)^2 = k^2 - y^2\)</p>
<p>(b) \(y^2\left[\frac{dy}{dx}\right]^2 = k^2 - y^2\)</p>
<p>(c) \(y\frac{dy}{dx} = k^2 - y^2\)</p>
<p>(d) \(y^2\left[\frac{dy}{dx}\right]^2 = k^2 + y^2\)</p>
Step-by-Step Solution
Key Concept: The length of the normal to a curve at any point is given by y√(1 + (dy/dx)²). Setting this equal to constant k and simplifying yields the differential equation.
<p><strong>Step 1: Recall the formula for length of normal</strong></p><p>For a curve y = f(x), at a point (x, y), the length of the normal from the point to the x-axis is:</p><p>L = y√(1 + (dy/dx)²)</p><p><strong>Step 2: Set the length equal to constant k</strong></p><p>According to the problem, the length of the normal equals k:</p><p>y√(1 + (dy/dx)²) = k</p><p><strong>Step 3: Square both sides to eliminate the square root</strong></p><p>y²[1 + (dy/dx)²] = k²</p><p><strong>Step 4: Expand the left side</strong></p><p>y² + y²(dy/dx)² = k²</p><p><strong>Step 5: Rearrange to isolate the differential term</strong></p><p>y²(dy/dx)² = k² - y²</p><p><strong>Step 6: Verify this matches option (b)</strong></p><p>The equation y²[dy/dx]² = k² - y² matches option (b) exactly.</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b