Applications of Derivatives
Orthogonal curves / slopes
Grade 12
Question:
<p>Consider the following curves and lines. Which of the following pairs of curves/lines are orthogonal (i.e., intersect at right angles)?</p><p>(a) \(32x + 2y\dfrac{dy}{dx} = 0\) and \(16y^{15}\dfrac{dy}{dx} = k\)</p><p>(b) \(\dfrac{dy}{dx} = 1 - ce^{-x}\) and the curve related by \(ke^{-y} = x - y + 2\)</p><p>(c) \(x^2 y^2 + xy_1 - 6y^2 = 0\)</p><p>(d) Other related differential equations on this page</p>
<p>(a) The curves \(32x + 2y\dfrac{dy}{dx}=0\) and \(16y^{15}\dfrac{dy}{dx}=k\) are orthogonal.</p>
<p>(b) \(\dfrac{dy}{dx}=1-ce^{-x}\) and \(\left[1-(x+2-y)\right]\dfrac{dy}{dx}=1\) are orthogonal.</p>
<p>(c) \(x^2y^2 + xyy_1 - 6y^2 = 0\) gives \(y = cx^2\) or \(x^3y = c\).</p>
<p>(d) All of the above.</p>
Step-by-Step Solution
Key Concept: Two curves are orthogonal at intersection if the product of their slopes is -1 (m₁ × m₂ = -1). You must find dy/dx for each curve, evaluate slopes at intersection points, and verify m₁ × m₂ = -1.
<p><strong>Step 1: Understanding Orthogonality</strong></p><p>Two curves are orthogonal if at their intersection point, their tangent lines are perpendicular, meaning: m₁ · m₂ = -1, or equivalently (dy/dx)₁ · (dy/dx)₂ = -1</p><p><strong>Step 2: Analyze Option (a)</strong></p><p>Curve 1: 32x + 2y(dy/dx) = 0 ⟹ dy/dx = -16x/y</p><p>Curve 2: 16y¹⁵(dy/dx) = k ⟹ dy/dx = k/(16y¹⁵)</p><p>Product: (-16x/y) · (k/(16y¹⁵)) = -kx/y¹⁶</p><p>For orthogonality at intersection, we need this = -1, giving kx = y¹⁶. This is possible for appropriate k values at specific intersection points. ✓</p><p><strong>Step 3: Analyze Option (b)</strong></p><p>Curve 1: dy/dx = 1 - ce⁻ˣ</p><p>Curve 2: ke⁻ʸ = x - y + 2. Differentiating: -ke⁻ʸ(dy/dx) = 1 - dy/dx</p><p>⟹ dy/dx = 1/(1 - ke⁻ʸ)</p><p>Product: (1 - ce⁻ˣ) · 1/(1 - ke⁻ʸ) = -1</p><p>⟹ (1 - ce⁻ˣ)(1 - ke⁻ʸ) = -(1 - ke⁻ʸ) = ke⁻ʸ - 1</p><p>This relationship can be satisfied for appropriate c and k values. ✓</p><p><strong>Step 4: Analyze Option (c)</strong></p><p>Curve: x²y² + xy₁ - 6y² = 0 (where y₁ = dy/dx)</p><p>Rearranging as quadratic in y₁: x·y₁ + x²y² - 6y² = 0</p><p>This represents a single differential equation. Two solutions from this would be orthogonal trajectories if they satisfy the orthogonality condition through the structure of this equation. ✓</p><p><strong>Step 5: Verification</strong></p><p>Options (a), (b), and (c) can each contain pairs of curves satisfying the orthogonality condition m₁ · m₂ = -1 at their intersection points with appropriate parameter choices.</p><p><strong>∴ Answer: A, B, C</strong></p>
Correct Answer: A, B, C, D