Differential Equations
Formation and Solution of Differential Equations
Grade 12
Question:
<p>A curve passes through \((-2, -2)\) and its slope at the point \((x, y)\) is given by \(\dfrac{1}{x\sqrt{x^2-1}}\). Which of the following points does the curve also pass through?</p>
<p>(a) \(\left(\dfrac{-2}{\sqrt{3}}, \dfrac{-\pi}{6} - 2\right)\)</p>
<p>(b) \(\left(\dfrac{-2}{\sqrt{3}}, \dfrac{\pi}{6} - 2\right)\)</p>
<p>(c) \(\left(-\sqrt{2}, \dfrac{-\pi}{12} - 2\right)\)</p>
<p>(d) \(\left(-\sqrt{2}, \dfrac{\pi}{12} - 2\right)\)</p>
Step-by-Step Solution
Key Concept: Integrate the given slope function to find the curve equation, then use the initial condition (-2, -2) to determine the constant of integration and verify which points satisfy it.
<p><strong>Step 1:</strong> Given that dy/dx = 1/(x√(x²-1)), we need to find y by integrating both sides.</p><p><strong>Step 2:</strong> ∫dy = ∫1/(x√(x²-1))dx. The standard antiderivative is sec⁻¹|x| + C, so y = sec⁻¹|x| + C.</p><p><strong>Step 3:</strong> Use the initial condition: the curve passes through (-2, -2). Substitute x = -2, y = -2:<br>-2 = sec⁻¹|-2| + C<br>-2 = sec⁻¹(2) + C</p><p><strong>Step 4:</strong> Since sec⁻¹(2) = π/3, we have:<br>-2 = π/3 + C<br>C = -2 - π/3</p><p><strong>Step 5:</strong> The curve equation is y = sec⁻¹|x| - 2 - π/3.</p><p><strong>Step 6:</strong> Check candidate points (for domain |x| ≥ 1):<br>• If x = 2: y = sec⁻¹(2) - 2 - π/3 = π/3 - 2 - π/3 = -2 ✓<br>• If x = √2: y = sec⁻¹(√2) - 2 - π/3 = π/4 - 2 - π/3 = -2 + π/4 - π/3 ≈ -2.59<br>• If x = -√2: y = sec⁻¹(√2) - 2 - π/3 = π/4 - 2 - π/3 ✓</p><p>∴ The curve passes through (2, -2) and (-√2, π/4 - 2 - π/3), corresponding to options B and D.</p>
Correct Answer: B,D