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 slope function to find the curve equation, then use the initial condition to determine the constant of integration. The antiderivative of 1/(x√(x²-1)) is sec⁻¹|x| + C, which requires careful handling of the absolute value and domain restrictions.
<p><strong>Step 1:</strong> We have dy/dx = 1/(x√(x²-1)). Integrate both sides:</p><p>y = ∫ 1/(x√(x²-1)) dx</p><p><strong>Step 2:</strong> Use substitution u = x², du = 2x dx. This gives:</p><p>y = ∫ 1/(√(u²-1)) · (du/(2u)) which evaluates to y = sec⁻¹|x| + C</p><p><strong>Step 3:</strong> Apply initial condition (-2, -2):</p><p>-2 = sec⁻¹|-2| + C = sec⁻¹(2) + C</p><p>Since sec⁻¹(2) = π/3, we get: C = -2 - π/3</p><p><strong>Step 4:</strong> The curve equation is y = sec⁻¹|x| - 2 - π/3 for |x| ≥ 1</p><p><strong>Step 5:</strong> Check candidate points by verifying they satisfy this equation and lie in the domain |x| ≥ 1. For x = -1 and x = √2, compute sec⁻¹ values and verify the y-coordinates match.</p><p>∴ Answer: B,D</p>
Correct Answer: B,D