Area Under the Curve
Area using parametric curves
Grade 12
Question:
<p>The area of the loop formed by the curve given by \(x = a(1-t^2)\), \(y = at(1-t^2)\), \(-1 \leq t \leq 1\) is</p>
<p>\(a^2\)</p>
<p>\(\dfrac{3a^2}{5}\)</p>
<p>\(\dfrac{5a^2}{8}\)</p>
<p>\(\dfrac{8a^2}{15}\)</p>
Step-by-Step Solution
Key Concept: The loop is traced parametrically as t varies from -1 to 1, and we must use the parametric area formula ∫y dx rather than treating x as a function of y. The curve is symmetric about the x-axis, so we can double the area for 0 ≤ t ≤ 1.
<p><strong>Step 1:</strong> Identify the parametric equations. We have x = a(1-t²), y = at(1-t²) for -1 ≤ t ≤ 1.</p><p><strong>Step 2:</strong> Verify the loop closes: At t = ±1, x = a(1-1) = 0 and y = 0, so both endpoints meet at the origin.</p><p><strong>Step 3:</strong> Use the parametric area formula: A = |∫₋₁¹ y(dx/dt) dt|. First, find dx/dt:</p><p>dx/dt = a·(-2t) = -2at</p><p><strong>Step 4:</strong> Substitute into the area integral:</p><p>A = |∫₋₁¹ at(1-t²)·(-2at) dt| = |∫₋₁¹ -2a²t²(1-t²) dt|</p><p><strong>Step 5:</strong> Expand and integrate:</p><p>A = 2a² ∫₀¹ t²(1-t²) dt = 2a² ∫₀¹ (t² - t⁴) dt</p><p>(Using symmetry: the integrand is even, so integrate from 0 to 1 and double)</p><p><strong>Step 6:</strong> Evaluate the integral:</p><p>A = 2a² [t³/3 - t⁵/5]₀¹ = 2a² (1/3 - 1/5) = 2a² · (5-3)/(15) = 2a² · (2/15) = <strong>4a²/15</strong></p><p>∴ Answer: D</p>
Correct Answer: D