Find the length of the loop of the curve given by $x=6t-2t^3$, $y=6t^2$.

• A. $48$
• B. $36$
• C. $12$
• D. $24$

Answer: (A)

Verified step by step guidance

1. Step 1: Recall the formula for the arc length of a parametric curve. The length of a curve defined by parametric equations x(t) and y(t) is given by: $L\; =\; \int$_a^b √((dx/dt)² + (dy/dt)²) dt. Here, we need to compute dx/dt and dy/dt first.
2. Step 2: Differentiate x = 6t - 2t³ with respect to t to find dx/dt. This gives: $dx/dt\; =\; 6\; -\; 6t2$. Similarly, differentiate y = 6t² with respect to t to find dy/dt. This gives: $dy/dt\; =\; 12t$.
3. Step 3: Substitute dx/dt and dy/dt into the arc length formula. The integrand becomes: $\surd ((6\; -\; 6t2)2+\; (12t)2)$. Simplify the expression inside the square root.
4. Step 4: Determine the limits of integration. The problem specifies the loop of the curve, which corresponds to the values of t where the curve intersects itself. Solve x = 6t - 2t³ = 0 to find the limits of integration. This gives t = 0 and t = √3.
5. Step 5: Set up the integral for the arc length using the simplified integrand and the limits of integration. Compute $L\; =\; \int$₀^√3 √((6 - 6t²)² + (12t)²) dt. Evaluate the integral to find the length of the loop.