1. Find the length of the parametric curve \[ x=\left(t^{2}-4\right)^{3 / 2}, \quad y=1-3 t^{2} \] from \( t=2 \) to \( t=3 \)

To find the length of the parametric curve given by the equations \[ x = (t^2 - 4)^{3/2}, \quad y = 1 - 3t^2 \] from \( t = 2 \) to \( t = 3 \), we use the formula for the length of a parametric curve: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] where \( a = 2 \) and \( b = 3 \). First, we need to compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). 1. **Calculate \( \frac{dx}{dt} \)**: \[ x = (t^2 - 4)^{3/2} \] Using the chain rule: \[ \frac{dx}{dt} = \frac{3}{2} (t^2 - 4)^{1/2} \cdot \frac{d}{dt}(t^2 - 4) = \frac{3}{2} (t^2 - 4)^{1/2} \cdot 2t = 3t(t^2 - 4)^{1/2} \] 2. **Calculate \( \frac{dy}{dt} \)**: \[ y = 1 - 3t^2 \] Differentiating with respect to \( t \): \[ \frac{dy}{dt} = -6t \] 3. **Substituting into the length formula**: Now we substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the length formula: \[ L = \int_{2}^{3} \sqrt{\left(3t(t^2 - 4)^{1/2}\right)^2 + (-6t)^2} \, dt \] Calculating the squares: \[ \left(3t(t^2 - 4)^{1/2}\right)^2 = 9t^2(t^2 - 4) \] \[ (-6t)^2 = 36t^2 \] Thus, we have: \[ L = \int_{2}^{3} \sqrt{9t^2(t^2 - 4) + 36t^2} \, dt \] Factoring out \( 9t^2 \): \[ L = \int_{2}^{3} \sqrt{9t^2((t^2 - 4) + 4)} \, dt = \int_{2}^{3} \sqrt{9t^2(t^2)} \, dt = \int_{2}^{3} 3t^2 \, dt \] 4. **Evaluating the integral**: Now we compute the integral: \[ L = 3 \int_{2}^{3} t^2 \, dt \] The integral of \( t^2 \) is: \[ \int t^2 \, dt = \frac{t^3}{3} \] Thus, \[ L = 3 \left[ \frac{t^3}{3} \right]_{2}^{3} = \left[ t^3 \right]_{2}^{3} = 3^3 - 2^3 = 27 - 8 = 19 \] Therefore, the length of the parametric curve from \( t = 2 \) to \( t = 3 \) is \[ \boxed{19} \]