To find the length of a curve given in parametric form, we can use the formula for the arc length of a parametric curve:

$L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$

Given the parametric equations:

$x(t) = t^3 - 3t \quad \text{and} \quad y(t) = 3t^2$

We first need to find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

1. $\frac{dx}{dt} = 3t^2 - 3$
2. $\frac{dy}{dt} = 6t$

Now, substitute these into the formula for arc length:

$L = \int_0^1 \sqrt{(3t^2 - 3)^2 + (6t)^2} \, dt$

Simplify the expression inside the square root:

$L = \int_0^1 \sqrt{(9t^4 - 18t^2 + 9) + 36t^2} \, dt$ $L = \int_0^1 \sqrt{9t^4 + 18t^2 + 9} \, dt$ $L = \int_0^1 \sqrt{9(t^4 + 2t^2 + 1)} \, dt$ $L = \int_0^1 3\sqrt{(t^2 + 1)^2} \, dt$ $L = \int_0^1 3(t^2 + 1) \, dt$

Now, compute the integral:

$L = 3 \int_0^1 (t^2 + 1) \, dt$ $L = 3 \left[ \frac{t^3}{3} + t \right]_0^1$ $L = 3 \left( \frac{1^3}{3} + 1 \right) = 3 \left( \frac{1}{3} + 1 \right) = 3 \times \frac{4}{3} = 4$

Thus, the length of the curve is:

$L = 4$

Would you like more details or explanations on any steps?

Here are 5 related questions to explore further:

1. How do you find the length of a curve in polar coordinates?
2. What is the formula for the surface area of a surface generated by revolving a parametric curve?
3. Can you derive the arc length formula for parametric curves?
4. How does the arc length change if the parametric range changes?
5. What is the relationship between curvature and the length of a curve?

Tip: Always verify your parametric derivatives before applying them to the arc length formula for accuracy.