To find the Cartesian equation of the path from the parametric equations:

$x(t) = \frac{3t}{\sqrt{1 + t^3}} \quad \text{and} \quad y(t) = \frac{3t^2}{1 + t^3}$

we want to eliminate the parameter $t$ and express $y$ directly as a function of $x$.

Step-by-step process:

1. Square the expression for $x(t)$:

   $x(t)^2 = \left( \frac{3t}{\sqrt{1 + t^3}} \right)^2 = \frac{9t^2}{1 + t^3}$

   Thus, we have:

   $x^2 = \frac{9t^2}{1 + t^3}$

2. Use the expression for $y(t)$:

   $y(t) = \frac{3t^2}{1 + t^3}$

3. Observe the relationship between $y(t)$ and $x^2$:

   From step 1, we have:

   $x^2 = 9 \cdot \frac{t^2}{1 + t^3}$

   From step 2:

   $y = 3 \cdot \frac{t^2}{1 + t^3}$

   Dividing these two equations, we get:

   $\frac{y}{x^2} = \frac{3 \cdot \frac{t^2}{1+t^3}}{9 \cdot \frac{t^2}{1+t^3}} = \frac{1}{3}$

   So, the Cartesian form of the equation is:

   $y = \frac{x^2}{3}$

Thus, the Cartesian equation of the path is:

$y = \frac{x^2}{3}$

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Would you like more details on any step? Here are 5 related questions:

1. How can we check if this is a parabola based on the equation?
2. How can we derive the velocity of the particle along this path?
3. What is the domain of this Cartesian equation based on the given parametric limits?
4. How can we visualize the path of the particle?
5. How can we use implicit differentiation to find the slope of the path at any point?

Tip: When converting from parametric to Cartesian form, look for relationships between $x(t)$ and $y(t)$ that can help eliminate the parameter $t$.