Given the parametric equations:

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

To find the second derivative $\frac{d^2y}{dx^2}$, we first need to determine $\frac{dy}{dx}$.

Step 1: Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$

$\frac{dx}{dt} = \frac{d(t^2)}{dt} = 2t$ $\frac{dy}{dt} = \frac{d(t^3)}{dt} = 3t^2$

Step 2: Find $\frac{dy}{dx}$

Using the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3t}{2}$

Step 3: Find $\frac{d^2y}{dx^2}$

To find the second derivative, we differentiate $\frac{dy}{dx}$ with respect to $t$, and then divide by $\frac{dx}{dt}$:

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t}$

Thus, the second derivative $\frac{d^2y}{dx^2}$ is:

$\boxed{\frac{3}{4t}}$

Would you like more details or have any further questions?

Here are five related questions for further exploration:

1. How do you interpret the physical meaning of $\frac{d^2y}{dx^2}$ in the context of parametric equations?
2. What would $\frac{d^2y}{dx^2}$ be if $x = t^3$ and $y = t^4$?
3. How does changing the parametric equations affect the curvature of the curve described by $y$ and $x$?
4. Can you find the point of inflection for the curve defined by these parametric equations?
5. What happens to $\frac{d^2y}{dx^2}$ as $t$ approaches zero?

Tip: When working with parametric equations, always remember to apply the chain rule carefully, especially when transitioning between different derivatives.