x is equal to t square, y is equal to t cube. Find double differentiation of dy upon dx.

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.

Explore the double differentiation of parametric equations x = t^2, y = t^3. Understand how to find the second derivative, \(\frac{d^2y}{dx^2}\), using the chain rule and differentiation principles. Suitable for advanced high school students.

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