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This problem is asking for the second derivative of the parametric curve given by \( x = t^2 \) and \( y = t^3 - 3t \). The goal is to compute the second derivative \( \frac{d^2 y}{dx^2} \) and determine the concavity of the curve (whether it is concave up or down) at different points.

Let's break it down:

### Step 1: First derivative \(\frac{dy}{dx}\)

We know that:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

Given:
- \( x = t^2 \), so \( \frac{dx}{dt} = 2t \)
- \( y = t^3 - 3t \), so \( \frac{dy}{dt} = 3t^2 - 3 \)

Now, we can compute the first derivative:
\[
\frac{dy}{dx} = \frac{3t^2 - 3}{2t}
\]

### Step 2: Second derivative \(\frac{d^2 y}{dx^2}\)

Using the relation for the second derivative:
\[
\frac{d^2 y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \div \frac{dx}{dt}
\]

First, we need to compute \( \frac{d}{dt} \left( \frac{dy}{dx} \right) \):
\[
\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{3t^2 - 3}{2t} \right)
\]

Using the quotient rule for differentiation:
\[
\frac{d}{dt} \left( \frac{3t^2 - 3}{2t} \right) = \frac{(2t)(6t) - (3t^2 - 3)(2)}{(2t)^2}
\]

Simplifying the numerator:
\[
= \frac{12t^2 - 2(3t^2 - 3)}{4t^2} = \frac{12t^2 - 6t^2 + 6}{4t^2} = \frac{6t^2 + 6}{4t^2} = \frac{6(t^2 + 1)}{4t^2}
\]

Now we can calculate \( \frac{d^2 y}{dx^2} \):
\[
\frac{d^2 y}{dx^2} = \frac{6(t^2 + 1)}{4t^2} \div 2t = \frac{6(t^2 + 1)}{8t^3}
\]

This is the second derivative.

### Step 3: Concavity

The sign of \( \frac{d^2 y}{dx^2} \) determines the concavity of the curve. If \( \frac{d^2 y}{dx^2} > 0 \), the curve is concave up, and if \( \frac{d^2 y}{dx^2} < 0 \), the curve is concave down.

- When \( t > 0 \), \( \frac{d^2 y}{dx^2} > 0 \) (concave up).
- When \( t < 0 \), \( \frac{d^2 y}{dx^2} < 0 \) (concave down).

Would you like more details or have any questions about the steps? Here are 5 related questions for you:

1. Can you verify the calculation of the first derivative \(\frac{dy}{dx}\)?
2. What is the geometric interpretation of the first and second derivatives for a curve?
3. How would the concavity change if the parametric equations were slightly different?
4. How do the values of \( t \) relate to inflection points on the curve?
5. How would you interpret concavity for this curve in the context of motion or physics?

**Tip:** When working with parametric curves, it's helpful to always differentiate with respect to the parameter \( t \), as it simplifies handling derivatives and analyzing curve properties.

Find the second derivative of the parametric curve defined by x = t^2 and y = t^3 - 3t, and determine its concavity.

Learn how to calculate the second derivative of the parametric curve defined by x = t^2 and y = t^3 - 3t, and determine the concavity of the curve at different points. This calculus problem covers parametric differentiation and concavity analysis, ideal for college-level calculus students.

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