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To solve this problem, we need to determine when the parametric curve described by:

\[
x(t) = 4 + 2t^2, \quad y(t) = t^2 + 2t^3
\]

is concave up. Concavity depends on the second derivative of the curve, so we will first need to compute the second derivative of \(y(t)\) with respect to \(x(t)\). 

### Step-by-step approach:

1. **Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)**:
   - \( \frac{dx}{dt} = \frac{d}{dt}(4 + 2t^2) = 4t \)
   - \( \frac{dy}{dt} = \frac{d}{dt}(t^2 + 2t^3) = 2t + 6t^2 \)

2. **Find \(\frac{dy}{dx}\)**:
   Using the chain rule:
   \[
   \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 6t^2}{4t} = \frac{2t(1 + 3t)}{4t} = \frac{1 + 3t}{2}
   \]
   So, \(\frac{dy}{dx} = \frac{1 + 3t}{2}\).

3. **Find the second derivative \(\frac{d^2y}{dx^2}\)**:
   We differentiate \(\frac{dy}{dx}\) with respect to \(t\) and divide by \(\frac{dx}{dt}\):
   \[
   \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{1 + 3t}{2}\right) = \frac{3}{2}
   \]
   Now, dividing by \(\frac{dx}{dt} = 4t\), we get:
   \[
   \frac{d^2y}{dx^2} = \frac{\frac{3}{2}}{4t} = \frac{3}{8t}
   \]

4. **Determine when the curve is concave up**:
   A curve is concave up when \(\frac{d^2y}{dx^2} > 0\). For this, we need:
   \[
   \frac{3}{8t} > 0
   \]
   This inequality holds when \(t > 0\).

Therefore, the curve is concave up for \(t > 0\).

### Correct answer: Option 1: \( t > 0 \)

Would you like further clarification on any part of this solution, or more details?

### Follow-up questions:
1. How would the concavity change if \( t < 0 \)?
2. What happens to the curve when \( t = 0 \)?
3. Could you describe the curve's behavior for large \( t \) values?
4. What does the first derivative \(\frac{dy}{dx}\) indicate about the slope of the curve?
5. How does changing the parameter \( t \) affect the shape of the curve?

### Tip:
When analyzing concavity for parametric curves, always remember to compute \(\frac{d^2y}{dx^2}\) and check its sign for different intervals of \( t \).

Find all values of t for which the curve given parametrically by x(t) = 4 + 2t^2, y(t) = t^2 + 2t^3 is concave up.

Determine when the curve described by parametric equations x(t) = 4 + 2t^2 and y(t) = t^2 + 2t^3 is concave up. This problem involves calculating first and second derivatives to check the concavity, applying the second derivative test. Suitable for college-level calculus students.

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