calculate the second derivative of the parametrix equation x(t)=t^3+2t^2 and y(t)=1-3t{2

To find the second derivative of the parametric equation \( x(t) = t^3 + 2t^2 \) and \( y(t) = 1 - 3t^2 \), we need to compute \( \frac{d^2 y}{dx^2} \).

### Step 1: Find the first derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \):
\[
\frac{dx}{dt} = \frac{d}{dt} \left(t^3 + 2t^2\right) = 3t^2 + 4t
\]
\[
\frac{dy}{dt} = \frac{d}{dt} \left(1 - 3t^2\right) = -6t
\]

### Step 2: Find the first derivative of \( y \) with respect to \( x \) (\( \frac{dy}{dx} \)):
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-6t}{3t^2 + 4t}
\]

### Step 3: Differentiate \( \frac{dy}{dx} \) with respect to \( t \) to find \( \frac{d^2y}{dx^2} \) using the chain rule:
\[
\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{-6t}{3t^2 + 4t}\right)
\]
Use the quotient rule: \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \).

Here, \( u = -6t \) and \( v = 3t^2 + 4t \).

First, compute the derivatives:
\[
\frac{du}{dt} = -6
\]
\[
\frac{dv}{dt} = 6t + 4
\]

Now apply the quotient rule:
\[
\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(3t^2 + 4t)(-6) - (-6t)(6t + 4)}{(3t^2 + 4t)^2}
\]
Simplify the numerator:
\[
= \frac{-18t^2 - 24t + (36t^2 + 24t)}{(3t^2 + 4t)^2}
= \frac{18t^2}{(3t^2 + 4t)^2}
\]

### Step 4: Find \( \frac{d^2y}{dx^2} \) by dividing the derivative of \( \frac{dy}{dx} \) by \( \frac{dx}{dt} \):
\[
\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{18t^2}{(3t^2 + 4t)^2} \cdot \frac{1}{3t^2 + 4t}
\]
Simplify the expression:
\[
\frac{d^2y}{dx^2} = \frac{18t^2}{(3t^2 + 4t)^3}
\]

This is the second derivative of \( y \) with respect to \( x \) for the given parametric equations.

Would you like to go over any part of this solution in more detail?

### Related Questions:
1. How does the second derivative \( \frac{d^2y}{dx^2} \) affect the curvature of the curve?
2. What are the conditions for inflection points in parametric curves?
3. How can you find the tangent and normal lines to the curve at a specific point \( t \)?
4. What role do parametric equations play in the context of vector calculus?
5. How can you generalize this method to three-dimensional parametric curves?

**Tip:** Always simplify your derivatives as much as possible before proceeding to the next step to make calculations easier.

Learn how to calculate the second derivative of parametric equations x(t) = t^3 + 2t^2 and y(t) = 1 - 3t^2 with a detailed step-by-step solution. This problem involves parametric equations, derivatives, the chain rule, and the quotient rule. Suitable for college-level mathematics.

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