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We are given the following parametric equations:
- \( y = t^3 - 12t \)
- \( x = t^2 + 4t \)

We are asked to find the value of \( \frac{d^2 y}{dx^2} \) when \( \frac{dy}{dx} = 0 \).

### Step 1: Find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

First, differentiate both \( y \) and \( x \) with respect to \( t \):

- \( \frac{dy}{dt} = 3t^2 - 12 \)
- \( \frac{dx}{dt} = 2t + 4 \)

### Step 2: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 12}{2t + 4}
\]

### Step 3: Set \( \frac{dy}{dx} = 0 \)

To find the values of \( t \) where \( \frac{dy}{dx} = 0 \), set the numerator equal to zero:
\[
3t^2 - 12 = 0
\]
\[
t^2 = 4
\]
\[
t = \pm 2
\]

### Step 4: Find \( \frac{d^2 y}{dx^2} \)

Now, differentiate \( \frac{dy}{dx} \) with respect to \( t \) to find \( \frac{d^2 y}{dx^2} \).

First, use the quotient rule on \( \frac{dy}{dx} = \frac{3t^2 - 12}{2t + 4} \):

\[
\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{(2t+4)(6t) - (3t^2 - 12)(2)}{(2t+4)^2}
\]

Simplify the numerator:
\[
= \frac{(2t+4)(6t) - (3t^2 - 12)(2)}{(2t+4)^2}
= \frac{12t^2 + 24t - 6t^2 + 24}{(2t+4)^2}
= \frac{6t^2 + 24t + 24}{(2t+4)^2}
\]

Now, divide by \( \frac{dx}{dt} = 2t + 4 \) to find \( \frac{d^2 y}{dx^2} \):
\[
\frac{d^2 y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} = \frac{6t^2 + 24t + 24}{(2t+4)^3}
\]

### Step 5: Evaluate at \( t = 2 \)

Substitute \( t = 2 \) into the expression for \( \frac{d^2 y}{dx^2} \):
\[
\frac{d^2 y}{dx^2} = \frac{6(2)^2 + 24(2) + 24}{(2(2) + 4)^3} = \frac{6(4) + 48 + 24}{(4 + 4)^3}
= \frac{24 + 48 + 24}{8^3} = \frac{96}{512} = \frac{3}{16}
\]

Therefore, \( \frac{d^2 y}{dx^2} = \frac{3}{16} \) when \( \frac{dy}{dx} = 0 \).

### Final Answer:
The correct option is (d) \( \frac{3}{16} \).

---

Would you like more details on any step? 

Here are 5 related questions:
1. What is the geometric interpretation of \( \frac{d^2 y}{dx^2} \)?
2. How do you apply the quotient rule in different scenarios?
3. Why is the condition \( \frac{dy}{dx} = 0 \) important for finding extrema?
4. Can you solve this problem using implicit differentiation instead of parametric?
5. How do parametric equations model real-world motion problems?

**Tip:** Always verify the steps of differentiation carefully, especially when using the chain or quotient rule!

إذا كانت المعادلة الوسطية y = t³ - 12t, x = t² + 4t فإن dy/dx تساوي 0 عندما d²y/dx² تساوي:

Learn how to solve the parametric differentiation problem where dy/dx = 0 and d²y/dx² is calculated using the given parametric equations y = t³ - 12t and x = t² + 4t. This problem involves applying the quotient rule and chain rule, suitable for advanced calculus students.

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