To solve for $\frac{d^2y}{dx^2}$ as a function of $t$ and find its value at $t = 6$, we need to follow these steps:

Step 1: Find $\frac{dy}{dx}$

From the problem, we have:

• $x = 4t^2$
• $y = -2t^3 + 18t^2$

The formula for $\frac{dy}{dx}$ in parametric equations is: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

From your work, we already have: $\frac{dy}{dx} = \frac{-3t + 18}{4}$

Step 2: Differentiate $\frac{dy}{dx}$ with respect to $t$

To find $\frac{d^2y}{dx^2}$, we differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide by $\frac{dx}{dt}$: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

Calculating $\frac{d}{dt}\left(\frac{dy}{dx}\right)$

Given $\frac{dy}{dx} = \frac{-3t + 18}{4}$, differentiate with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{-3}{4}$

Calculating $\frac{dx}{dt}$

From $x = 4t^2$, we have: $\frac{dx}{dt} = 8t$

Thus, $\frac{d^2y}{dx^2} = \frac{\frac{-3}{4}}{8t} = \frac{-3}{32t}$

Step 3: Evaluate $\frac{d^2y}{dx^2}$ at $t = 6$

Substitute $t = 6$: $\frac{d^2y}{dx^2}(6) = \frac{-3}{32 \cdot 6} = \frac{-3}{192} = -\frac{1}{64}$

Final Answer

• $\frac{d^2y}{dx^2} = \frac{-3}{32t}$
• $\frac{d^2y}{dx^2}(6) = -\frac{1}{64}$

Would you like further details on any part of the solution?

Here are five related questions for deeper exploration:

1. How do we interpret concavity for parametric equations in terms of $\frac{d^2y}{dx^2}$?
2. What happens to $\frac{d^2y}{dx^2}$ as $t$ approaches zero?
3. How would you find the third derivative $\frac{d^3y}{dx^3}$ in parametric form?
4. Can you find the point on the curve where $\frac{d^2y}{dx^2} = 0$?
5. How would this solution change if $x$ and $y$ were functions of a different parameter?

Tip: When differentiating parametric equations, remember that each differentiation step with respect to $t$ affects both the numerator and denominator in the context of $\frac{dy}{dx}$ and higher derivatives.