Let $x=2 t^{2}, y=-4 t^{2}+36 t$. Determine $\frac{d y}{d x}$ as a function of $t$, then find the slope of the parametric curve at $t=6$. \[ \begin{array}{l} \frac{d y}{d x}= \\ \frac{d y}{d x}(6)= \end{array} \] Determine $\frac{d^{2} y}{d x^{2}}$ as a function of $t$, then find the concavity of the parametric curve at $t=6$. (Hint: The Second Derivative Test for Extrema could help.) \[ \begin{array}{l} \frac{d^{2} y}{d x^{2}}= \\ \frac{d^{2} y}{d x^{2}}(6)= \end{array} \]

Step 1 ：First, we need to find the derivatives of \(x\) and \(y\) with respect to \(t\).

Step 2 ：\(x = 2t^2\) and \(y = -4t^2 + 36t\), so \(\frac{dx}{dt} = 4t\) and \(\frac{dy}{dt} = 36 - 8t\).

Step 3 ：Then, we can use the chain rule to find \(\frac{dy}{dx}\) as a function of \(t\).

Step 4 ：\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{36 - 8t}{4t}\).

Step 5 ：After that, we can substitute \(t=6\) into the function to find the slope of the parametric curve at \(t=6\).

Step 6 ：\(\frac{dy}{dx}(6) = -\frac{1}{2}\).

Step 7 ：Next, we need to find the second derivative of \(y\) with respect to \(x\) as a function of \(t\). This involves finding the derivative of \(\frac{dy}{dx}\) with respect to \(t\) and then dividing by \(\frac{dx}{dt}\).

Step 8 ：\(\frac{d^2y}{dx^2} = \frac{-2/t - (36 - 8t)/(4t^2)}{4t}\).

Step 9 ：After that, we can substitute \(t=6\) into the function to find the concavity of the parametric curve at \(t=6\).

Step 10 ：\(\frac{d^2y}{dx^2}(6) = -\frac{1}{96}\).

Step 11 ：Final Answer: The derivative of \(y\) with respect to \(x\) as a function of \(t\) is \(\frac{36 - 8t}{4t}\), and the slope of the parametric curve at \(t=6\) is \(\boxed{-\frac{1}{2}}\). The second derivative of \(y\) with respect to \(x\) as a function of \(t\) is \(\frac{-2/t - (36 - 8t)/(4t^2)}{4t}\), and the concavity of the parametric curve at \(t=6\) is \(\boxed{-\frac{1}{96}}\).