Let $x=4 t^{2}, y=-5 t^{2}+45 t$ Determine $\frac{d y}{d x}$ as a function of $t$, then find the slope of the parametric curve at $t=4.5$. \[ \begin{array}{l} \frac{d y}{d x}=\frac{-10 t+45}{8 t} \\ \frac{d y}{d x}(4.5)=0 \end{array} \] Determine $\frac{d^{2} y}{d x^{2}}$ as a function of $t$, then find the concavity of the parametric curve at $t=4.5$. (Hint: The Second Derivative Test for Extrema could help.) \[ \frac{d^{2} y}{d x^{2}}= \] \[ \frac{d^{2} y}{d x^{2}}(4.5)= \]

Step 1 ：Given the parametric equations \(x = 4t^2\) and \(y = -5t^2 + 45t\), we first need to find the derivative of \(y\) with respect to \(x\) as a function of \(t\). This is given by the formula \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Step 2 ：Differentiating \(x\) and \(y\) with respect to \(t\), we get \(dx/dt = 8t\) and \(dy/dt = 45 - 10t\).

Step 3 ：Substituting these into the formula for \(\frac{dy}{dx}\), we get \(\frac{dy}{dx} = \frac{45 - 10t}{8t}\).

Step 4 ：To find the slope of the parametric curve at \(t=4.5\), we substitute \(t=4.5\) into the expression for \(\frac{dy}{dx}\), which gives us \(\frac{dy}{dx}(4.5) = 0\).

Step 5 ：Next, we need to find the second derivative of \(y\) with respect to \(x\), again as a function of \(t\). This is given by the formula \(\frac{d^2y}{dx^2} = \frac{d(dy/dx)/dt}{dx/dt}\).

Step 6 ：Differentiating \(\frac{dy}{dx}\) with respect to \(t\), we get \(d(\frac{dy}{dx})/dt = -5/(4t) - (45 - 10t)/(8t^2)\).

Step 7 ：Substituting these into the formula for \(\frac{d^2y}{dx^2}\), we get \(\frac{d^2y}{dx^2} = \frac{-5/(4t) - (45 - 10t)/(8t^2)}{8t}\).

Step 8 ：To find the concavity of the parametric curve at \(t=4.5\), we substitute \(t=4.5\) into the expression for \(\frac{d^2y}{dx^2}\), which gives us \(\frac{d^2y}{dx^2}(4.5) = -0.00771604938271605\).

Step 9 ：Final Answer: The slope of the parametric curve at \(t=4.5\) is \(\boxed{0}\) and the concavity of the parametric curve at \(t=4.5\) is \(\boxed{-0.00771604938271605}\).