5. (15 points) A particle moves clockwise along the ellipse $9 x^{2}+16 y^{2}=144$ in the first quadrant. Find $\frac{d y}{d t}$ at the moment when $x=2$ if $\frac{d x}{d t}=8 \mathrm{~m} / \mathrm{s}$. 6. (20 points) An ant is moving along the $\mathrm{x}$-axis following the law of motion $s(t)=t^{3}-12 t^{2}+36 t$ where

Step 1 ：Given the equation of the ellipse is \(9x^{2}+16y^{2}=144\), we can rewrite it as \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\).

Step 2 ：Taking the derivative of both sides with respect to \(t\), we get \(\frac{2x}{16}\frac{dx}{dt}+\frac{2y}{9}\frac{dy}{dt}=0\).

Step 3 ：Simplify the equation to get \(\frac{x}{8}\frac{dx}{dt}+\frac{y}{9}\frac{dy}{dt}=0\).

Step 4 ：We are given that \(\frac{dx}{dt}=8\) and \(x=2\), substitute these values into the equation, we get \(\frac{2}{8}*8+\frac{y}{9}\frac{dy}{dt}=0\).

Step 5 ：Simplify the equation to get \(\frac{y}{9}\frac{dy}{dt}=-1\).

Step 6 ：We need to find \(y\) when \(x=2\), substitute \(x=2\) into the equation of the ellipse, we get \(\frac{2^{2}}{16}+\frac{y^{2}}{9}=1\).

Step 7 ：Solve the equation for \(y\), we get \(y=\sqrt{9*(1-\frac{2^{2}}{16})}=\sqrt{7}\).

Step 8 ：Substitute \(y=\sqrt{7}\) into the equation \(\frac{y}{9}\frac{dy}{dt}=-1\), we get \(\frac{\sqrt{7}}{9}\frac{dy}{dt}=-1\).

Step 9 ：Solve the equation for \(\frac{dy}{dt}\), we get \(\frac{dy}{dt}=-\frac{9}{\sqrt{7}}\).

Step 10 ：So, the rate of change of \(y\) with respect to \(t\) when \(x=2\) is \(\boxed{-\frac{9}{\sqrt{7}}}~m/s\).