Given:
[tex]y=-\frac{1}{110}x^2+127[/tex]
The horizontal component of velocity is 5 ft/s.
[tex]\frac{dx}{dt}=5[/tex]
To find the rate of change,
[tex]\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}(-\frac{1}{110}x^2+127) \\ \frac{dy}{dx}=-\frac{1}{110}(2x)\frac{dx}{dt}+0 \\ \frac{dy}{dx}=-\frac{1}{110}(2x)(5) \\ \frac{dy}{dx}=-\frac{x}{11} \\ at\text{ x=22} \\ \frac{dy}{dx}=-\frac{22}{11}=-2 \end{gathered}[/tex]
So, the rate of chnage of elevation is -2 ft/s.
That means decreasing by 2 ft/s.
