Step 1:
Write the displacement function
[tex]y\text{ = }-\frac{1}{130}x^2\text{ }+\text{ 180}[/tex]
Step 2:
[tex]\frac{d.\text{ x}}{d\mathrm{}y}\text{ = 5 ft/s}[/tex]
Step 3
[tex]\begin{gathered} \frac{d.y}{d\mathrm{}x}\text{ = }\frac{-2}{130}x\text{ } \\ d.y\text{ = }\frac{-1}{65}xd.x \end{gathered}[/tex]
Step 4
[tex]\begin{gathered} \frac{d.y}{d\mathrm{}t}\text{ = }\frac{-1}{65}\text{ }\times\text{ }26\text{ }\times\text{ }\frac{d.x}{d.t} \\ \\ \frac{d.y}{d\mathrm{}t}\text{ = }\frac{-1}{65}\text{ }\times\text{ 26 }\times\text{ 5} \\ =\text{ }\frac{-130}{65} \\ =\text{ -2 ft/s} \end{gathered}[/tex]
Final answer
Rate of change of the elevation = -2 ft/s
