Answer:0.210 ft/min
Explanation:
Given
Length of trough [tex]L=16 ft[/tex]
width of base [tex]b=2 ft[/tex]
height of triangle[tex]h=1 ft[/tex]
From Similar triangles property
[tex]\frac{4}{2x}=\frac{1}{y}[/tex]
[tex]2y=x[/tex]
volume of water in time t
[tex]V=\frac{1}{2}\times (2x\cdot y)\cdot [/tex]
[tex]V=16xy[/tex]
[tex]V=32y^2[/tex]
differentiating
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=32\times 2\times y\times \frac{\mathrm{d} y}{\mathrm{d} t}[/tex]
at [tex]y=8 in.\approx 0.667 ft[/tex]
[tex]9=64\times 0.667\times \frac{\mathrm{d} y}{\mathrm{d} t}[/tex]
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{9}{64\times 0.667}[/tex]
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=0.210 ft/min[/tex]
