Answer:
[tex]\frac{25}{8\pi}ft/min[/tex]
Explanation:
We are given that
[tex]\frac{dV}{dt}=8 ft^3/min[/tex]
Height of tank=h=10 ft
Radius of tank, r=4 feet
We have to find the dh/dt when the water is 4 feet deep.
[tex]\frac{r}{h}=\frac{4}{10}=\frac{2}{5}[/tex]
[tex]r=\frac{2}{5}h[/tex]
Volume of cone , [tex]V=\frac{1}{3}\pi r^2 h[/tex]
Substitute the values
Volume of cone , [tex]V=\frac{1}{3}\pi(\frac{2}{5}h)^2h=\frac{4}{75}\pi h^3[/tex]
Differentiate w.r.t t
[tex]\frac{dV}{dt}=\frac{12}{75}\pi\times h^2\times \frac{dh}{dt}[/tex]
Substitute the values
[tex]8=\frac{12}{75}\times\pi (4)^2\times \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}=\frac{8\times 75}{12\times 16\pi}=\frac{25}{8\pi}ft/min[/tex]
