A reservoir has the shape of a right circular cone. The altitude is 10 feet, and the radius of the base is 4 feet. Water is poured into the reservoir at a constant rate of 5 cubic feet per minute. How fast is the water level rising when the depth of the water is 5 feet?

Answer:Step-by-step explanation:Given that a reservoir has the shape of a right circular cone. The altitude is 10 feet, and the radius of the base is 4 feet.We know in a cone at any time of water level, r/h will not changei.e. at any level of water we have[tex]\frac{r}{h} =\frac{4}{10} \\=0.4[/tex]Volume of the cone = [tex]V=\frac{1}{3} \pi r^2 h\\V=\frac{1}{3} \pi(0.4h)^2 h\\V = \frac{0.16\pi }{3} h^3[/tex]Differentiate with respect to t[tex]\frac{dV}{dt } =\frac{0.16\pi}{3} (3h^2) \frac{dh}{dt}[/tex]Substitute the values for dv/dt as 5, h =5 and r = 0.4 (5) = 2[tex]5=\frac{0.16\pi}{3} (3)(5)^2 \frac{dh}{dt} \\ \frac{dh}{dt}=\frac{1}{0.8\pi} \\=\frac{0.125}{\pi}[/tex] ft/sec