Question

Water is flowing out of a conical funnel through its apex at a rate of 12 cubic inches per minute. If the tunnel is initially full, how long will it take for it to be one-third full? What is the height of the water level? Assume the radius to be 3 inches and the altitude of the cone to be 4 inches.

Answer 1

check the picture below.

so, bearing in mind that, the radius and height are two sides in a right-triangle, thus both are at a ratio of each other, thus the radius is at 3:4 ratio in relation to the height.


`\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\quad \begin{cases} r=3\\ h=4 \end{cases}\implies \stackrel{full}{V}=\cfrac{\pi \cdot 3^2\cdot 4}{3}\implies \stackrel{full}{V}=12\pi \\\\\\ \stackrel{(2)/(3)~full}{V}=12\pi \cdot \cfrac{2}{3}\implies \stackrel{(2)/(3)~full}{V}=8\pi ~in^3 \\\\\\ \textit{is draining water at a rate of }12~(in^3)/(min)\qquad \cfrac{8\pi ~in^3}{12~(in^3)/(min)}\implies \cfrac{2\pi }{3}~min`


`\bf \textit{it takes }(2\pi )/(3)\textit{ minutes to drain }(2)/(3)\textit{ of it, leaving only }(1)/(3)\textit{ in it}\\\\ -------------------------------\\\\ \stackrel{(1)/(3)~full}{V}=12\pi \cdot \cfrac{1}{3}\implies \stackrel{(1)/(3)~full}{V}=4\pi\impliedby \textit{what's \underline{h} at this time?} \\\\\\ V=\cfrac{\pi r^2 h}{3}\quad \begin{cases} r=(3h)/(4)\\ V=4\pi \end{cases} \implies 4\pi =\cfrac{\pi \left( (3h)/(4) \right)^2 h}{3}`


`\bf 12\pi =\cfrac{\pi \cdot 3^2h^2 h}{4^2}\implies 12\pi =\cfrac{9\pi h^3}{16}\implies \cfrac{192\pi }{9\pi }=h^3 \\\\\\ \sqrt[3]{\cfrac{192\pi }{9\pi }}=h\implies \sqrt[3]{\cfrac{64}{3}}=h\implies \cfrac{4}{\sqrt[3]{3}}=h \\\\\\ \cfrac{4}{\sqrt[3]{3}}\cdot \cfrac{\sqrt[3]{3^2}}{\sqrt[3]{3^2}}=h\implies \cfrac{4\sqrt[3]{9}}{\sqrt[3]{3^3}}=h\implies \cfrac{4\sqrt[3]{9}}{3}=h`