A conical tank has height 9 m and radius 3 m at the base. Water flows at a rate of \( 2 \mathrm{~m}^{3} / \mathrm{min} \). How fast is the water level rising when the level is 1 m and 2 m ? (Use symbolic notation and fractions where needed.) When the water level is 1 m , the water level is rising at a rate of \( \frac{18}{\pi} \frac{\mathrm{~m}}{\mathrm{~min}} \) When the water level is 2 m , the water level is rising at a rate of \( \frac{9}{2 \pi} \frac{\mathrm{~m}}{\mathrm{~min}} \) Note: You can earn partial credit on this problem. Preview My Answers Submit Answers

The volume of water in a conical tank is given by \[ V=\frac{1}{3}\pi r^2h. \] Since the cone and the water are similar, the ratio of the radius to the height is constant: \[ \frac{r}{h}=\frac{3}{9}=\frac{1}{3} \quad \Longrightarrow \quad r=\frac{h}{3}. \] Substitute \( r=\frac{h}{3} \) into the volume formula: \[ V=\frac{1}{3}\pi \left(\frac{h}{3}\right)^2h=\frac{1}{3}\pi \frac{h^2}{9}h=\frac{1}{27}\pi h^3. \] Differentiate both sides with respect to time \( t \): \[ \frac{dV}{dt}=\frac{d}{dt}\left(\frac{1}{27}\pi h^3\right)=\frac{1}{27}\pi \cdot 3h^2\frac{dh}{dt}=\frac{1}{9}\pi h^2\frac{dh}{dt}. \] We know that water flows at the rate \[ \frac{dV}{dt}=2 \, \text{m}^3/\text{min}. \] Thus, solving for \(\frac{dh}{dt}\) we have \[ \frac{dh}{dt}=\frac{9}{\pi h^2}\frac{dV}{dt}=\frac{9}{\pi h^2}(2)=\frac{18}{\pi h^2}. \] Now, compute \(\frac{dh}{dt}\) for the two water levels: 1. When \( h=1 \, \text{m} \): \[ \frac{dh}{dt}=\frac{18}{\pi (1)^2}=\frac{18}{\pi}\,\frac{\mathrm{m}}{\mathrm{min}}. \] 2. When \( h=2 \, \text{m} \): \[ \frac{dh}{dt}=\frac{18}{\pi (2)^2}=\frac{18}{4\pi}=\frac{9}{2\pi}\,\frac{\mathrm{m}}{\mathrm{min}}. \]