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 When the water level is 2 m , the water level is rising at a rate of

To solve the problem, we need to relate the volume of the cone to the height of the water. The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Since the tank's radius and height maintain a specific ratio due to the conical shape, we can express the radius \( r \) in terms of the height \( h \). Given the dimensions of the tank, we have the proportion: \[ \frac{r}{h} = \frac{3}{9} \quad \Rightarrow \quad r = \frac{h}{3} \] Substituting \( r \) into the volume formula gives: \[ V = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{9} h = \frac{\pi}{27} h^3 \] Now, we can differentiate the volume with respect to time \( t \): \[ \frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt} \] Given that \( \frac{dV}{dt} = 2 \, \mathrm{m}^3/\mathrm{min} \), we need to find \( \frac{dh}{dt} \) when \( h = 1 \) m and \( h = 2 \) m. 1. For \( h = 1 \) m: \[ 2 = \frac{\pi}{9} (1)^2 \frac{dh}{dt} \] Solving for \( \frac{dh}{dt} \): \[ 2 = \frac{\pi}{9} \frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{2 \cdot 9}{\pi} = \frac{18}{\pi} \, \mathrm{m/min} \] 2. For \( h = 2 \) m: \[ 2 = \frac{\pi}{9} (2)^2 \frac{dh}{dt} \] Solving for \( \frac{dh}{dt} \): \[ 2 = \frac{\pi}{9} \cdot 4 \frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{2 \cdot 9}{4\pi} = \frac{18}{4\pi} = \frac{9}{2\pi} \, \mathrm{m/min} \] So, summarizing the findings: When the water level is 1 m, the water level is rising at a rate of \( \frac{18}{\pi} \, \mathrm{m/min} \). When the water level is 2 m, the water level is rising at a rate of \( \frac{9}{2\pi} \, \mathrm{m/min} \).