1. **State the problem:** We have a paper cup shaped like an inverted right circular cone with height $h=6$ inches and diameter $6$ inches (so radius $r=3$ inches). Water is being poured in at a rate of $\frac{dV}{dt}=2$ cubic inches per minute. We want to find how fast the water level (height $h$ of water) is rising when the water depth is $2$ inches. 2. **Relevant formulas and rules:** - Volume of a cone: $$V=\frac{1}{3}\pi r^2 h$$ - Since the cup is a cone, the radius of the water surface changes with height. Using similar triangles, $$\frac{r}{h} = \frac{3}{6} = \frac{1}{2} \Rightarrow r = \frac{h}{2}$$ - Substitute $r$ in volume formula: $$V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{\pi}{12} h^3$$ 3. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$ 4. **Plug in known values:** - $\frac{dV}{dt} = 2$ - $h = 2$ $$2 = \frac{\pi}{4} (2)^2 \frac{dh}{dt} = \frac{\pi}{4} \cdot 4 \frac{dh}{dt} = \pi \frac{dh}{dt}$$ 5. **Solve for $\frac{dh}{dt}$:** $$\frac{dh}{dt} = \frac{2}{\pi}$$ inches per minute. **Final answer:** The water level is rising at a rate of $\frac{2}{\pi}$ inches per minute when the water is 2 inches deep.