1. **State the problem:** Water is flowing out of a conical tank at a rate of 4 cm³/s. The tank has radius 4 cm and height 10 cm. We want to find how fast the radius of the water surface is changing when the water height is 6 cm. 2. **Relevant formula:** The volume of a cone is given by $$V = \frac{1}{3} \pi r^2 h$$ where $r$ is the radius of the water surface and $h$ is the height of the water. 3. **Relate radius and height:** Since the tank is a cone with fixed dimensions, the radius and height of the water are proportional: $$\frac{r}{h} = \frac{4}{10} = \frac{2}{5} \implies r = \frac{2}{5}h$$ 4. **Express volume in terms of $h$ only:** Substitute $r = \frac{2}{5}h$ into the volume formula: $$V = \frac{1}{3} \pi \left(\frac{2}{5}h\right)^2 h = \frac{1}{3} \pi \frac{4}{25} h^3 = \frac{4\pi}{75} h^3$$ 5. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt} = \frac{4\pi}{25} h^2 \frac{dh}{dt}$$ 6. **Given:** $\frac{dV}{dt} = -4$ cm³/s (negative because water is flowing out), and $h=6$ cm. Substitute these values: $$-4 = \frac{4\pi}{25} \times 6^2 \times \frac{dh}{dt} = \frac{4\pi}{25} \times 36 \times \frac{dh}{dt} = \frac{144\pi}{25} \frac{dh}{dt}$$ 7. **Solve for $\frac{dh}{dt}$:** $$\frac{dh}{dt} = \frac{-4 \times 25}{144\pi} = \frac{-100}{144\pi} = \frac{-25}{36\pi} \approx -0.221$$ cm/s 8. **Find $\frac{dr}{dt}$:** Recall $r = \frac{2}{5} h$, so $$\frac{dr}{dt} = \frac{2}{5} \frac{dh}{dt} = \frac{2}{5} \times \frac{-25}{36\pi} = \frac{-50}{180\pi} = \frac{-5}{18\pi} \approx -0.088$$ cm/s **Final answer:** The radius of the water surface is decreasing at approximately $0.088$ cm/s when the water height is 6 cm.