1. **State the problem:** A conical vase has a height of 20 cm and a radius of 3 cm at the top. It is being filled with liquid at a rate of 10 cm³/s. We need to find how fast the height $h$ of the liquid is changing. 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 liquid surface and $h$ is the height of the liquid. 3. **Relate radius and height:** Since the vase is a cone with fixed dimensions, the radius and height are proportional: $$\frac{r}{h} = \frac{3}{20} \implies r = \frac{3}{20}h$$ 4. **Express volume in terms of $h$ only:** Substitute $r = \frac{3}{20}h$ into the volume formula: $$V = \frac{1}{3} \pi \left(\frac{3}{20}h\right)^2 h = \frac{1}{3} \pi \frac{9}{400} h^3 = \frac{3\pi}{400} h^3$$ 5. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{3\pi}{400} \cdot 3h^2 \frac{dh}{dt} = \frac{9\pi}{400} h^2 \frac{dh}{dt}$$ 6. **Given:** $\frac{dV}{dt} = 10$ cm³/s. Substitute and solve for $\frac{dh}{dt}$: $$10 = \frac{9\pi}{400} h^2 \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{10 \cdot 400}{9\pi h^2} = \frac{4000}{9\pi h^2}$$ 7. **Find $\frac{dh}{dt}$ when the height $h$ is at the top (20 cm):** $$\frac{dh}{dt} = \frac{4000}{9\pi (20)^2} = \frac{4000}{9\pi \cdot 400} = \frac{4000}{3600\pi} = \frac{10}{9\pi} \approx 0.354 \text{ cm/s}$$ **Final answer:** The height of the liquid is increasing at approximately $0.354$ cm/s when the vase is full height 20 cm.