1. **State the problem:** We have a cylindrical water bottle with height $15$ cm and inner radius $4$ cm. The water height is currently $8$ cm, and the bottle will be full (height $15$ cm) in $10$ seconds. We need to find the rate at which the bottle is being filled, i.e., the rate of change of volume with respect to time $\frac{dV}{dt}$. 2. **Formula for volume of a cylinder:** $$V = \pi r^2 h$$ where $r$ is the radius and $h$ is the height of the water. 3. **Given:** - Radius $r = 4$ cm (constant) - Initial height $h_0 = 8$ cm - Final height $h_f = 15$ cm - Time to fill from $8$ cm to $15$ cm is $\Delta t = 10$ seconds 4. **Calculate the change in height:** $$\Delta h = h_f - h_0 = 15 - 8 = 7 \text{ cm}$$ 5. **Calculate the change in volume:** $$\Delta V = \pi r^2 \Delta h = \pi \times 4^2 \times 7 = \pi \times 16 \times 7 = 112\pi \text{ cm}^3$$ 6. **Calculate the rate of change of volume:** $$\frac{dV}{dt} = \frac{\Delta V}{\Delta t} = \frac{112\pi}{10} = 11.2\pi \text{ cm}^3/\text{s}$$ 7. **Final answer:** The bottle is being filled at a rate of $$\boxed{11.2\pi \text{ cm}^3/\text{s}}$$ which is approximately $35.2$ cm$^3$/s.