1. **State the problem:** A clay cylinder has a fixed volume of 4 cubic inches. Its height $h$ is decreasing at a rate of $\frac{1}{5}$ inches per second. We need to find how fast the radius $r$ is increasing when the height is $\frac{9}{10}$ inches. 2. **Relevant formulas and rules:** The volume $V$ of a cylinder is given by $$V = \pi r^2 h.$$ Since the volume is constant at 4 cubic inches, we have $$\pi r^2 h = 4.$$ We will differentiate both sides with respect to time $t$ to relate the rates of change of $r$ and $h$. 3. **Differentiate the volume equation:** Differentiating implicitly, $$\frac{d}{dt}(\pi r^2 h) = \frac{d}{dt}(4)$$ Using the product rule, $$\pi \left(2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}\right) = 0.$$ 4. **Solve for $\frac{dr}{dt}$:** $$2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0$$ $$2r h \frac{dr}{dt} = -r^2 \frac{dh}{dt}$$ $$\frac{dr}{dt} = \frac{-r^2 \frac{dh}{dt}}{2r h} = \frac{-r \frac{dh}{dt}}{2 h}.$$ 5. **Find $r$ when $h = \frac{9}{10}$:** Using the volume formula, $$4 = \pi r^2 \times \frac{9}{10}$$ $$r^2 = \frac{4}{\pi} \times \frac{10}{9} = \frac{40}{9\pi}$$ $$r = \sqrt{\frac{40}{9\pi}} = \frac{2\sqrt{10}}{3\sqrt{\pi}}.$$ 6. **Substitute values:** Given $\frac{dh}{dt} = -\frac{1}{5}$ (height decreasing), $$\frac{dr}{dt} = \frac{-r \times (-\frac{1}{5})}{2 \times \frac{9}{10}} = \frac{r \times \frac{1}{5}}{\frac{18}{10}} = \frac{r}{5} \times \frac{10}{18} = \frac{r}{9}.$$ 7. **Calculate $\frac{dr}{dt}$ numerically:** $$\frac{dr}{dt} = \frac{1}{9} \times \frac{2\sqrt{10}}{3\sqrt{\pi}} = \frac{2\sqrt{10}}{27\sqrt{\pi}}.$$ **Final answer:** The radius is increasing at a rate of $$\boxed{\frac{2\sqrt{10}}{27\sqrt{\pi}} \text{ inches per second}}$$ when the height is $\frac{9}{10}$ inches.