1. **State the problem:** The height $h$ of a cone is increasing at a rate of $\frac{dh}{dt} = 10$ cm/sec, and the radius $r$ is changing so that the volume $V$ remains constant. We need to find the rate of change of the radius $\frac{dr}{dt}$ when $r=4$ cm and $h=10$ cm. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ Since $V$ is constant, $\frac{dV}{dt} = 0$. 3. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right) = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) = 0$$ 4. **Plug in known values:** $$r=4, \quad h=10, \quad \frac{dh}{dt} = 10$$ 5. **Substitute and solve for $\frac{dr}{dt}$:** $$\frac{1}{3} \pi \left( 2 \times 4 \times \frac{dr}{dt} \times 10 + 4^2 \times 10 \right) = 0$$ $$\frac{1}{3} \pi \left( 80 \frac{dr}{dt} + 160 \right) = 0$$ Multiply both sides by $3/\pi$: $$80 \frac{dr}{dt} + 160 = 0$$ Solve for $\frac{dr}{dt}$: $$80 \frac{dr}{dt} = -160$$ $$\frac{dr}{dt} = -2$$ 6. **Interpretation:** The radius is decreasing at a rate of 2 cm/sec when $r=4$ cm and $h=10$ cm. **Final answer:** $$\boxed{\frac{dr}{dt} = -2 \text{ cm/sec}}$$