1. **State the problem:** A ladder 25 feet long leans against a barn. The top slides down at 3 ft/sec. Find how fast the foot moves away when it is 7 feet from the wall. 2. **Identify variables:** Let $x$ = distance from wall to foot of ladder (ft), $y$ = height of ladder on barn (ft). 3. **Relation:** By Pythagoras, $$x^2 + y^2 = 25^2 = 625$$ 4. **Differentiate w.r.t time $t$:** $$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$ 5. **Simplify:** $$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$ 6. **Given:** $\frac{dy}{dt} = -3$ ft/sec (negative because sliding down), $x=7$ ft. 7. **Find $y$ when $x=7$:** $$y = \sqrt{625 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$$ 8. **Plug values into differentiated equation:** $$7 \frac{dx}{dt} + 24 (-3) = 0$$ 9. **Solve for $\frac{dx}{dt}$:** $$7 \frac{dx}{dt} = 72 \Rightarrow \frac{dx}{dt} = \frac{72}{7} \approx 10.286$$ 10. **Answer:** The foot of the ladder moves away at approximately **10.286** ft/sec.