1. **State the problem:** A ladder 5 m long leans against a wall. The bottom is pulled away from the wall at 2 cm/s (0.02 m/s). We want to find how fast the height of the ladder on the wall is decreasing when the foot is 4 m from the wall. 2. **Set variables:** Let $x$ = distance from wall to foot of ladder (m), $y$ = height of ladder on wall (m). 3. **Relation:** By Pythagoras, $$x^2 + y^2 = 5^2 = 25$$ 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:** $$x = 4, \quad \frac{dx}{dt} = 0.02$$ (since 2 cm/s = 0.02 m/s) 7. **Find $y$ when $x=4$:** $$y = \sqrt{25 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$$ 8. **Plug values into differentiated equation:** $$4 \times 0.02 + 3 \times \frac{dy}{dt} = 0$$ 9. **Solve for $\frac{dy}{dt}$:** $$3 \frac{dy}{dt} = -0.08$$ 10. **Divide both sides by 3:** $$\cancel{3} \frac{dy}{dt} = \frac{-0.08}{\cancel{3}}$$ 11. **Final answer:** $$\frac{dy}{dt} = -\frac{0.08}{3} \approx -0.0267 \text{ m/s}$$ The height is decreasing at approximately 0.0267 m/s when the foot is 4 m from the wall.