A 10-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 6 ft from the house, the base is moving away at the rate of \( 16 \mathrm{ft} / \mathrm{sec} \). a. What is the rate of change of the height of the top of the ladder then? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle between the ladder and the ground changing then?

We start with the ladder of length L = 10 ft leaning against the wall so that its base is a distance x from the house and its top is a height y above the ground. The relationship between x and y is given by the Pythagorean theorem:   x² + y² = 10² = 100. At the moment of interest, the ladder’s base is x = 6 ft, hence   y = √(100 – 6²) = √(100 – 36) = √64 = 8 ft. The base is sliding away at a rate of dx/dt = 16 ft/sec. We now proceed step by step. ──────────────────────────── Part (a): Rate of Change of the Height (dy/dt) Differentiate the relation x² + y² = 100 with respect to time t:   2x (dx/dt) + 2y (dy/dt) = 0. Solve for dy/dt:   y (dy/dt) = – x (dx/dt)  ⟹  dy/dt = – (x/y) (dx/dt). Plug in x = 6 ft, y = 8 ft, and dx/dt = 16 ft/sec:   dy/dt = – (6/8) · 16 = – (0.75) · 16 = –12 ft/sec. Thus, the height of the ladder top is decreasing at 12 ft/sec. ──────────────────────────── Part (b): Rate of Change of the Area of the Triangle The triangle formed by the ladder, the wall, and the ground has a base x and height y. Its area A is given by:   A = (1/2) x y. Differentiate A with respect to time:   dA/dt = (1/2) (x dy/dt + y dx/dt). Using x = 6 ft, y = 8 ft, dx/dt = 16 ft/sec, and dy/dt = –12 ft/sec:   dA/dt = (1/2)(6 · (–12) + 8 · 16)        = (1/2)(–72 + 128)        = (1/2)(56)        = 28 ft²/sec. So, the area of the triangle is increasing at 28 ft²/sec. ──────────────────────────── Part (c): Rate of Change of the Angle Between the Ladder and the Ground Let θ be the angle between the ladder and the ground. Since the ladder’s length is constant, one convenient relation is to use:   cos θ = x / 10. Differentiate both sides with respect to time t:   –sin θ (dθ/dt) = (1/10)(dx/dt). Solve for dθ/dt:   dθ/dt = – (1/10) (dx/dt) / sin θ. At the moment of interest, x = 6 ft and y = 8 ft so we have:   sin θ = y / 10 = 8/10 = 0.8. Now substitute dx/dt = 16 ft/sec:   dθ/dt = – (16/10) / 0.8 = – (1.6) / 0.8 = –2 rad/sec. The negative sign indicates that the angle is decreasing at 2 rad/sec. ──────────────────────────── Summary of Answers: a. The top of the ladder is lowering at 12 ft/sec (dy/dt = –12 ft/sec). b. The area of the triangle is increasing at 28 ft²/sec. c. The angle between the ladder and the ground is decreasing at a rate of 2 rad/sec (dθ/dt = –2 rad/sec).