a 10 - ft ladder is leaning against a house when its base starts to slide away. by the time the base is 8 ft…

Explanation:

Step1: Establish relationship for part a

Let (x) be the distance of the base from the house. Given (x = 8) ft, (dx/dt=12) ft/sec, and the length of the ladder (L = 10) ft. We know (\cos\theta=\frac{x}{L}) (since (\cos\theta=\text{adjacent}/\text{hypotenuse}) in a right - triangle). Differentiate both sides with respect to (t): (-\sin\theta\frac{d\theta}{dt}=\frac{1}{L}\frac{dx}{dt}). First, find (\sin\theta) using (\sin\theta=\sqrt{1-\cos^{2}\theta}). Since (\cos\theta=\frac{x}{L}=\frac{8}{10}=\frac{4}{5}), then (\sin\theta=\frac{3}{5}).

Step2: Solve for (d\theta/dt)

Substitute (\sin\theta = \frac{3}{5}), (L = 10), and (\frac{dx}{dt}=12) into (-\sin\theta\frac{d\theta}{dt}=\frac{1}{L}\frac{dx}{dt}). We get (-\frac{3}{5}\frac{d\theta}{dt}=\frac{1}{10}\times12). Then (\frac{d\theta}{dt}=-\frac{12\times5}{10\times3}=- 2) rad/sec.

Step3: Establish relationship for part b

Let (y) be the height of the top of the ladder. Using the Pythagorean theorem (x^{2}+y^{2}=L^{2}) (where (L = 10)). Differentiate with respect to (t): (2x\frac{dx}{dt}+2y\frac{dy}{dt}=0), so (x\frac{dx}{dt}+y\frac{dy}{dt}=0). When (x = 8), (y=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=6).

Step4: Solve for (dy/dt)

Substitute (x = 8), (\frac{dx}{dt}=12), and (y = 6) into (x\frac{dx}{dt}+y\frac{dy}{dt}=0). We have (8\times12+6\frac{dy}{dt}=0). Then (6\frac{dy}{dt}=-96), and (\frac{dy}{dt}=-16) ft/sec.

Answer:

a. (-2) b. (-16)