(14) 4. a 13 foot ladder is leaning against a wall. the bottom of the ladder is sliding away from the wall…

Explanation:

Step1: Establish relationship for part a

We know that $\sin\theta=\frac{y}{13}$. Differentiate both sides with respect to time $t$ using the chain - rule. $\cos\theta\frac{d\theta}{dt}=\frac{1}{13}\frac{dy}{dt}$. Given $\frac{dy}{dt} = 2$ ft/min. When $y = 5$ ft, we can find $x=\sqrt{13^{2}-5^{2}}=12$ ft, and $\cos\theta=\frac{x}{13}=\frac{12}{13}$.

Step2: Solve for $\frac{d\theta}{dt}$

Substitute $\cos\theta=\frac{12}{13}$ and $\frac{dy}{dt}=2$ into $\cos\theta\frac{d\theta}{dt}=\frac{1}{13}\frac{dy}{dt}$. We get $\frac{12}{13}\frac{d\theta}{dt}=\frac{1}{13}\times2$. Then $\frac{d\theta}{dt}=\frac{1}{6}$ rad/min.

Step3: Establish relationship for part b

By the Pythagorean theorem, $x^{2}+y^{2}=13^{2}$. Differentiate both sides with respect to time $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. We know that $\frac{dy}{dt} = 2$ ft/min, when $y = 5$ ft, $x = 12$ ft.

Step4: Solve for $\frac{dx}{dt}$

Substitute $x = 12$, $y = 5$ and $\frac{dy}{dt}=2$ into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We have $12\frac{dx}{dt}+5\times2=0$. Then $12\frac{dx}{dt}=- 10$, and $\frac{dx}{dt}=-\frac{5}{6}$ ft/min. The negative sign indicates that the top of the ladder is moving down.

Answer:

a) $\frac{1}{6}$ rad/min b) $\frac{5}{6}$ ft/min