4. a 13 foot ladder is leaning against a wall. the bottom of the ladder is sliding away from the wall at 2…

Explanation:

Step1: Relate variables using trigonometry (for part a)

We know that (\sin\theta=\frac{y}{13}), where (y) is the distance of the bottom of the ladder from the wall. Differentiate both sides with respect to time (t): (\cos\theta\frac{d\theta}{dt}=\frac{1}{13}\frac{dy}{dt}) Given (y = 5) ft, using the Pythagorean theorem (x=\sqrt{13^{2}-5^{2}} = 12) ft. So (\cos\theta=\frac{x}{13}=\frac{12}{13}) and (\frac{dy}{dt}=2) ft/min.

Step2: Solve for (\frac{d\theta}{dt}) (for part a)

Substitute (\cos\theta=\frac{12}{13}) and (\frac{dy}{dt}=2) into (\cos\theta\frac{d\theta}{dt}=\frac{1}{13}\frac{dy}{dt}): (\frac{12}{13}\frac{d\theta}{dt}=\frac{1}{13}\times2) (\frac{d\theta}{dt}=\frac{2}{12}=\frac{1}{6}) rad/min

Step3: Relate variables using Pythagorean theorem (for part b)

We have (x^{2}+y^{2}=13^{2}). Differentiate both sides with respect to (t): (2x\frac{dx}{dt}+2y\frac{dy}{dt}=0) We want to find (\frac{dx}{dt}), given (y = 5) ft, (x = 12) ft and (\frac{dy}{dt}=2) ft/min.

Step4: Solve for (\frac{dx}{dt}) (for part b)

Substitute into (2x\frac{dx}{dt}+2y\frac{dy}{dt}=0): (2\times12\times\frac{dx}{dt}+2\times5\times2 = 0) (24\frac{dx}{dt}+20 = 0) (24\frac{dx}{dt}=- 20) (\frac{dx}{dt}=-\frac{20}{24}=-\frac{5}{6}) ft/min

Answer:

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