Question

1. a 13 foot ladder is leaning against a wall. the bottom of the ladder is sliding away from the wall at 2 feet per minute.

a) how fast is the angle \\( \theta \\) between the wall and the ladder increasing when the bottom of the ladder is 5 feet from the wall?
b) how fast is the top of the ladder moving down the wall when the foot of the ladder is 5 feet from the wall?

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