Question

1. a 12 foot ladder stands against a vertical wall. the lower end of the ladder is being pulled away from the wall at the rate of 2 ft/sec.

a.) how fast is the top of the ladder moving down the wall at the instant it is 6 feet above the ground?
b.) how fast is the angle of elevation of the ladder changing at the same instant?

Explanation:

Step1: Establish the relationship

Let \(x\) be the distance of the lower - end from the wall and \(y\) be the height of the upper - end on the wall. By the Pythagorean theorem, \(x^{2}+y^{2}=12^{2}=144\). Differentiating both sides with respect to time \(t\), we get \(2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0\), or \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\). Given \(\frac{dx}{dt}=2\) ft/sec. When \(y = 6\) ft, then \(x=\sqrt{144 - 36}=\sqrt{108}=6\sqrt{3}\) ft.

Step2: Solve for \(\frac{dy}{dt}\) (a)

Substitute \(x = 6\sqrt{3}\), \(y = 6\) and \(\frac{dx}{dt}=2\) into \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\).

$$6\sqrt{3}\times2+6\times\frac{dy}{dt}=0$$

$$12\sqrt{3}+6\frac{dy}{dt}=0$$

$$\frac{dy}{dt}=- 2\sqrt{3}\text{ ft/sec}$$

Step3: Define the angle \(\theta\) (b)

Let \(\theta\) be the angle of elevation. Then \(\sin\theta=\frac{y}{12}\). Differentiating both sides with respect to \(t\): \(\cos\theta\frac{d\theta}{dt}=\frac{1}{12}\frac{dy}{dt}\). When \(y = 6\), \(\sin\theta=\frac{6}{12}=\frac{1}{2}\), so \(\theta = 30^{\circ}\) and \(\cos\theta=\frac{\sqrt{3}}{2}\). We know \(\frac{dy}{dt}=-2\sqrt{3}\) from part (a).
Substitute into \(\cos\theta\frac{d\theta}{dt}=\frac{1}{12}\frac{dy}{dt}\)

$$\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=\frac{1}{12}\times(- 2\sqrt{3})$$

$$\frac{d\theta}{dt}=-\frac{1}{3}\text{ rad/sec}$$

Answer:

a. The top of the ladder is moving down at a rate of \(-2\sqrt{3}\text{ ft/sec}\) (the negative sign indicates the downward direction).
b. The angle of elevation is changing at a rate of \(-\frac{1}{3}\text{ rad/sec}\) (the negative sign indicates the angle is decreasing).