Question

the top of a 15 - foot - long ladder rests against a vertical wall with the bottom of the ladder on level ground, as shown above. the ladder is sliding down the wall at a constant rate of 2 feet per second. at what rate, in radians per second, is the acute angle between the bottom of the ladder and the ground changing at the instant the bottom of the ladder is 9 feet from the base of the wall?
a $-\frac{2}{9}$
b $-\frac{1}{6}$
c $-\frac{2}{25}$
d $\frac{2}{25}$
e $\frac{1}{9}$

Explanation:

Step1: Set up the trigonometric relationship

Let the angle between the ladder and the ground be \(\theta\), the distance of the bottom of the ladder from the wall be \(x\), and the length of the ladder \(L = 15\). We know that \(\sin\theta=\frac{y}{L}\) (where \(y\) is the height of the top - of the ladder on the wall). By the Pythagorean theorem \(x^{2}+y^{2}=L^{2}=225\), so \(y=\sqrt{225 - x^{2}}\). Also, \(\sin\theta=\frac{\sqrt{225 - x^{2}}}{15}\).
Differentiate both sides with respect to time \(t\) using the chain rule. \(\cos\theta\frac{d\theta}{dt}=\frac{-x}{15\sqrt{225 - x^{2}}}\frac{dx}{dt}\).
We could also use \(\cos\theta=\frac{x}{L}=\frac{x}{15}\) (from the right - triangle definition \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)).

Step2: Find the value of \(y\)

Given \(x = 9\), then from \(x^{2}+y^{2}=225\), we have \(y=\sqrt{225 - 81}=\sqrt{144} = 12\). So \(\cos\theta=\frac{x}{15}=\frac{9}{15}=\frac{3}{5}\).
We know that \(\frac{dy}{dt}=- 2\) (negative because \(y\) is decreasing). From \(x^{2}+y^{2}=225\), differentiating with respect to \(t\) gives \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). Substitute \(x = 9\), \(y = 12\), and \(\frac{dy}{dt}=-2\) into \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\).
\(9\frac{dx}{dt}+12\times(-2)=0\), so \(9\frac{dx}{dt}=24\), and \(\frac{dx}{dt}=\frac{8}{3}\).

Step3: Solve for \(\frac{d\theta}{dt}\)

From \(\cos\theta\frac{d\theta}{dt}=\frac{-x}{15\sqrt{225 - x^{2}}}\frac{dx}{dt}\) or using the relationship from the derivative of \(\sin\theta=\frac{y}{15}\) (\(\cos\theta\frac{d\theta}{dt}=\frac{1}{15}\frac{dy}{dt}\)).
Since \(\cos\theta=\frac{3}{5}\), substituting into \(\cos\theta\frac{d\theta}{dt}=\frac{1}{15}\frac{dy}{dt}\) (another way: from \(\sin\theta=\frac{y}{15}\), \(y = 15\sin\theta\), \(\frac{dy}{dt}=15\cos\theta\frac{d\theta}{dt}\)).
Substitute \(\cos\theta=\frac{3}{5}\) and \(\frac{dy}{dt}=-2\) into \(15\cos\theta\frac{d\theta}{dt}=\frac{dy}{dt}\).
\(15\times\frac{3}{5}\frac{d\theta}{dt}=-2\), \(9\frac{d\theta}{dt}=-2\), \(\frac{d\theta}{dt}=-\frac{2}{9}\).

Answer:

A. \(-\frac{2}{9}\)