The rate of change of the acute angle made by the ladder with the ground is -.021
According to the question,
Given: length of ladder(L)=30ft
Sliding down the wall at the rate of [tex]$=\frac{d y}{d t}=\frac{-1}{2} \mathrm{ft} / \mathrm{sec}$[/tex]
To find: Rate of change of the acute angle made by the ladder when upper end is 18 ft above the ground.
Acute angle measure less than 90 degrees. right angle measures 90 degrees. obtuse angle measures more than 90 degrees
Now, we know that
[tex]$\sin \theta=\frac{y}{30}$[/tex]
let the time be (t) at general height (h) and at angle(θ)
[tex]$\begin{aligned}& \Rightarrow \quad \frac{d y}{d t}=-\frac{1}{2} \mathrm{ft} / \mathrm{sec} \\& \Rightarrow \quad \sin \theta=\frac{y}{30}\end{aligned}$[/tex]
Differentiate above equation
[tex]$\begin{aligned}\Rightarrow & (\cos \theta)\left(\frac{d \theta}{d t}\right)=\left(\frac{1}{30}\right)\left(\frac{d y}{d t}\right) \\& \text { Substitute } \frac{d y}{d t}=-\frac{1}{2} \mathrm{ft} / \mathrm{sec} . \\\Rightarrow & \frac{d \theta}{d t}=\left(\frac{1}{30}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{\cos \theta}\right)\end{aligned}$[/tex]
[tex]$\Rightarrow \frac{d \theta}{d t}=\left(\frac{-1}{60}\right)\left(\frac{1}{\cos \theta}\right)$\\[/tex]
So, when h=18, from (1)
[tex]$\begin{aligned}& \Rightarrow \quad \sin \theta=\frac{18}{30} \\& \Rightarrow \sin \theta=\frac{3}{5}\end{aligned}$[/tex]
As we know that,
[tex]$\begin{aligned}& \sin ^2 \theta+\cos ^2 \theta=1 \\& =\cos \theta=\sqrt{1-\sin ^2 \theta} \\& =\cos \theta=\sqrt{1-\left(\frac{3}{5}\right)^2} \\& \Rightarrow \cos \theta=\frac{\sqrt{25-9}}{5} \\& \Rightarrow \cos \theta=\frac{4}{5}\end{aligned}$[/tex]
let the time be (t) at general height (h) and at angle(θ)
2) Substitute [tex]$\left(\cos \theta=\frac{4}{5}\right)$[/tex] from (4) eqn in.
(3) equ,
Then, [tex]$\quad \frac{d \theta}{d t}=\left(-\frac{1}{60}\right)\left(\frac{5}{4}\right) \mathrm{rad} / \mathrm{sec}$[/tex]
[tex]$\frac{d \theta}{d t}=\frac{-1}{48} \mathrm{rad} / \mathrm{sec}$[/tex]
Therefore, the rate of change of the acute angle made by the ladder with the ground is [tex]$\frac{d \theta}{d t}=\frac{-1}{48} \mathrm{rad} / \mathrm{sec}$[/tex].
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