Answer
$\frac{\sqrt 6}{48}$
Work Step by Step
Let height of the wall $= y $ ft
Let $x$ be the distance of the bottom of the plank from the wall.
$x=2$ ft
Length of the plank $=10$ ft
Reference to the figure, plank is the hypotenuse of the right-angled triangle.
By Pythagoras theorem
$x^2+y^2=100$
Putting $x=2$ in the above equation
$y^2=96 \Longrightarrow$ $y=4\sqrt 6$ ft
Let $\theta $ be the acute angle that plank makes with the ground.
Reference to the figure
$\cos\theta=\frac{x}{10} $ Or $ x=10\cos\theta$
Taking derivative with respect to t
$\frac{dx}{dt}=\frac{d(10 \cos\theta)} {dt}= 10 \frac{d\cos\theta)}{d\theta} \frac{d\theta}{dt}=-10\sin\theta \frac{d\theta}{dt}$............... eq (1)
Now from figure
$\sin\theta=\frac{y}{10} $
Putting $ y=4\sqrt 6$
$\sin\theta=\frac{4\sqrt 6}{10}= \frac{2\sqrt 6}{5}$ ............... eq (2)
$\frac{dx}{dt}=-6 $ in/s $=-\frac{1}{2} ft/s$ .................... eq (3)
Putting equation (2) and equation (3) in equation (1)
$ -\frac{1}{2}=-10 \times \frac{2\sqrt 6}{5} \frac{d\theta}{dt} = -4 \sqrt 6 \frac{d\theta}{dt} $
$\frac{d\theta}{dt}=\frac{ \frac{1}{2}}{ 4\sqrt 6}=\frac{\sqrt 6}{48}$
