Question

A 10 ft plank is leaning against a wall. If at a certain instant the bottom of the plank is 2 ft from the wall and is being pushed toward the wall at the rate of 6 in/s , how fast is the acute angle that the plank makes with the ground increasing?

Answer 1

Solution:

$\dfrac{dx}{dt}=-6\ in/sec=-0.5\ ft/sec\ ...(i) \\ \dfrac{d \theta}{dt}=? \\ \cos \theta=\dfrac{x}{10}$

On differentiating both sides w.r.t $t$,

$-\sin \theta \dfrac{d \theta}{dt} =\dfrac1{10}\dfrac{dx}{dt}$ ...(ii)

Using pythagoras theorem,

$h^2 +x^2 =10^2$

When $x=2$

$h^2 +2^2 =10^2 \\ \Rightarrow h^2 =100-4=96 \\ \Rightarrow h=\sqrt{96}=4\sqrt 6\ ft$

Now, $\sin \theta=\dfrac{h}{10}=\dfrac{4\sqrt6}{10}=\dfrac{2\sqrt6}{5}$ ...(iii)

Using (i) and (iii) in (ii),

$-(\dfrac{2\sqrt6}{5}) \dfrac{d \theta}{dt} =\dfrac1{10}\times (-0.5) \\ \Rightarrow \dfrac{d \theta}{dt}=0.051\ rad/sec$