Question

A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall.

Answer 1

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Then,

\[\tan \theta = \frac{y}{x} \text { and } \]
\[ x^2 + y^2 = \left( 13 \right)^2 \]
\[ \Rightarrow x^2 \left( 1 + \tan^2 \theta \right) = 169\]
\[ \Rightarrow \sec^2 \theta = \frac{169}{x^2}\]
\[ \Rightarrow 2 \sec^2 \theta \tan \theta\frac{d\theta}{dt} = 169 \left( \frac{- 2}{x^3} \right)\frac{dx}{dt}\]
\[ \Rightarrow \frac{d\theta}{dt} = \frac{- 338 \times 1 . 5}{\left( 12 \right)^3 2 \sec^2 \theta \tan \theta} . . . \left( 1 \right)\]
\[\text { When } x = 12, y = \sqrt{169 - 144} = 5  m  \]
\[So, \]
\[\sec \theta = \frac{13}{12} \text { and } \tan \theta = \frac{12}{5}\]
\[\text { From eq } . \left( 1 \right), \text { we get }\]
\[\frac{d\theta}{dt} = \frac{- 338 \times 1 . 5}{\left( 12 \right)^3 \times 2 \times \left( \frac{13}{12} \right)^2 \times \frac{5}{12}} = \frac{- 338 \times 1 . 5}{10 \times 169} = - 0 . 3 \text { rad }/\text { sec }\]