Question

The top of a ladder 6 metres long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance?
How far is the foot from the wall when it and the top are moving at the same rate?

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.

Here,

\[x^2 + y^2 = 36\]
\[ \Rightarrow 2x\frac{dx}{dt} = - 2y\frac{dy}{dt} . . . \left( 1 \right)\]
\[\text{When }x  = 4, y = \sqrt{36 - 16} = 2\sqrt{5}\]
\[ \Rightarrow 2 \times 4 \times 0 . 5 = - 2 \times 2\sqrt{5}\frac{dy}{dt} \left[ \because \frac{dx}{dt} = 0 . 5 m/\sec \right]\]
\[ \Rightarrow \frac{dy}{dt} = \frac{- 1}{\sqrt{5}}m/\sec\]
\[\text { From eq }. (1), \text { we get }\]
\[2x\frac{dx}{dt} = - 2y\frac{dy}{dt} \left[ \because \frac{dx}{dt} = \frac{dy}{dt} \right] \]
\[ \Rightarrow x = - y\]
\[\text { Substituting } x=-\text { y in } x^2 + y^2 =36, \text { we get }\]
\[ x^2 + x^2 = 36\]
\[ \Rightarrow x^2 = 18\]
\[ \Rightarrow x = 3\sqrt{2} m\]