A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Explanation

Then, by Pythagoras theorem, we have:



`x^{2}+y^{2}=25 text { [Length of the ladder }=5 {m} text { ] } `


`Rightarrow y=sqrt{25-x^{2}}`



Then, the rate of change of height `(y)` with respect to time `(t)` is given by,



`frac{d y}{d t}=frac{-x}{sqrt{25-x^{2}}} cdot frac{d x}{d t}`


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