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consider the ladder of 13m length of AB.
![\therefore \frac{dy}{dt}]_{when x= 5m}\; = \frac{-5}{\sqrt{169-5^{2}}}\times 2= \frac{-5}{6}\, cm/sec](https://learn.careers360.com/latex-image/?%5Ctherefore%20%5Cfrac%7Bdy%7D%7Bdt%7D%5D_%7Bwhen%20x%3D%205m%7D%5C%3B%20%3D%20%5Cfrac%7B-5%7D%7B%5Csqrt%7B169-5%5E%7B2%7D%7D%7D%5Ctimes%202%3D%20%5Cfrac%7B-5%7D%7B6%7D%5C%2C%20cm/sec)
Also set at any line t, the lower end of the laddet be at a distance of xm from the wall BC and at that moment, its height on the wall be ym.
consider the figer show.
we have
As
Here the height on the wall is decreasing at the rate of