Question

# 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?

Solution

## The height is decreasing at the rate of $$2.67$$cms$$^{-1}$$For this we  have to use Pythagoras theorem:$$x^{2}+y^{2}=h^{2}$$     ...............(1)differentiate equition 1 with respect to 't'.$$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$$Since we have$$\dfrac{dx}{dt}=2$$cms$$^{-1}$$$$x=4m$$so,The length of wall is when the foot of the ladder is $$400$$cm away is:$$y=\sqrt{5^{2}-4^{2}}=\sqrt{25-16}=\sqrt{9}=3m=300cm$$Thus,The decreasing of the wall is:Erom equation (1) we get$$\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}$$$$=-\dfrac{400cm}{300cm}\times2$$cms$$^{-1}$$$$=2.67$$cms$$^{-1}$$This is the required answer.Mathematics

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