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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.

951793_879490_ans_9545f599291648e3b6ce51cd6a2fa2c2.PNG

Mathematics

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