Question

# A ladder 24 ft long leans against a vertical wall. The lower end is moving away at the rate of 3 ft\sec. Find the rate at which the top of the ladder is moving downwards, if its foot is 8 ft from the wall.

Solution

## We know that $$x^2+y^2=24^2$$  ...... $$(1)$$$$\implies 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$$$$\implies x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0$$$$\implies x\dfrac{dx}{dt}+y(-3)=0$$$$\implies x\dfrac{dx}{dt}-3y=0$$ ...... $$(2)$$Given $$x=8$$ ft$$\implies 8^2+y^2=24^2$$$$\implies y=22.62$$From $$(2)$$$$8\dfrac{dx}{dt}-3(22.62)=0$$$$8\dfrac{dx}{dt}=67.88$$$$\implies \dfrac{dx}{dt}=8.4$$ ft\secMathematics

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