A ladder 5 m in length is resting against a | vedclass

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(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the ground.Given that the leng

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$2$

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(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the ground.Given that the length of the ladder is $5 \ m$,by Pythagoras theorem,we have $x^2 + y^2 = 5^2 = 25$ .....$(i)$Differentiating equation $(i)$ with respect to time $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ .....$(ii)$Given that $\frac{dx}{dt} = 1.5 \ m/sec$ and at the instant $x = 4 \ m$:From $(i)$,$4^2 + y^2 = 25 \Rightarrow 16 + y^2 = 25 \Rightarrow y^2 = 9 \Rightarrow y = 3 \ m$.Substituting these values in $(ii)$:$4(1.5) + 3 \frac{dy}{dt} = 0$$6 + 3 \frac{dy}{dt} = 0$$3 \frac{dy}{dt} = -6$$\frac{dy}{dt} = -2 \ m/sec$.The negative sign indicates that the height is decreasing. Thus,the height of the top of the ladder decreases at the rate of $2 \ m/sec$.

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