A ladder 10 , m long rests against a vertical | vedclass

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(B) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the upper end of the ladder from the ground.Given th

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$6$

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(B) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the upper end of the ladder from the ground.Given that the length of the ladder is $10 \, m$,we have the relation $x^2 + y^2 = 10^2 = 100$.Differentiating both sides with respect to time $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$Given $\frac{dx}{dt} = 3 \, cm/sec$ and $\frac{dy}{dt} = -4 \, cm/sec$ (since the height is decreasing).Substituting these values,we get:$x(3) + y(-4) = 0$$3x = 4y \implies x = \frac{4}{3}y$Substituting $x = \frac{4}{3}y$ into the original equation $x^2 + y^2 = 100$:$(\frac{4}{3}y)^2 + y^2 = 100$$\frac{16}{9}y^2 + y^2 = 100$$\frac{25}{9}y^2 = 100$$y^2 = 100 	imes \frac{9}{25} = 36$$y = 6 \, m$.Thus,the height of the upper end is $6 \, m$.

$A$ ladder $10 \, m$ long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of $3 \, cm/sec$. The height of the upper end while it is descending at the rate of $4 \, cm/sec$ is (in $, m$)

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