A ladder of 5 m long rests against a vertica | vedclass

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(C) 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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$3$

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(C) 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 the length of the ladder is $5 \ m$,by the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.We are given $\frac{dx}{dt} = 3 \ m/sec$ and the ladder is descending at $\frac{dy}{dt} = -4 \ m/sec$.Substituting these values into the differentiated equation: $x(3) + y(-4) = 0$,which gives $3x = 4y$,or $x = \frac{4}{3}y$.Substitute $x = \frac{4}{3}y$ into the original equation $x^2 + y^2 = 25$:$(\frac{4}{3}y)^2 + y^2 = 25$$\frac{16}{9}y^2 + y^2 = 25$$\frac{25}{9}y^2 = 25$$y^2 = 9$$y = 3 \ m$ (since height must be positive).Thus,the height of the upper end is $3 \ m$.

$A$ ladder of $5 \ m$ 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 $3 \ m/sec$. The height of the upper end (in meters) while it is descending at the rate of $4 \ m/sec$,is

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