A ladder of length 13 m has one end resting | vedclass

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(C) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the upper end from the ground. The ladder forms a right-angled tri

Vedclass · Accepted Answer

$\frac{5}{6}$

Vedclass · Answer

(C) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the upper end from the ground. The ladder forms a right-angled triangle with the wall and the ground,so $x^2 + y^2 = 13^2 = 169$.Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.Given that the lower end moves away from the wall at a speed of $\frac{dx}{dt} = 2 \ m/min$.When $x = 5 \ m$,we find $y$ using $x^2 + y^2 = 169$: $5^2 + y^2 = 169 \Rightarrow 25 + y^2 = 169 \Rightarrow y^2 = 144 \Rightarrow y = 12 \ m$.Substituting these values into the differentiated equation: $5(2) + 12 \frac{dy}{dt} = 0$.$10 + 12 \frac{dy}{dt} = 0 \Rightarrow 12 \frac{dy}{dt} = -10 \Rightarrow \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \ m/min$.The negative sign indicates that the upper end is falling. Thus,the speed at which the upper end falls is $\frac{5}{6} \ m/min$.

$A$ ladder of length $13 \ m$ has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed of $2 \ m/min$,then the speed (in $m/min$) at which the upper end falls when the bottom is $5 \ m$ away from the wall is

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