A ladder 20 ft long leans against a vertical | vedclass

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(A) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the top end of the ladder from the floor.Given the l

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$\frac{8}{3} \ ft/sec$

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(A) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the top end of the ladder from the floor.Given the length of the ladder is $20 \ ft$,by the Pythagorean theorem,we have:$x^2 + y^2 = 20^2 = 400$Differentiating both sides with respect to time $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$Given that the top end slides downwards at a rate of $2 \ ft/sec$,we have $\frac{dy}{dt} = -2 \ ft/sec$.When $x = 12 \ ft$,we find $y$ using the equation $x^2 + y^2 = 400$:$12^2 + y^2 = 400$$144 + y^2 = 400$$y^2 = 256 \Rightarrow y = 16 \ ft$Now,substitute $x = 12$,$y = 16$,and $\frac{dy}{dt} = -2$ into the differentiated equation:$12 \left(\frac{dx}{dt}\right) + 16(-2) = 0$$12 \left(\frac{dx}{dt}\right) - 32 = 0$$12 \left(\frac{dx}{dt}\right) = 32$$\frac{dx}{dt} = \frac{32}{12} = \frac{8}{3} \ ft/sec$Thus,the lower end moves away from the wall at a rate of $\frac{8}{3} \ ft/sec$.

$A$ ladder $20 \ ft$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \ ft/sec$. The rate at which the lower end moves on a horizontal floor when it is $12 \ ft$ from the wall is

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