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(D) 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.According to the Py

Vedclass · Accepted Answer

$0.075$

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(D) 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.According to the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.Given that the top slides downwards at a rate of $10 \ cm/sec$,so $\frac{dy}{dt} = -10 \ cm/sec = -0.1 \ m/sec$.When $x = 4 \ m$,we have $4^2 + y^2 = 25$,which gives $y^2 = 25 - 16 = 9$,so $y = 3 \ m$.Differentiating $x^2 + y^2 = 25$ with respect to $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$x \frac{dx}{dt} = -y \frac{dy}{dt}$Substituting the values $x = 4$,$y = 3$,and $\frac{dy}{dt} = -0.1$:$4 \frac{dx}{dt} = -3(-0.1)$$4 \frac{dx}{dt} = 0.3$$\frac{dx}{dt} = \frac{0.3}{4} = 0.075 \ m/sec$.Thus,the foot of the ladder is sliding at the rate of $0.075 \ m/sec$.

$A$ ladder $5 \ m$ long rests against a vertical wall. If its top slides downwards at the rate of $10 \ cm/sec$,then the foot of the ladder is sliding at the rate of . . . . . . $m/sec$,when it is $4 \ m$ away from the wall.

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