A ladder 5 m in length is leaning against a | vedclass

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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 on the wall.Given that the length o

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

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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 on the wall.Given that 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$$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$Given $\frac{dx}{dt} = 2 \ m/sec$ and at the instant when $x = 4 \ m$,we have $y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$.Substituting these values into the derivative equation:$\frac{dy}{dt} = -\frac{4}{3} 	imes 2 = -\frac{8}{3} \ m/sec$.The negative sign indicates that the height is decreasing.Therefore,the height on the wall is decreasing at the rate of $\frac{8}{3} \ m/sec$.

$A$ ladder $5 \ m$ in length is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $2 \ m/sec$. How fast is the height on the wall decreasing when the foot of the ladder is $4 \ m$ away from the wall?

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