A ladder 5 m long is leaning against a wall. | vedclass

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Question

(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the ladder on the wall. Given the length of the ladder is

Vedclass · Accepted Answer

$\frac{8}{3} \ cm/s$

Vedclass · Answer

(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the ladder on the wall. Given the length of the ladder is $5 \ m$,by Pythagoras theorem: $x^2 + y^2 = 5^2 = 25$. Differentiating with respect to time $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$. Given $\frac{dx}{dt} = 2 \ cm/s = 0.02 \ m/s$. When $x = 4 \ m$,$y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$. Substituting the values: $\frac{dy}{dt} = -\frac{4}{3} 	imes 2 = -\frac{8}{3} \ cm/s$. The negative sign indicates that the height is decreasing. Thus,the height is decreasing at the rate of $\frac{8}{3} \ cm/s$.

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

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