A ladder 5 m long rests against a vertical w | vedclass

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Question

(B) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the top of the ladder from the floor. The length of the ladder is

Vedclass · Accepted Answer

$0.025$

Vedclass · Answer

(B) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the top of the ladder from the floor. The length of the ladder is $L = 5 \ m$.From the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.We are given that the top slides downwards at $10 \ cm/s = 0.1 \ m/s$,so $\frac{dy}{dt} = -0.1 \ m/s$.We want to find the rate of change of the angle $	heta$ between the ladder and the floor,where $\sin 	heta = \frac{y}{5}$.Differentiating with respect to $t$: $\cos 	heta \frac{d	heta}{dt} = \frac{1}{5} \frac{dy}{dt}$.When $x = 4 \ m$,$y = \sqrt{5^2 - 4^2} = 3 \ m$.Then $\cos 	heta = \frac{x}{5} = \frac{4}{5}$.Substituting the values: $\frac{4}{5} \frac{d	heta}{dt} = \frac{1}{5} (-0.1)$.$\frac{d	heta}{dt} = -\frac{0.1}{4} = -0.025 \ rad/s$.The rate of decrease is $0.025 \ rad/s$.

$A$ ladder $5 \ m$ long rests against a vertical wall. If its top slides downwards at the rate of $10 \ cm/s$,then the angle between the ladder and the floor is decreasing at the rate of how many $rad/s$ when its lower end is $4 \ m$ away from the wall?

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