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Question
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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Solution
If the foot of the ladder is at distance x from the wall and the top is at a height y at any instant of ti,e t, then
(5 m)2 = x2 + y2 ....(i)
Differentiating (i) w.r.t. t, we have
`= d/dt (25 m^2) = 2x dx/dt + 2y dy/dt` ....(ii)
we have `dx/dt = 0.02 m //sec`
x = 4 m and y = `sqrt(25-4^2) m = 3`
`(∵ x^2 + y^2 = 25 m^2, y = sqrt (25 = x^2)) m`
Hence from (ii), we get,
`0 = 2 xx 4 m xx 0.02 m// sec + 2 xx 3 dy/dt`
`= dy/dt = -0.16/6` m/sec
∴ Rate of decrease of height on the wall
`= 16/600 m// sec = 1600/600 cm//sec = 8/3 cm//sec.`
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