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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

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प्रश्न

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?

योग
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उत्तर

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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अध्याय 6: Application of Derivatives - Exercise 6.1 [पृष्ठ १९८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.1 | Q 10 | पृष्ठ १९८

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