Question

A ladder 10 meter long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 meters per seconds, find how fast the top of the ladder is sliding down the wall when the bottom is 6 meters away from the wall

Answer 1

Let AC be the ladder. BC = x be the distance of the bottom of the ladder from the wall and AB = y be the distance of the top of the ladder from the floor.

Then,  `("d"x)/"dt"` = 1.2 m/sec, AC =10 m, BC = 6 m .......[Given]

By Pythagoras theorem, we get

x² + y² = AC²

∴ y² = AC² – x²

∴ y² = (10)² – x²    .......(i)

Differentiating w.r.t. t, we get

`2y ("d"y)/"dt" = -2x ("d"x)/"dt"`

∴ `("d"y)/"dt" = (-x)/y*("d"x)/"dt"`

= `(-6(1.2))/y`   .......(ii)

Substituting x = 6 in (i), we get

y² = (10)² – (6)²

= 100 – 36

= 64

∴ y = 8

Substituting y = 8 in (ii), we get

`("d"y)/"dt" = ((-6)(1.2))/8`

= – 0.9 metre/sec

Thus, the top of the ladder is sliding down at the rate of 0.9 meters/sec.