Question

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is ______.

विकल्प

• 1/10` radian/sec

• 1/20 radian/sec

• 20 radian/sec

• 10 radian/sec

Answer 1

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is 1/20 radian/sec.

Explanation:

Length of ladder = 5 m

Let AB = y m and BC = x m

∴ In right ΔABC,

AB² + BC² = AC²

⇒ x² + y² = (5)² 

⇒ x² + y² = 25

Differentiating both sides w.r.t x, we have

`2x * "dx"/"dt" + 2y * "dy"/"dt"` = 0

⇒ `x "dx"/"dt" + y * "dy"/"dt"` = 0

⇒ `2 * "dx"/"dt" + y xx (-0.1)` = 0  ....[∵ x = 2m]

⇒ `2 * "dx"/"dt" + sqrt(25 - x^2) xx (-0.1)` = 0

⇒ `2 * "dx"/"dt" + sqrt(25 - 4) xx (-0.1)` = 0

⇒ `2 * "dx"/"dt" - sqrt(21)/10` = 0

⇒ `"dx"/"dt" = sqrt(21)/20`

Now cos θ = `"BC"/"AC"`  ....(θ is in radian)

⇒ cos θ = `x/5`

Differentiating both sides w.r.t. t, we get

`"d"/"dt" cos theta = 1/5 * "dx"/"dt"`

⇒ `- sin theta ("d"theta)/"dt" = 1/5 * sqrt(21)/20`

⇒  `("d"theta)/"dt" = sqrt(21)/100 xx (- 1/sin theta)`

= `sqrt(21)/100 xx -(1/("AB"/"AC"))`

= `- sqrt(21)/100 xx "AC"/"AB"`

= `- sqrt(21)/100 xx 5/sqrt(21)`

= `- 1/20` radian/sec

[(–) sign shows the decrease of change of angle]

Hence, the required rate = `1/20` radian/sec