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 ______.

Options

• 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

${\displaystyle 2x\cdot \frac{\text{dx}}{\text{dt}}+2y\cdot \frac{\text{dy}}{\text{dt}}}$ = 0

⇒ ${\displaystyle x\frac{\text{dx}}{\text{dt}}+y\cdot \frac{\text{dy}}{\text{dt}}}$ = 0

⇒ ${\displaystyle 2\cdot \frac{\text{dx}}{\text{dt}}+y\times (-0.1)}$ = 0  ....[∵ x = 2m]

⇒ ${\displaystyle 2\cdot \frac{\text{dx}}{\text{dt}}+\sqrt{25-x^2}\times (-0.1)}$ = 0

⇒ ${\displaystyle 2\cdot \frac{\text{dx}}{\text{dt}}+\sqrt{25-4}\times (-0.1)}$ = 0

⇒ ${\displaystyle 2\cdot \frac{\text{dx}}{\text{dt}}-\frac{\sqrt{21}}{10}}$ = 0

⇒ ${\displaystyle \frac{\text{dx}}{\text{dt}}=\frac{\sqrt{21}}{20}}$

Now cos θ = ${\displaystyle \frac{\text{BC}}{\text{AC}}}$  ....(θ is in radian)

⇒ cos θ = ${\displaystyle \frac{x}{5}}$

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

${\displaystyle \frac{\text{d}}{\text{dt}}\cos \theta =\frac{1}{5}\cdot \frac{\text{dx}}{\text{dt}}}$

⇒ ${\displaystyle -\sin \theta \frac{\text{d}\theta }{\text{dt}}=\frac{1}{5}\cdot \frac{\sqrt{21}}{20}}$

⇒  ${\displaystyle \frac{\text{d}\theta }{\text{dt}}=\frac{\sqrt{21}}{100}\times (-\frac{1}{\sin \theta })}$

= ${\displaystyle \frac{\sqrt{21}}{100}\times -\left(\frac{1}{\frac{\text{AB}}{\text{AC}}}\right)}$

= ${\displaystyle -\frac{\sqrt{21}}{100}\times \frac{\text{AC}}{\text{AB}}}$

= ${\displaystyle -\frac{\sqrt{21}}{100}\times \frac{5}{\sqrt{21}}}$

= ${\displaystyle -\frac{1}{20}}$ radian/sec

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

Hence, the required rate = ${\displaystyle \frac{1}{20}}$ radian/sec