Question

A leader 20 ft long leans against a vertical wall.the top end slides downwards at the rate of 2 ft per second.the rate at which the lower end moves on a horizontal floor when it is 12 ft from the wall is​

Answer 1

Step-by-step explanation:

Assume the angle at an instant is \theta (see diagram) . given

$\frac{d}{dt}( l \sin( \theta) ) = l \cos( \theta) \frac{d \theta}{dt} = 2 = const.$

Given at a instant

$l ( \cos \theta ) =12 \implies \\ \cos \theta = {12 \over20} = {3 \over5} \: \: and \: \: \sin( \theta) = {4 \over \: 5}$

use first formula . at this instant

$\frac{d \theta}{dt} = \frac{2}{12 \cos( \theta) } = {5 \over 18}$

$\frac{d}{d \theta} (12 \cos( \theta) ) = - 12 \sin( \theta) \frac{d \theta}{d \theta} = - 12 \times {4 \over5} \times \frac{5}{18} = \frac{8}{3} ftps$

we need to find

$\frac{d}{d \theta} (12 \cos( \theta) ) = - 12 \sin( \theta) \frac{d \theta}{d \theta}$

Negative sign shows opposite direction.

*****M 2

Velocity component along ladder must be same since it length won't change so let velocity v then

$v \cos( \theta) =2 \cos( \frac{\pi}{2} - \theta ) = 2 \sin( \theta) \\ v = 2 \tan( \theta) = 2 \times \frac{ \sqrt{ {20}^{2} - {12}^{2} } }{12} = 2 \times \frac{16}{12} = \frac{8}{3}$