Question

A ladder 10 ft long rests against vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.7 of ft/s, now fast (in rad/s) is

the angle (in radians) between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the
angle's rate of change when the bottom of the ladder 8 ft from the wall.)
Wall

Answer 1

Check the picture below.

`cos(\theta )=\cfrac{x}{10}\implies \stackrel{chain~rule}{-sin(\theta )\cdot \cfrac{d\theta }{dt}}=\cfrac{1}{10}\cdot \cfrac{dx}{dt} -sin(\theta )\cdot \cfrac{d\theta }{dt}=\cfrac{1}{10}(0.7)`

`\cfrac{d\theta }{dt}=\cfrac{0.07}{-sin(\theta )}~\hfill \stackrel{\textit{when the ladder's bottom is 8ft, x = 8}}{sin(\theta )=\cfrac{8}{10}\implies sin(\theta )=\cfrac{4}{5}} \\\\\\ \cfrac{d\theta }{dt}=-\cfrac{0.07}{~~ (4 )/(5 ) ~~}\implies \implies \cfrac{d\theta }{dt}=-0.07\cdot \cfrac{5}{4}\implies {\Large \begin{array}{llll} \cfrac{d\theta }{dt}=-0.0875~(rad)/(s) \end{array}}`

the rate is negative because the angle is decreasing as the ladder slides outwards.