Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast is the angle 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 is 8 ft from the wall.) rad/s

Answer 1

Answer:

$-\frac{d\theta}{dt} = 0.18 rad/s$

Explanation:

As we know that the length of the ladder is given as

$L = 10 ft$

now at any instant of time let the ladder is at distance "x" from the vertical wall

then the angle made with the horizontal for the ladder is given as

$cos\theta = \frac{x}{L}$

now differentiate both sides with respect to time

$-sin\theta (\frac{d\theta}{dt}) = \frac{1}{L} \frac{dx}{dt}$

so here we have

$-\frac{d\theta}{dt} = \frac{1}{Lsin\theta}(\frac{dx}{dt})$

given that

$\frac{dx}{dt} = 1.1 ft/s$

$cos\theta = \frac{8}{10}$

$\theta = 37 degree$

now we have

$-\frac{d\theta}{dt} = \frac{1}{10 sin37}(1.1)$

$-\frac{d\theta}{dt} = 0.18 rad/s$