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 1.3 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.)

Answer 1

Check the picture below.

so, when r = 10 and x = 8, the vertical distance is 6, namely in pythagorean theorem lingo, b = 6.

let's keep in mind that, the ladder is not growing any longer or shrinking, and therefore is constantly always just 10, that matters, since is just a scalar value and also because the derivative of a constant is 0.

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

$\bf -sin(\theta )\cfrac{d\theta }{dt}=-\cfrac{1}{10}\cdot \cfrac{dx}{dt}\implies \cfrac{d\theta }{dt}=-\cfrac{1}{-10sin(\theta )}\cdot \cfrac{dx}{dt} \\\\\\ \begin{cases} \frac{dx}{dt}=1.3\\ sin(\theta )=\frac{6}{10} \end{cases}\implies \cfrac{d\theta }{dt}=-\cfrac{1}{10\cdot \frac{6}{10}}\cdot 1.3\implies \cfrac{d\theta }{dt}=-\cfrac{1.3}{6}~radians$