• Oct 25th 2009, 02:00 PM
kashifzaidi
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 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)
• Oct 25th 2009, 02:19 PM
skeeter
Quote:

Originally Posted by kashifzaidi
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 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)

make a sketch of the problem.

let $\displaystyle x$ = horizontal distance ... $\displaystyle \frac{dx}{dt} = 1.5$ ft/sec

let $\displaystyle \theta$ = angle

determine the trigonometric relationship equation using $\displaystyle x$ , $\displaystyle \theta$, and the 10 ft ladder.

take the time derivative, sub in your given/calculated values, then determine $\displaystyle \frac{d\theta}{dt}$
• Oct 25th 2009, 02:43 PM
You want to find: $\displaystyle \frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt}$
$\displaystyle x=10cos(\theta)$
$\displaystyle \frac{dx}{d\theta}=-10sin(\theta)$
$\displaystyle \frac{d\theta}{dx}=\frac{-csc(\theta)}{10}$
I think that's the trick you're looking for. Just plug this into the equation for $\displaystyle \frac{d\theta}{dt}$.
To find the numerical answer, you can use the fact that you are given $\displaystyle 6=10cos(\theta)$ at this point.