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#### karush

##### Well-known member

- Jan 31, 2012

- 2,929

If the foot of the ladder is pulled away from the building

at a constant rate of $\displaystyle \frac{2 in}{sec}$

how fast is the angle formed by the ladder and the ground changing in $\displaystyle\frac {rad}{sec}$

at the instant when the top of the ladder is $12 ft$ above the ground.

View attachment 1511

I started with

$\displaystyle\theta = cos^{-1}\left(\frac{\sqrt{169-x^2}}{13}\right)$

$\displaystyle\frac{d\theta}{dt}=\frac{dx}{dt} cos^{-1}\left(\frac{\sqrt{169-x^2}}{13}\right)$

I didn't know how to use the $\displaystyle \frac{2 in}{sec}$ after this