Question

A $13ft$. ladder is leaning against a wall when its base starts to slide away At the instant when the base is $12ft$. away from the wall, the base is moving away from the wall at the rate of $5ft/sec$. The rate at which the angle $\theta$ between the ladder and the ground is changing is

• A. $-\frac{12}{13}rad/sec.$
• B. $-1rad/sec.$
• C. $-\frac{13}{12}rad/sec.$
• D. $-\frac{10}{13}rad/sec.$

Answer 1

Answer: (B)

$-1rad/sec.$

The base
$b=13\cos \theta$
Hence
$\frac{db}{dt}=-13\sin \theta .\frac{d\theta }{dt}$
Now
$\sin \theta =\frac{5}{13}$ at that instant.
Hence
$5=-13.\left(\frac{5}{13}\right).\frac{d\theta }{dt}$
Or
$\frac{d\theta }{dt}=-1rads/sec$.