Mathematics

Problem

An observer walks directly away from the base of a 15-ft streetlight at a constant rate of 2 ft/s along level ground. The angle of elevation

\theta

(in radians) from the observer to the top of the streetlight satisfies

\tan\theta = 15/x

, where

x

is the observer’s distance (in feet) from the base of the streetlight. When

\theta = \pi/6

, what is

d\theta/dt

? a)

-\dfrac{2}{45}\text{ rad/s}

\dfrac{1}{30}\text{ rad/s}

-\dfrac{1}{30}\text{ rad/s}

-\dfrac{1}{15}\text{ rad/s}