39 - Rate of increase of angle of elevation of the line of sight

$\theta = \arctan \frac{1}{6}t$

$\dfrac{d\theta}{dt} = \dfrac{\frac{1}{6}}{1 + \frac{1}{36}t^2}$
 

after t = 8 sec
$\dfrac{d\theta}{dt} = \dfrac{\frac{1}{6}}{1 + \frac{1}{36}(8^2)}$

$\dfrac{d\theta}{dt} = \dfrac{\frac{1}{6}}{1 + \frac{64}{36}}$

$\dfrac{d\theta}{dt} = 0.06 \, \text{ rad/sec}$           answer