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

**Problem 39**

A balloon, leaving the ground 60 ft from an observer, rises 10 ft/sec. How fast is the angle of elevation of the line of sight increasing, after 8 seconds?

**Solution 39**

$\theta = \arctan (10t/60)$

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

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

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*

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