A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is Ο€/3 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

Accepted Solution

check the picture below.

notice, the "x" distance, whilst the balloons are moving up up and up, is not changing, thus, is a constant.

[tex]\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{3}\implies tan(\theta )=\cfrac{1}{3}y \\\\\\ \stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{3}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot \cfrac{d\theta }{dt}=\cfrac{1}{3}\cdot \cfrac{dy}{dt}[/tex]

[tex]\bf \cfrac{3\cdot \frac{d\theta }{dt}}{cos^2(\theta )}=\cfrac{dy}{dt}\quad \begin{cases} \frac{d\theta }{dt}=0.1\\\\ \theta =\frac{\pi }{3} \end{cases}\implies \cfrac{3\cdot 0.1}{cos^2\left( \frac{\pi }{3} \right)}=\cfrac{dy}{dt}\implies \cfrac{0.3}{\left( \frac{1}{2} \right)^2}=\cfrac{dy}{dt} \\\\\\ \cfrac{0.3}{\frac{1}{4}}=\cfrac{dy}{dt}\implies \stackrel{rad/min}{1.2}=\cfrac{dy}{dt}[/tex]