Question

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 π3π3 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

Answer 1

Answer:

$\frac{dy}{dt}=1.2\frac{mi}{min}$

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

$tan\theta=\frac{y}{x}\\y=xtan\theta(1)$

Differentiating (1) with respect to time, we get:

$\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}$

$\frac{dx}{dt}=0$ since x is a constant value. Replacing:

$\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}$