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

`(dy)/(dt)=1.2(mi)/(min)`

Step-by-step explanation:

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

`tan\theta=(y)/(x)\\y=xtan\theta(1)`

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

`(dy)/(dt)=tan\theta(dx)/(dt)+xsec^(2)\theta(d\theta)/(dt)\\`

`(dx)/(dt)=0` since x is a constant value. Replacing:

`(dy)/(dt)=3mi(sec^(2)(\pi)/(3))0.1(rad)/(min)\\(dy)/(dt)=1.2(mi)/(min)`