MATH SOLVE

4 months ago

Q:
# 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

A:

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]

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]