Related rates - How fast is the balloon rising?
A hot air balloon rising vertically is tracked by an observer located 4 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is π/5 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
1 Answer
Let $\theta$ be the angle between the observer's line-of-sight and the horizontal, and $h$ be the height of the balloon. Then
\[\tan \theta= \frac{h}{4}\]
\[\Rightarrow h=4\tan \theta (\text{in miles}).\]
Hence
\[\frac{dh}{dt}=4 (\tan \theta) \frac{d\theta}{dt}=4 \sec^2 \theta \frac{d\theta}{dt}.\]
So we have
\[\frac{dh}{dt}=4 \sec^2 \theta \frac{d\theta}{dt}=4 \sec^2 (\frac{\pi}{5}) \times 0.1 \]
\[=4 (1.2360)^2 \times 0.1=0.6110 \text{mile/min}.\]
So the baloon is rising with an speed about $0.61$ mile/min.
![Daniel90](https://matchmaticians.com/storage/user/100012/thumb/matchmaticians-c8zjsb-file-4-avatar-512.jpg)
-
Please leave a comment if you have any questions. Also this took me about 15 minutes to answer. Please consider offering bounties for your future posts.
- 1 Answer
- 230 views
- Pro Bono
Related Questions
- Calculus 2 / Calculate the surface of F
- Rainbow Vectors
- Prove that $\int _0^{\infty} \frac{1}{1+x^{2n}}dx=\frac{\pi}{2n}\csc (\frac{\pi}{2n})$
- Applications of Integration [Calculus 1 and 2]
- Reduction formulae
- Custom Solutions to Stewart Calculus Problems, 9th Edition
- Basic calc question
- Does $\lim_{(x,y)\rightarrow (0,0)}\frac{(x^2-y^2) \cos (x+y)}{x^2+y^2}$ exists?