# Related Rates Hot Air Balloon

I have a related rates problem on a hot air balloon that is rising and I am asked to determine the rate of change in the angle. I'm having difficulties developing a relationship.

Here is the question:

So far my attempt at this question is the following and I am unsure if it's correct or not.

$$Tan\theta = \frac{y}{450}$$ $$\frac{d}{dt} (Tan\theta) = (\frac{d}{dt})(\frac{y}{450})\cdot\frac{dh}{dt}$$

$$\frac{d}{d\theta}\cdot sec^2\theta=\frac{1}{450}\cdot 2$$

I found $sec^2$ through the pythagorean thereom; $$sec^2\theta = (\frac{h}{a})^2 = 1.444$$

placing it back into

$$\frac{d}{d\theta}\cdot 1.444=\frac{1}{450}\cdot 2$$

$$\frac{d}{d\theta} = \frac{2}{450\cdot(1.444)}$$

giving a final answer of $0.003077 radians$

In degrees $Tan^{-1}(0.003077) = 0.176^\circ$

• If you differentiate both sides you get $\frac{d\theta}{dt} \sec^2 (\theta)=(1/450)\frac{dy}{dt}=2(\frac{1}{450})$. If you use pythag and the fact that sec is 1 over cosine you get $\sec^2(\theta)=1.44$ So you are correct. If you go the path which I showed you will get the same thing. However, you converting to degrees is incorrect. To convert to degrees multiply by $\frac{180 \text{degrees}}{\pi \text{rad}}=1$. – Ahmed S. Attaalla Aug 21 '16 at 3:45
• Actually you got approx. the correct answer in degrees, you just showed an incorrect method for finding it. You got lucky. – Ahmed S. Attaalla Aug 21 '16 at 3:53
• For example $\tan^{-1}(\pi/4)=\sqrt{2}/2$ yet we know $\frac{\pi}{4}$ radians is 45 degrees. Why you got lucky is that $\arctan(x) \approx x$ in radians $x$, for $x$ close to $0$, and your calculator interpreted it as a degree. – Ahmed S. Attaalla Aug 21 '16 at 4:03

Call the height of the ballon $h$ by simple trigonometry we have:
$$\tan(\theta)=\frac{h}{450}$$
$$\theta=\arctan(\frac{h}{450})$$
Differentiate both sides with respect to time $t$.Remember you are trying find $\frac{d\theta}{dt}$ when $h=300$.
Also if you want to write $\frac{d\theta}{dt}$ completely in terms of $t$ can you write $h$ in terms of $t$? Hint: distance over time is your average speed, in our case we have a constant vertical speed.