# Thread: Rate of Change With Chief Sohcahtoa

1. ## Rate of Change With Chief Sohcahtoa

A hot air balloon is rising straight up from a level field and is tracked by a range finder 500ft from the lift point. At the moment the range finder's elevation angle is pi/4, the angle is increasing by .14 radians per minute. How fast is the balloon rising at that instant.

So using the the trigonometric laws I know that the current height is equal to tan θ = op/adj which for this problem would be tan pi/4 = opp/500. I can use that to get the height, but can I also use that for the derivative to calculate the height?

I guess I'm not really sure how to relate the change in angle towards the change in height necessarily.

I'm also a little thrown off by the use of radians. Would it be advantageous to convert to degrees and then convert back to radians at the end?

So far the only guess I have at a derivative would be sec^2 θ * dθ/dt = dh/dt / 500, but I'm pretty sure that's pretty wrong.

2. You're on the right track.

$\frac{y}{500}=tan{\theta}$

$y=500tan{\theta}$

$\frac{dy}{dt}=500sec^{2}{\theta}\frac{d{\theta}}{d t}$

$\frac{dy}{dt}=500sec^{2}(\frac{\pi}{4})\cdot\frac{ 7}{50}$

3. Originally Posted by Menudo
A hot air balloon is rising straight up from a level field and is tracked by a range finder 500ft from the lift point. At the moment the range finder's elevation angle is pi/4, the angle is increasing by .14 radians per minute. How fast is the balloon rising at that instant.
Why do you doubt yourself?

Code:
                 .
/|
/  |
/    |
/      h
/        |
/          |
/  θ         |
--o-------500----+
As you observed, we have

$h = 500\tan\theta$

So, we differentiate implicitly with respect to time $t$,

$\frac{dh}{dt} = 500\sec^2\theta\frac{d\theta}{dt}$

When the angle is $\frac\pi4$, we have $\frac{d\theta}{dt} = 0.14$, so we substitute:

$\frac{dh}{dt} = 500\cdot0.14\sec^2\left(\frac\pi4\right)$

$= 500\left(\frac{14}{100}\right)\left[\frac1{\cos^2\left(\frac\pi4\right)}\right]$

$= 5\cdot14\cdot\left(\sqrt2\right)^2 = 70\cdot2 = 140$

If you carry through the units, you should see that we have feet per minute, so the balloon is rising at a rate of $140\text{ ft/min}$.

4. Wow, thank you so much. I'm glad to know I was on the right track. I'll try to trust my instincts a little better when I suspect they're well founded.

### a hot air balloon rises straight up from a level field

Click on a term to search for related topics.