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**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:

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/|
/ |
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/ h
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/ θ |
--o-------500----+

As you observed, we have

$\displaystyle h = 500\tan\theta$

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

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

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

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

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

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

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