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I have the following problem:

A hot air balloon rising straight up from a level field is tracked by a range finder $150$ meters from the liftoff point. At the moment that the range finder’s elevation angle is $\frac{\pi}{4}$, the angle is increasing at the rate of $0.14$ rad/min. How fast is the balloon rising at that moment?

My development was:

Let $h$ the altitude of the hot air balloon, $\theta$ the angle.

Using trigonometry, i got: $\sin(\theta) \cdot 150\sqrt{2} = h$, where $150\sqrt{2}$ is the hypotenuse.

Using implicit derivation respect to the time or moment $t$, to get:

$\frac{d}{dt}\sin(\theta) \cdot 150\sqrt{2}=\frac{d}{dt}h$

Since $\sin(\theta)$ is a composition of the functions $\sin(x)$ and $\theta(t)$ i need to use chain rule, so i have: $\cos(\theta) \cdot \frac{d}{dt}\theta \cdot 150\sqrt{2}=\frac{d}{dt}h \implies \frac{d}{dt}h=21$.

But the correct answer is $42$ that is exactly the double of my answer, what is wrong with my development? Thanks in advance.

ESCM
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1 Answers1

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When you write $\sin(\theta) \cdot 150\sqrt{2} = h$ you are implicitly assuming that the hypotenuse isn't changing, which is false. If you were going to set it up like this you would need to say $\sin(\theta)\cdot c = h$ where here $c$ is the hypotenuse. You can then use the fact that $c^2 = 150^2 + h^2$ to eliminate $c$ and then proceed with differentiating with respect to $t$.

I should also note that another route you could take is to say $\tan(\theta) = \frac{h}{150}$ and go from there. In that case you wouldn't need to worry about the hypotenuse.

DMcMor
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