Hello again, Raj!

3. A boy is 20 m away from a rising balloon.

The angle of elevation is increasing at a rate of 3 degrees/sec.

At the instant the balloon is 8 m above the ground, find its speed of ascent. Code:

*
* |
* |
* | h
* |
* θ |
* - - - - - *
20

We have: .$\displaystyle \tan\theta \:=\:\frac{h}{20} \quad\Rightarrow\quad h \:=\:20\tan\theta$

Differentiate with respect to time: .$\displaystyle \frac{dh}{dt} \:=\:20\!\cdot\!\sec^2\!\theta\!\cdot\!\frac{d\the ta}{dt}$ .[1]

We are given: .$\displaystyle \frac{d\theta}{dt} \:=\:3\text{ deg/sec} \:=\:\frac{\pi}{60}\text{ radians/sec}$

We are told that $\displaystyle h = 8$.

The right triangle looks like this: Code:

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* |
* | 8
* θ |
* - - - - - - - *
20

We have: .$\displaystyle opp = 8,\;adj = 20$

Pythagorus says: .$\displaystyle hyp = \sqrt{464} = 4\sqrt{29}$

Hence: .$\displaystyle \sec\theta \:=\:\frac{hyp}{adj} \:=\:\frac{4\sqrt{29}}{20} \:=\:\frac{\sqrt{29}}{5} $

Substitute into [1]: .$\displaystyle \frac{dh}{dt}\;=\;20\left(\frac{\sqrt{29}}{5}\righ t)^2\!\left(\frac{\pi}{60}\right) \;=\;\frac{29\pi}{75}\text{ m/sec} $