Okay, so first we need to draw a picture. You can turn radians into degrees by multiplying 180/π and degrees into radians by multiplying by π/180.
So, here's the picture
O
- []
| |
| |
4m/s | \O/
| | |
| |____________________Θ/\
|----200m--------|
Just imagine there is a diagonal between the balloon and the person and you have the triangle.
Notice that the balloon is traveling up at a constant rate of 4m/sec, so we can use this to calculate the height of the balloon at 30sec using this equation
speed = distance/time, speed * time = distance, 4m/sec * 30sec = 120m
We will need this in order to solve for the rate of change in the angle.
Now, we can use the tan(Θ) = Opposite/Adjacent to solve for what the angle is at 30sec as well.
\[\tan (\Theta)=O/A, \tan(\Theta)=120m/200m, \tan^{-1}(120m/200m) = \Theta \]
So, Θ=30.9638 degrees @ t=30sec
Now, will make a little change to the diagram by making the opposite side of the angle a variable, like so:
O
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| |
| |
y | \O/
| | |
| |____________________Θ/\
|----200m--------|
because we know as the balloon rises its height relative to the ground is changing as well as the angle. So, now we have this equation:
\[\tan(\Theta)=y/200m\]Take the derivative
\[\sec ^{2}(Θ)Θ\prime=y\prime/200m\]We know y'=4m/sec and theta = 30.9638 degrees at t=30sec so all we do is plug those in and solve for the rate of change in theta, i.e. Θ'
\[Θ\prime=(4m/s) / (200m*\sec^{2}(30.9638))=(4m/s * \cos^{2}(30.9638))/200m\]
because sec^2(Θ) = 1 / cos^2(Θ)
So,\[Θ'=.0147059\]degrees/sec or \[Θ'=.0147059*π/180=0.257\]radians/sec