A weather balloon V, rising vertically from a point P on the ground, is observed from a point O on the ground, 90 m from P. Let θ be the angle between OP and OV (i.e. the angle between the ground and the sightline to the balloon). At what rate is the balloon rising when θ=45˚ and θ is increasing at 1˚/second?
So, I'm thinking the job is to find:
d(VP)/dt at the time when dθ/dt=1.
We already know that
So I take the derivative of both functions, accounting for the fact that both VP and θ are functions of time as well as of each other:
Knowing that dθ/dt=1 and OP=90, I get that:
Which is 1/180. I notice that if I flip this over and solve for d(VP)/dt in the other equation, I get that
Is this right?? If so, should I have been able to see the from the start? A lot of steps in this one, and I feel like I'm kinda guessing as to what I should do.
However, this statement is only true when x is expressed in radians. The derivatives/anti-derivatives of the trig functions are actually different if you express them in degrees.
For example, if you REALLY wanted to take a derivative in degrees, you'd actually get this:
You could use this to solve this problem, but instead it's generally a lot easier to convert everything to radians first!