Ok, so I'm going over a review worksheet for an exam next week and I'm not sure if I solved this problem correctly.
You observe a model rocket launch. You are standing at a position 19meters from the point of where the rocket launches from, and you track the angle of elevation of the rocket as it flies. At one point the angle of elevation is 53degrees and increasing at .3deg/s. Find the rocket's speed at that time.
Ok, so this is what I have. ${d\theta/dt=.3^{\circ}/s}$ and ${x=19}$. I want to find ${dy/dt}$ when ${\theta=53^{\circ}}$.
So I know that ${tan(\theta)=y/x}$ and the derivative of that in regards to t is ${sec^2(\theta)*d\theta/dt = 1/19 * dy/dt}$. Thus, ${dy/dt = 19sec^2(53)*.3^{\circ}/s = 15.7379733639629367}$.
Does this look correct? I'm also a bit confused about finding the rocket's "speed" since I am under the impression that velocity is speed and a direction, not just speed.
Best Answer
The unstated assumption is that the rocket travels vertically. If you have watched model rockets fly, that is not a good assumption, but we will accept it. This links speed to velocity as you have the direction. You should define your axes and variables-it appears $y$ is the altitude and $D$ is the elevation angle where you observe the rocket. It would be more common for $D$ to be the distance to the rocket, which is why it is good to define your variables.
You are correct that $\tan D =\frac yx$ but your formula for the derivative of the tangent assumes that the argument is radians, not degrees. As long as you evaluate the secant in degrees, that is not a problem, but your $\frac {dD}{dt}$ is off by a factor $\frac {180}\pi$