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.


1 Answer 1


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 $\frac {180}\pi$

  • $\begingroup$ So basically I have to convert my degrees to radians for this to be correct, and I would have to do this for both D and dD/dt? $\endgroup$
    – boidkan
    Commented Apr 7, 2014 at 4:10
  • $\begingroup$ You need you units to be consistent. For $D$, that can stay in degrees-you are evaluating trig functions of $D$ and as long as you do that in degrees you are fine. The formulas for derivatives of trig functions assume the argument is in radians. You can define $d=\frac \pi{180}D$ for the angle in radians and use the chain rule. I wouldn't checkmark my answer when you have a question this significant. An upvote is fine, but a checkmark can scare away other answers. With this explanation, I hope you get it. If not, ask again. $\endgroup$ Commented Apr 7, 2014 at 5:01

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