# Related rates

1. Jun 12, 2010

### nefliege

1. The problem statement, all variables and given/known data
A rocket has launched straight up, and its altitude is h = 10t2
feet after t seconds. You are on the ground 1000 feet from the launch site. The line
of sight from you to the rocket makes an angle θ with the horizontal. By how many
Radians per second is θ changing ten seconds after the launch?

It's a problem form MIT OpenCourseWare, and I know the solution is given, but I did it my way and (suprisingly) it doesn't work. What did I wrong?

It's not a homework question, but I didn't want to spoil your Mathematics section :)
2. Relevant equations
$$\frac{dh}{dt}=20t$$
3. The attempt at a solution
$$\theta=arctan\frac{h}{1000}$$
$$\frac{d}{dt}(\theta=arctan\frac{h}{1000})$$
$$\frac{d\theta}{dt}=\frac{1}{1+\frac{h^{2}}{1000^2}}\frac{dh}{dt}$$
which is 100 ;/

Thanks for any help :)

2. Jun 12, 2010

### Staff: Mentor

No, $d \theta/dt$ is a variable quantity. At t = 10 sec. dh/dt = 200 ft/sec, h = 1000 ft, and theta = pi/4. By your calculation, d(theta)/dt is 100 when t = 10 sec.

Your mistake is in your derivative of arctan(h/1000). You forgot to use the chain rule, so your value for the derivative is too large by a factor of 1000.

3. Jun 13, 2010

### nefliege

I'm so stupid ! the last part (dh/dt) should be:
$$\frac{d}{dt}(\frac{h}{1000})=\frac{1}{1000}\frac{dh}{dt}$$
Right ?
And thank you for help :) I'd been already really upset; I thought the whole solution was wrong and didn't know why.

And of course "100" was only when t=10s. I just hadn't written it.

4. Jun 13, 2010

### Staff: Mentor

Yes, that's right.

5. Jun 13, 2010

thanks :)