Angle of Change: Rocket Launch 10 Seconds Later

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Homework Help Overview

The problem involves a rocket's altitude described by the equation h = 10t² feet after t seconds, with the observer positioned 1000 feet from the launch site. The goal is to determine the rate of change of the angle θ between the observer's line of sight to the rocket and the horizontal, specifically ten seconds after launch.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of derivatives to the angle θ using the arctangent function. There is an exploration of the chain rule in differentiation and its implications on the calculations.

Discussion Status

Some participants have offered guidance regarding the use of the chain rule in differentiation, suggesting that a mistake was made in the original calculation of the derivative. There is acknowledgment of the specific values of h and dh/dt at t = 10 seconds, which are relevant to the discussion.

Contextual Notes

Participants note that the problem is sourced from MIT OpenCourseWare and emphasize that it is not strictly a homework question, indicating a desire to engage with the material without directly seeking a solution.

nefliege
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Homework Statement


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?
The answer is 1/10

It's not a homework question, but I didn't want to spoil your Mathematics section :)

Homework Equations


\frac{dh}{dt}=20t

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 :)
 
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nefliege said:

Homework Statement


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?
The answer is 1/10

It's not a homework question, but I didn't want to spoil your Mathematics section :)

Homework Equations


\frac{dh}{dt}=20t

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 :)
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.
 
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.
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.
 
thanks :)
 

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