Rocket Angle Change Rate Calculation

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Homework Help Overview

The problem involves calculating the rate of change of the angle between a bystander and a rocket that is ascending vertically. The rocket is currently at a height of 1/2 miles and is moving at a speed of 40 miles per hour, while the bystander is positioned 1 mile horizontally from the launch point.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up a triangle to relate the height of the rocket and the distance from the bystander. There are attempts to differentiate the angle with respect to time, and questions arise about the necessity of knowing the angle itself. Some suggest using the Pythagorean Theorem to find relationships without directly calculating the angle.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants are questioning the adequacy of the variables defined, while others are reiterating similar methods. There is no explicit consensus, but guidance has been offered regarding the use of trigonometric identities and differentiation.

Contextual Notes

Participants note the challenge of solving the problem without a calculator and the need to express the rate of change in radians per second. There is also mention of the specific values involved, such as the height of the rocket and the horizontal distance from the bystander.

physstudent1
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Homework Statement


A rocket is 1/2 miles in the air going 40mi/h a bystander is standing 1 mile away from where the rocket took off straight up. What is the rate of change of the angle that the bystander makes with the rocket.


Homework Equations





The Attempt at a Solution



I set up a triangle and got tan(theta)=y. Then I took the derivative and got dtheta/dt * sec^2(theta) = dy/dt. from here I'm kinda lost it seems like i won't be able to do it now because I have to do this without a calculator so how would I calculate the theta at the time where y=1/2 its not a common angle...?
 
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You're right that it's not a common angle, but do you really need the angle \theta? You're looking for \sec(\theta) so why don't you use the Pythagorean Theorem, figure out what your hypotenuse is, and then you can find \sec(\theta) without having to know \theta
 
Well, one of your problems is that you do not have enough variables. Try defining the 1/2 miles part x and set up a tan theta= x/1 equation to differentiate.
 
How do you figure there aren't enough variables? With the derived equation one can very easily solve for \frac{d\theta}{dt} which is what we desire. Furthermore, you'll notice that what you suggested, is EXACTLY what physstudent1 did.

physstudent1 said:
I set up a triangle and got tan(theta)=y. Then I took the derivative and got dtheta/dt * sec^2(theta) = dy/dt.
 
Ok,
define the 1/2 mile part to be x. Now set up the equation tan(theta)=x/1. Differentiate and get sec^2(theta)*dtheta/dt=dx/dt. Solve the triangle at the time when x = 1/2, and you find that the hypotenuse is sqrt(5)/2. Solve for sec(theta)^2. Plug in your 40 mph for dx/dt and rearrange the varibles.

Does that make sense, Kerizhn? I don't know that I did it right, but I get an answer for the problem.
 
Oh,
Remember to divide your dtheta/dt by 60 to get radians per second change.
 
Younglearner, I'm not sure if you're trying to show us a different way of doing things here, but you emulated precisely what physstudent 1 did in the original post, only with a different variable name.

Furthermore, you just proved that physstudent1 did indeed have enough variables. I'm not exactly sure what the point of just repeating everything we've already done is...
 

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