Deriving the Rate of Change of Angle in a Rocket Launch Using Differentiation

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Homework Help Overview

The problem involves a rocket launch where the altitude is defined by a quadratic function of time, and participants are tasked with determining the rate of change of the angle of elevation from a fixed point on the ground. The subject area includes calculus, specifically differentiation and related rates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their differentiation processes and question the validity of their results. There are attempts to clarify the relationships between the variables involved, particularly the angle θ and the height h of the rocket. Some express uncertainty about the magnitude of their calculated rates of change.

Discussion Status

The discussion is active with participants sharing their calculations and expressing doubts about their results. There is a recognition of the need to differentiate correctly and convert units appropriately, indicating a collaborative effort to refine their understanding of the problem.

Contextual Notes

Some participants mention past experiences with grading that emphasize the importance of showing work clearly. There is also a note about the potential for point deductions if differentiation steps are not adequately presented.

Bostonpancake0
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The question is as follows:

A rocket was launched straight up, and its altitude is given by h = 10 t2 m after t
seconds. You are on the ground 300 m from the launch site watching the rocket going
up. 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 10 seconds after the launch?


I guess my question is whether my understanding of the problem and working is correct. I'm not 100% confident on my process and my answer and am just wondering if I have missed anything important.

By working and answer is tagged below.
 

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Looks okay when my head is turned sideways. If you want to be picky, you could show working out for how you differentiated to get dh/dt. It's not necessary but the grader might deduct points off that, in which I had suffered first hand experience from that.
 
Ahhh yes thanks for that missed it, alright its just that 11degrees a second seems rather large? I tried it another way and got d(theta)/d(t)=60*t/t^4+900...then subbing t gives me d(theta)/d(t)=0.055046 degrees/sec? doesn't this seem like a more likely answer?
 
I think the left side of your page should be dθ/dh

When t = 5 secs, h = 10(5)^2 = 250m
When t = 10 secs, h = 1000m
When t = 15 secs, h = 2250m
When t = 20 secs, h = 4000m
 
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If you post an image of your work, at least do us the courtesy of posting it right side up instead of rotated 90 deg.
 
Hi I think I am doing the same assignment as you. I got the same values that dh/dt= 20t, d(theta)/d(h)= 60t/t^4+900 and multiplying them gets d(theta)/d(t)= 1200/109. But I agree, I thought it was too high. But then i tried differentiating d(theta)/d(h) and then subing in h=10t^2 afterward and it got a much lower answer which i believe is correct. Because the original way, we were differentiating dh/dt at the same time as d(theta)/d(h). DON'T FORGET TO CONVERT INTO RADIANS THOUGH! You didn't do this on the sheet. Good luck!
 
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