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

In summary, when t = 5 secs, h = 10(5)^2 = 250m, when t = 10 secs, h = 1000m, when t = 15 secs, h = 2250m, and when t = 20 secs, h = 4000m.
  • #1
Bostonpancake0
42
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
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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  • #5
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.
 
  • #6
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 origional 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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1. What is differentiation and why is it important in science?

Differentiation is a mathematical process that calculates the rate of change of a variable. It is important in science because it allows us to analyze and understand the behavior and relationships between variables in a system.

2. How is differentiation applied in real-world scenarios?

Differentiation has many real-world applications, including in physics, biology, economics, and engineering. It is used to model and analyze various phenomena such as motion, growth, optimization, and rates of change.

3. What are the different types of differentiation?

The two main types of differentiation are implicit differentiation, which involves finding the derivative of a function that is not explicitly defined, and explicit differentiation, which involves finding the derivative of a function that is explicitly defined.

4. What are the limitations of differentiation?

One limitation of differentiation is that it assumes a continuous, smooth relationship between variables, which may not always be the case in real-world scenarios. It also cannot take into account sudden changes or discontinuities in a system.

5. How does differentiation relate to integration?

Differentiation and integration are inverse operations, meaning that they are two sides of the same coin. Integration is used to find the original function from its derivative, while differentiation is used to find the rate of change of a function from its original function.

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