Tips for Solving the Doomsday Equation

  • Thread starter Laxer42
  • Start date
In summary: Right, but for ##\alpha > 1## I would introduce a new constant ##\beta = \alpha -1 > 0## to tidy things up. What can you then observe as ##t\to...##?
  • #1
Laxer42
8
0
Homework Statement
Solve the “Doomsday equation”

dN/dt = N^α

and show that for α > 1, the solutions approach a vertical asymptote in finite time. What happens for α < 1?
Relevant Equations
n/a
dN/dt = N^α
int(N^-a)=int(dt)
[N^(1-a)]/[1-a]+c=tI believe this is a solution, but it does not produce the results I am told to be looking for.
 
Physics news on Phys.org
  • #2
Laxer42 said:
but it does not produce the results I am told to be looking for.
Did you try solving for N?
 
  • #3
Orodruin said:
Did you try solving for N?
Yes. I didn't think that helped though.
 
  • #4
Laxer42 said:
Yes. I didn't think that helped though.
It also does not help that you do not tell us what you got.
 
  • #5
Orodruin said:
It also does not help that you do not tell us what you got.
Thanks for being nice about it.

N=(t-at)^(1/(1-a))
 
  • #6
Laxer42 said:
N=(t-at)^(1/(1-a))

What happened to the integration constant?

Laxer42 said:
Thanks for being nice about it.
You are welcome for having a stranger trying to help you for free. The least you can do when asking people to help you for free is to make it easier for those people to help you. If you just say "it did not help me" you are not furthering your own cause.
 
  • #7
Orodruin said:
What happened to the integration constant?You are welcome for having a stranger trying to help you for free. The least you can do when asking people to help you for free is to make it easier for those people to help you. If you just say "it did not help me" you are not furthering your own cause.

I assumed the constant was equal to zero.

If you want to be rude because I've made honest mistakes, I don't think your help will be productive for either of us.
 
  • Skeptical
Likes sysprog
  • #8
Laxer42 said:
I assumed the constant was equal to zero.
You cannot arbitrarily set integration constants to zero. What do you get if you do not assume this?

Laxer42 said:
If you want to be rude because I've made honest mistakes, I don't think your help will be productive for either of us.
Your problem here is your own arrogance, not me being rude. You are the one who wrongly interpreted a factual statement about the process (it does not help your learning process to keep people trying to help you in the dark regarding what you have done so far) for rudeness and decided to respond with a snarky remark.

FYI, people helping here generally know the answer but nobody here believes in just providing answers. We will want you to show your effort and we will help guide your thinking. We believe this is much more beneficial to you long term. However, in order to help you think, we need to know what you have bern thinking so far and you have been very reluctant in providing this information. It took me two additional points to get information from you that should have been provided in the OP. I am sure your mistakes are honest and I have no problem with that. What I do have a problem with is the impression you give of somehow being offended when I have been asking questions about your progress that we have no way of knowing unless you provide it.

Had you actually provided all the necessary information in your OP, this thread would have been over by now.
 
  • Like
Likes pinball1970 and SammyS
  • #9
Laxer42 said:
If you want to be rude because I've made honest mistakes
How exactly should someone correct your mistakes without being seen as rude?
 
  • Like
Likes Orodruin
  • #10
Borg said:
How exactly should someone correct your mistakes without being seen as rude?
"Did you solve for N?"
"Yes. I didn't think that helped though."
"Well show me what you got anyway."
...
"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."
 
  • #11
Orodruin said:
You cannot arbitrarily set integration constants to zero. What do you get if you do not assume this?
N=(t-at)^(1/(1-a))+c
 
  • #12
The constant appears correctly in your original post. If you solve for N, it will not be the same thing as first solving for N with the constant equal to zero and then adding a constant. I suggest you rename the constant ##t_0## and put it on the other side of the equation. If that does not help, please show us each step of when you solve for N.
 
  • Like
Likes sysprog
  • #13
Orodruin said:
The constant appears correctly in your original post. If you solve for N, it will not be the same thing as first solving for N with the constant equal to zero and then adding a constant. I suggest you rename the constant ##t_0## and put it on the other side of the equation. If that does not help, please show us each step of when you solve for N.

N=[(t-##t_0##)(1-α)]^(1/(1-α))
 
  • #14
Right, but for ##\alpha > 1## I would introduce a new constant ##\beta = \alpha -1 > 0## to tidy things up. What can you then observe as ##t\to t_0##?
 
  • Like
Likes sysprog
  • #15
Orodruin said:
Right, but for ##\alpha > 1## I would introduce a new constant ##\beta = \alpha -1 > 0## to tidy things up. What can you then observe as ##t\to t_0##?
N=[-(t-##t_0##)(B)]^(1/(1-α))

It would go to zero as ##t\to t_0##
 
  • #16
Laxer42 said:
So N=[-(t-##t_0##)(##\beta )]^(1/(1-α))##

It would go to zero as ##t\to t_0##
Would it? Introduce ##\beta## in the exponent too and remember that ##\beta > 0## so also ##1/\beta > 0##. I also suggest writing ##-(t-t_0)\beta## as ##\beta(t_0-t)## (the relevant times for the finite solution are ##t < t_0##).
 
Last edited by a moderator:
  • Like
Likes sysprog
  • #17
Laxer42 said:
"Did you solve for N?"
"Yes. I didn't think that helped though."

So let me tell you how that comes through when you do not provide your actual work:
"Did you solve for N?"
"Yes, of course I did, why are you not helping?"

The better way (faster and more informative) would have been:
"Did you solve for N?"
"Yes. I got X. I did not think it helped because Y"

Again, how you expect people to be able to help you without providing your actual work is beyond me and it is also against forum rules. You provided your actual work up until ##N^{1-\alpha}/(1-\alpha) + t_0 = t## so naturally the next question is about whether you tried to solve for ##N## or not. If you did you should then provide your work. It is not strange that people become irritated with you when they have to drag information out of you over the course of several posts rather than getting straight to the point.

Laxer42 said:
"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."
This should not even need pointing out to you. It is quite obvious that we cannot help you unless we do not undertand where your problem lies. Try to imagine how things seem from the other side.
 
  • Like
Likes pinball1970
  • #18
Laxer42 said:
"Did you solve for N?"
"Yes. I didn't think that helped though."
"Well show me what you got anyway."
...
"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."

None of those remarks are rude; they are meant to be helpful, and you would find them to be so if you were less defensive and more willing to work with helpers.
 

1. What is the Doomsday equation?

The Doomsday equation, also known as the Drake equation, is a mathematical formula created by astronomer Frank Drake to estimate the number of intelligent civilizations in our galaxy that we could potentially communicate with.

2. How do you solve the Doomsday equation?

The Doomsday equation cannot be solved definitively as it relies on many unknown variables. However, it can be used as a framework for estimating the number of potential intelligent civilizations in our galaxy by making educated guesses and assumptions about these variables.

3. What are the variables in the Doomsday equation?

The variables in the Doomsday equation include the average rate of star formation in our galaxy, the fraction of stars that have planets, the number of planets per star that could potentially support life, the fraction of those planets where life actually evolves, the fraction of those planets where intelligent life evolves, and the average length of time that these civilizations last.

4. Is the Doomsday equation accurate?

The Doomsday equation is not a definitive or accurate way to determine the number of intelligent civilizations in our galaxy. It is simply a way to think about and discuss the possibility of extraterrestrial life. The accuracy of the equation is highly dependent on the assumptions and guesses made for each variable.

5. What does the Doomsday equation tell us about the existence of extraterrestrial life?

The Doomsday equation does not provide a definitive answer about the existence of extraterrestrial life. It is simply a tool for considering the potential number of intelligent civilizations in our galaxy. However, the fact that there are so many unknown variables and uncertainties in the equation suggests that the existence of extraterrestrial life is still a mystery and requires further exploration and research.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
551
  • Calculus and Beyond Homework Help
Replies
16
Views
547
  • Calculus and Beyond Homework Help
Replies
1
Views
520
  • Calculus and Beyond Homework Help
Replies
2
Views
362
  • Calculus and Beyond Homework Help
Replies
3
Views
300
  • Calculus and Beyond Homework Help
Replies
7
Views
790
  • Calculus and Beyond Homework Help
Replies
8
Views
749
  • Calculus and Beyond Homework Help
Replies
6
Views
740
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top