Determine what happens to the rate of change of the population over time

  • Thread starter ttpp1124
  • Start date
  • #1
110
4
Homework Statement:
I'm not sure how to start question b.) I understand that I have the denominator powered to the square, it's function "grows faster" than the function in the numerator.
Relevant Equations:
n/a
IMG_4247.jpg
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,970
7,209
it's function "grows faster" than the function in the numerator.
Does it grow? What happens to e-t as t increases?
Btw, you have a sign error in part a. Correction: no I had a sign error.
 
Last edited:
  • #3
epenguin
Homework Helper
Gold Member
3,873
897
I am not seeing this sign mistake and think your answer is correct. However it was unnecessary to use the full formula for derivative of u(x)/v(x) - since u is just a constant you only needed that for 1/v(x).

You are not thinking about the rest in quite the right way. What happens to e-t as t increases and as it becomes very large? Alternatively you get something that might be mere self-evident to you if you divide top and bottom of the fraction by e-t.
 
  • #4
110
4
I am not seeing this sign mistake and think your answer is correct. However it was unnecessary to use the full formula for derivative of u(x)/v(x) - since u is just a constant you only needed that for 1/v(x).

You are not thinking about the rest in quite the right way. What happens to e-t as t increases and as it becomes very large? Alternatively you get something that might be mere self-evident to you if you divide top and bottom of the fraction by e-t.
So you'll have lim t---> infinite , the function in the denominator will grow faster, so as t grows, P'(x) approaches zero. I believe another way to see this is to note that for t---> inf, P(x)---> 24, therefore reaching a constant value, with zero rate of change. My rate would be zero, right?
 
  • #5
epenguin
Homework Helper
Gold Member
3,873
897
Not very convincing, in fact wrong. Firstly, rather than the denominator, what about the numerator??
 
Last edited:
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,970
7,209
I am not seeing this sign mistake
You are right - my mistake.
the function in the denominator will grow faster,
I refer you again to my question in post #2.
 
  • #7
35,291
7,148
The work in post #1 has an error.
You (@ttpp1124) have ##P'(x) = \frac{0(2t + 10e^{-t}) - 48(-10e^{-t})}{(2 + 10e^{-t})^2}##.
The error is in the 2nd term in the numerator.
What's the derivative with respect to t, of ##2t + 10e^{-t}##?

Also, both P and P' are functions of time t, not x.

Edit: The type in the problem statement is so small that I didn't notice that ##\frac{48}{2 + 10e^{-t}}## there had turned into ##\frac{48}{2t + 10e^{-t}}## in the written work.
 
Last edited:
  • #8
benorin
Homework Helper
Insights Author
1,328
122
There is a typo/error in the denominator of ##P(t)##, an extra ##t##, not that it matters much.
 

Related Threads on Determine what happens to the rate of change of the population over time

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
8K
Replies
6
Views
1K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
4
Views
1K
Replies
3
Views
5K
Replies
10
Views
3K
Replies
4
Views
2K
Top