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

[ttpp1124]

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

 

[haruspex]

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.

 

[epenguin]

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.

 

[ttpp1124]

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?

 

[epenguin]

Not very convincing, in fact wrong. Firstly, rather than the denominator, what about the numerator??

 

[haruspex]

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.

 

[Mark44]

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.

 

[benorin]

There is a typo/error in the denominator of $P(t)$, an extra $t$, not that it matters much.