# Growth problem

1. Feb 11, 2007

### glid02

Here is the first question:
A population P obeys the logistic model. It satisfies the equation

Assume that P(0)=3. Find P(66)

First I multiplied both sides by dt and integrated, giving:
P=6/700Pt(7-P)+c
If P(0)=3 then c=3
P=6/700Pt(7-P)+3

Then I divided everything by P and had
1=6/700t(7/P-1)+3/P

Now to find P(66)
1=6/700*66(7/P-1)+3/P
1=396/700(7/P-1)+3/P
1=2772/700P-396/700+3/P
4872/700P=1096/700
P=4.445

That's not right, what am I missing?
Thanks.

2. Feb 11, 2007

### cristo

Staff Emeritus
Your first equation is not visible. Perhaps you could rewrite it?

3. Feb 11, 2007

### ranger

You have to accept some sort of web certificate to view the equation. It seems to be located on some university's website. Heres the equation that I get:

$$\frac{dP}{dt} = \frac {6}{700}P(7-P)$$

4. Feb 11, 2007

### cristo

Staff Emeritus
Ahh, ok, thanks for that, ranger. I must have clicked no automatically!

You have this equation: $$\frac{dP}{dt} = \frac {6}{700}P(7-P)$$. You cannot simply multiply by dt and integrate, since you have not integrated the terms including P wrt P! You must rearrange the equation to give: $$\int \frac{dP}{P(7-P)}=\int\frac{6}{700}dt +C$$

Do you know how to solve this?

5. Feb 11, 2007

### ranger

You did put +C accidentally right?

Last edited: Feb 11, 2007
6. Feb 11, 2007

### glid02

Yeah, I can solve that. Didn't think to seperate variables for some retarded reason. Thanks for the help.

7. Feb 11, 2007

### cristo

Staff Emeritus
Yea, I guess I haven't really integrated anything yet, so strictly the constant doesn't appear until the next line!

You're welcome!

8. Feb 11, 2007

### glid02

OK I lied, I'm still not getting the right answer.

dP/P(7-P)=6/700dt
1/7log(P)-1/7log(7-P)=6t/700+c
log(P)-log(7-P)=6t/100+c
Using e
P-7+P=e^(6t/100)+c
P=(e^(6t/100)+7)/2+c
P(0)=3 so c=-1
Subbing 66 for t, I get 28.729

Still not right, what am I doing wrong now?
Thanks again.

9. Feb 11, 2007

### cristo

Staff Emeritus
What you've done here is wrong. You must collect the logarithmic terms before you can take the exponential of both sides.

10. Feb 11, 2007

### glid02

You lost me with the collecting. Can you give me another example?

11. Feb 11, 2007

### cristo

Staff Emeritus
Well, have you come across the general rule: log(a)+log(b)=log(ab) ?

12. Feb 11, 2007

### glid02

If I had I'd forgotten it. I should be able to solve from here (again). Thanks again for helping.