Solving the Logistic Model: Find P(66) with P(0)=3

[glid02]

Here is the first question:
A population P obeys the logistic model. It satisfies the equation
https://webwork.math.uga.edu/webwork2_files/tmp/equations/11/02d0a645c053f1f6002d746c78143f1.png

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.

 

[cristo]

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

 

[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)

 

[cristo]

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)

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

Here is the first question:
A population P obeys the logistic model. It satisfies the equation
https://webwork.math.uga.edu/webwork2_files/tmp/equations/11/02d0a645c053f1f6002d746c78143f1.png

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.

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?

 

[ranger]

You did put +C accidentally right?

 

[glid02]

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

 

[cristo]

You did put +C accidentally right?

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

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

You're welcome!

 

[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.

 

[cristo]

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

dP/P(7-P)=6/700dt
1/7log(P)-[/color]1/7log(7-P)=6t/700+c
log(P)- log(7-P)=6t/100+c

The - sign in red should be a +

Using e
P-7+P=e^(6t/100)+c

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

 

[glid02]

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

 

[cristo]

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

 

[glid02]

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