- #1

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I started out by trying to do dy/dt = p^.5. I am used to the population problems where I use y=Ce^(rt) but am having trouble making the jump to this kind. How do you account for the increase of 20/month?

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- Thread starter cue928
- Start date

- #1

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I started out by trying to do dy/dt = p^.5. I am used to the population problems where I use y=Ce^(rt) but am having trouble making the jump to this kind. How do you account for the increase of 20/month?

- #2

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The population always increases at a rate of 20 per month, or is that an initial value?

- #3

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Yes, always increases at rate of 20/month and initial population is 100.

- #4

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- #5

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The time rate of change of a rabbit population P is proportional to the square root of P. At time t = 0 (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many after one year? The answer I guess is 484.

How do you set up logistics equations?

- #6

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So 20/month is an initial value.

The rate of change of P, i.e. it's derivative with respect to P is proportional to the square root of P:

[tex] \frac{dP(t)}{dt} = \alpha P^{1/2} [/tex]

where alpha is a constant with respect to. You also know that

[tex] \frac{dP(t)}{dt} \bigg|_{t=0}} = 20 \ rabbits/month [/tex]

and that

[tex]P(0) = 100 \ rabbits[/tex]

Which is all the info you need to determine P(t)

The rate of change of P, i.e. it's derivative with respect to P is proportional to the square root of P:

[tex] \frac{dP(t)}{dt} = \alpha P^{1/2} [/tex]

where alpha is a constant with respect to. You also know that

[tex] \frac{dP(t)}{dt} \bigg|_{t=0}} = 20 \ rabbits/month [/tex]

and that

[tex]P(0) = 100 \ rabbits[/tex]

Which is all the info you need to determine P(t)

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- #7

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- #8

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I appreciate what you wrote but I'm still unsure on where to start. I see that I was wrong to have a p^.5 on the right

OOOPSSS, my mistake. I corrected it in the previous post.

Should I go ahead and start separating variables and then moving forward?

Of course you should. Even in the case where you shouldn't, equations don't bite you know.

- #9

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- #10

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I came up with this equation after sep of variables: 2p^.5 = Bt + C.

1) With t = 0 and initial population of 100, I came up with a C value of 100.

2) With t = 1, population = 120, so I came up with a "B" value of 120.

3) But then when I do time = 12, I don't get anywhere near the 484 rabbits that I should have.

- #11

Matterwave

Science Advisor

Gold Member

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Er, did you solve for p in your equation so you have some p(t) on the left side instead of 2p^.5?

- #12

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2) With t = 1, population = 120, so I came up with a "B" value of 120.

This is wrong. What you have is P'(0) = 20. So use the derivative of P to find B. Do you realize that the rate of change in the population is the derivative of P?

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