Question

Suppose that rabbits are introduced onto an island where they have no natural enemies.
Because of natural conditions, the island can support a maximum of 1000 rabbits. Let P(t)
denote the number of rabbits at time t (measured in months), and suppose that the population
varies in size (due to births and deaths) at a rate proportional to both P(t) and 1000 − P(t).
That is, suppose that P(t) satisfies the differential equation dP/dt = kP(1000−P), where k is
a positive constant.
a) Find the value of P when the rate of change of the rabbit population is maximized.

Answer 1

im not sure if i should take the second derivative with respect to t, or take the derivative with respect to P

Answer 2

dP(t)/dt is a function, just like P(t). Since you are asked for maximum, just act like you would with any function. Find it's derivative ,make it equal to 0 and find values of t which gives you that.

Answer 3

yes, but with respect to what? with respect to P or with respect to t?

Answer 4

d/dP [ dP/dt ] = 0, or d/dt [ dP/dt ] = 0 ?

Answer 5

\[\frac {d^2 P}{dt^2}=k\frac{dP}{dt}(1000-P)-kP=0\]

Answer 6

ok then you took with respect to, my book did d/dP [ dP/dt ] instead

Answer 7

it's first order diferential equation

Answer 8

it's the same

Answer 9

my book did it differently d/dP [ kP (1000 - P ) ] = k(1000-P) + kP( -1)

Answer 10

in your book it is clearly diffrentiated respect P.

Answer 11

yes, why did they do that?

Answer 12

but i still say it's the same, because:
\[\frac {rabit's population rate}{dt}=\frac {rabit's population rate}{dP}\frac{dP}{dt}\]
it's just chane rule

Answer 13

I think its like a chain rule. you can set u = kP(1000-P) , then du/dt = du/dP *dP/dt

there is also a part b)
b) When is the rate of change of the rabbit population a minimum? Discuss your answers.

Answer 14

, and the dP/dt you can divide out , since it is equal to zero

Answer 15

we are multiplying by dP/dt

Answer 16

actually dP/dt is not 0, d^2P/dt^2=0

Answer 17

, im saying
the only difference between the two are
-2k (500 - P ) = 0 , or
-2k (500 - P ) *dP/dt = 0

Answer 18

d^2P/dt^2= -2k (500 - P ) *dP/dt = 0 or
d^2P/dP = -2k(500 - P ) = 0

Answer 19

so we get 500 - P = 0, P = 500

Answer 20

sry, got to go now

Answer 21

woops, i didnt mean to imply that dP/dt = 0, not necessarily

Answer 22

cya

Answer 23

im confused about part b)

Answer 24

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Answer 25

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Answer 26

For the graph, which are possible functions u0192 and g?

https://dlap.gradpoint.com/Resz/~0.FkGA30oIxT4yUA6.1YNrvQzsdujk2YjDsaLgVJUV0yHrt6gIST8n91mip18/7234256,D8E/Assets/questions/quiz/precalculus/pcalexpo/IMG73438.png

Answer 27

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Answer 28

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Answer 29

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Answer 30

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Answer 31

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Answer 32

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Answer 33

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Answer 34

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Answer 35

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Answer 36

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Answer 37

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Answer 38

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Answer 39

In this excerpt from William Dean Howells's "Editha," which sentences contain sar

Answer 40

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Answer 41

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Answer 42

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The image below shows two dilated figures with lines KO and Ku2019Ou2019 drawn.

Answer 43

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