OpenStudy (perl):

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.

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

it's first order diferential equation

5 years ago
OpenStudy (anonymous):

it's the same

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

in your book it is clearly diffrentiated respect P.

5 years ago
OpenStudy (perl):

yes, why did they do that?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (perl):

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.

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

we are multiplying by dP/dt

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

sry, got to go now

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

cya

5 years ago
OpenStudy (perl):