In desperate need of Calculus help.... plz help?

Question

A wild animal preserve can support no more than 120 elephants. 33 elephants were known to be in the preserve in 1980. Assume that the rate of growth of the population is

dP/dt = 0.0007P(120 - P)

where time is in years. Find a formula for the elephant population in terms of t.

idk if this stuff will help but here are some equations...

dP/dt = kP(M - P)

P = M / 1+Ae^-(Mk)t

Let 1980 = t = 0

plz explain if u can and steps would be awesome.... thank you

Answer 1

"Edgar" dropped the "-" in the exponent while "mohanroa d" rounded the value of ln (33/87) to "-1" rather than "-0.9694". The latter did a good review of how the answer came about.....P = 120 / [ 1 + (87/33) e^{ -0.084t}]

Answer 2

here is how it goes in this setup your are given that

k=.0007 and M=120

okay, and that the equation P=M/(1+Ae^-(Mk)t) has the point (0,33) that is at t=0 P=33.

So you plug the values of M and k into the equation above and solve for A which comes out to be about 2.636

so your final equation is P=120/(1+2.636*e^(120*.0007)t)

Answer 3

dP/dt = 0.0007P(120-P)

dP/dt = 7P(120-P)/10000

(10000) (dP/dt) = 7P(120-P)

separate the variables

(10000) dP / P (120 - P) = 7dt

resolving LHS into partial fractions

10000dP(1/120(P) + 1/120(120-P) = 7dt

(10000/120)[dP/P + dP/(120-P)] = 7dt

[dP/P + dP/(120-P)] = (840/10000)dt

integrating on both sides

[ln(P) - ln(120-P) = (21/250)t + c

when t = 0, P = 33

[ln(33) - ln(120-33)] = c

c = -1

ln(P) - ln(120-P) = (21/250)t -1

ln(P/120-P) = (21t - 250)/250