Exponential Growth and rates

intervade

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Ok, my problem is: Say we start with 1000 birds. Let B represent the bird population, and t = time in days. After 10 days the population is growing at a rate of 25 birds per day. Find B(t).

Now, can I assume that B(0) = 1000? If thats the case, and we are using P(t) = Ce^(kt), then C is 1000? Is this the correct thinking? I'm not really sure how to solve this problem. Help is much appreciated!
 
EXPONENTIAL growth:\displaystyle EXPONENTIAL \ growth:

Given: dBdt = 25 when t > 10 days and B(0) = 1000, find B(t).\displaystyle Given: \ \frac{dB}{dt} \ = \ 25 \ when \ t \ > \ 10 \ days \ and \ B(0) \ = \ 1000, \ find \ B(t).

B(t) = Aekt, exponential growth formula, k > 0.\displaystyle B(t) \ = \ Ae^{kt}, \ exponential \ growth \ formula, \ k \ > \ 0.

B(0) = 1000 = Aek(0), A = 1000\displaystyle B(0) \ = \ 1000 \ = \ Ae^{k(0)}, \ A \ = \ 1000

Hence, B(t) = 1000ekt\displaystyle Hence, \ B(t) \ = \ 1000e^{kt}

d[B(t)]dt = k1000ekt, d[B(11)]dt = 25 = k1000e11k, k = .02005158807.\displaystyle \frac{d[B(t)]}{dt} \ = \ k1000e^{kt}, \ \frac{d[B(11)]}{dt} \ = \ 25 \ = \ k1000e^{11k}, \ k \ = \ .02005158807.

Ergo, B(t) = 1000e.02005158807t.\displaystyle Ergo, \ B(t) \ = \ 1000e^{.02005158807t}.
 
Afterthought:

Observed that dBdt = kB, The rate of growth of the population is proportional to\displaystyle Observed \ that \ \frac{dB}{dt} \ = \ kB, \ The \ rate \ of \ growth \ of \ the \ population \ is \ proportional \ to

 the size of the population. This is find for small populations in a large environment, but as\displaystyle \ the \ size \ of \ the \ population. \ This \ is \ find \ for \ small \ populations \ in \ a \ large \ environment, \ but \ as

 time marches on, a more reasonable equation is the logistic equation, to wit:\displaystyle \ time \ marches \ on, \ a \ more \ reasonable \ equation \ is \ the \ logistic \ equation, \ to \ wit:

dBdt = k(1BN)B, where N = carrying capacity of the bird population.\displaystyle \frac{dB}{dt} \ = \ k(1-\frac{B}{N})B, \ where \ N \ = \ carrying \ capacity \ of \ the \ bird \ population.
 
NON-EXPONENTIAL GROWTH:

dBdt = 25 after 10 days and B(0) = 1000, (given).\displaystyle \frac{dB}{dt} \ = \ 25 \ after \ 10 \ days \ and \ B(0) \ = \ 1000, \ (given).

dBdt = 25      B(t) = 25t+C, t > 10, t a counting number.\displaystyle \frac{dB}{dt} \ = \ 25 \ \implies \ B(t) \ = \ 25t+C, \ t \ > \ 10, \ t \ a \ counting \ number.

This gives B(t) = 750+25t, t > 10, t a counting number.\displaystyle This \ gives \ B(t) \ = \ 750+25t, \ t \ > \ 10, \ t \ a \ counting \ number.

This is what you express in your original statement, which has nothing to do with\displaystyle This \ is \ what \ you \ express \ in \ your \ original \ statement, \ which \ has \ nothing \ to \ do \ with

exponential growth, as the bird population increases 25 birds a day (after 10 days), a linear\displaystyle exponential \ growth, \ as \ the \ bird \ population \ increases \ 25 \ birds \ a \ day \ (after \ 10 \ days), \ a \ linear

 function.\displaystyle \ function.

So, the correct answer is B(t) = 750+25t, B(11) = 1025, B(12) = 1050, etc.\displaystyle So, \ the \ correct \ answer \ is \ B(t) \ = \ 750+25t, \ B(11) \ = \ 1025, \ B(12) \ = \ 1050, \ etc.
 
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