Exponential Growth and rates

I

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!
 
\(\displaystyle EXPONENTIAL \ growth:\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\displaystyle \ function.\)

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