bio 1 - Biomatemtica y Ecolog a a 26 de marzo de 2011 Dado:...

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Biomatem´ atica y Ecolog´ ıa 26 de marzo de 2011 Dado: P t +1 = P t ± 1 + 0 , 7 ± 1 - P t 10 ²² (1) 1. Hallar el valor (tama˜no de la poblaci´ on) que corresponde al lugar donde la par´ abola cruza el eje horizontal Para hallarlo hacemos: P t +1 = 0 Reemplazando en (1): P t ± 1 + 0 , 7 ± 1 - P t 10 ²² = 0 P t + 0 , 7 P t ± 1 - P t 10 ² = 0 P t + 0 , 7 P t - 0 , 07 P 2 t = 0 1 , 7 P t - 0 , 07 P 2 t = 0 P t (1 , 7 - 0 , 07 P t ) = 0 P t = 0 o 1 , 7 - 0 , 07 P t = 0 0 , 07 P t = 1 , 7 P t = 24 , 3 Cuando P t = 24 , 3 cruza el eje horizontal 2. Qu´ e sucede si P 0 es mayor a dicho valor? Supongamos P 0 = 25 Reemplazando en (1): P 1 = P 0 ± 1 + 0 , 7 ± 1 - P 0 10 ²² = 25 ± 1 + 0 , 7 ± 1 - 25 10 ²² = 25 ± 1 + 0 , 7 - 0 , 7 × 25 10 ² = 42 , 5 - 43 , 75 = - 1 , 25 1
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Puesto que es un valor negativo la poblaci´ on se extingue. 3. Donde est´ a el m´ aximo de la par´abola? P t +1 = P t ± 1 + r ± 1 - P t k ²² dP t +1 dP t = 1 + r - 2 P t k = 0 P t = (1+ r ) k 2 Y es m´ aximo puesto que:
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This note was uploaded on 03/27/2011 for the course MATHEMATIC 504 taught by Professor Carlostrujillo during the Winter '09 term at Buena Vista.

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bio 1 - Biomatemtica y Ecolog a a 26 de marzo de 2011 Dado:...

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