AMATH_383_Solution_1 -...

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AMATH 383 Solution to Assignment #1 Winter 2010 Problem 2.4 a. 3/4 0 () ()/ . . (). cc c c Nt m t m E dd EN m dt m dt YYm Therefore the equation c c d YY N E N dt  Becomes: 0 . YE d Ym m m mm d t This is the same as: 0 ( ) Y d dt E E  . m So 0 / and / . aYmE bYE  b. m=M when dm/dt=0. From the above differential equation we find: 0. aM bM Solving for M : 4 (/) . M ab Now back to the time-dependent differential equation, we replace b/a by M -1/4 to get: 1/4 1( / ) . d ma m mM dt   Let  1/4 1/4 1/4 1/4 ( / ) , and 1 . . 4 1. 4 . 4 rm M R r dm d dt M dt da rr dt M RR dt M Solve for R(t) :
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1/4 () (0 )exp . 4 a R tR t M   Therefore 1/4 ln( ( ) / (0)) . 4 a R t M A plot of ln (R(t)/R(0)) vs at/4M 1/4 is a straight line with slope -1 for any mammal of regardless of its size. d. Since the nondimensional time at/4M 1/4 for the life development of any mammal is independent of M , and a is also independent of M , then time must scale with M 1/4 . It then
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This note was uploaded on 10/21/2011 for the course APPLIED MA AMATH 383 taught by Professor Hongqian during the Spring '10 term at University of Washington.

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AMATH_383_Solution_1 -...

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