00:01
So in this question we're told that a boosts growth rate, sorry, reproduction rate by a factor s.
00:20
Then if p is the proportion of a, then we have d .p by dt is one -half, s -p, one minus p, and p of 0.
00:37
Is p -0, and the integral of 1 over p, 1 minus p, dp, is the integral of 1 -5 -s d t, when t is 0, p -nought.
00:50
Then we can find out p of t by integrating up to t.
00:56
Now 1 over p 1 -1 -p is the same as 1 over p, plus 1 over 1 -p, because then we're going to have 1 -p plus p over p 1 minus p and this is going to be equal to s t over 2 so now we can integrate on the p side log p of t over p nought minus log 1 minus p of t over 1 minus p nought is equal to s t over 2 now let's combine the logs because remember that the difference of two logs is equal to the log of the quotient so this is log of p of t 1 minus p p -0 over p -0 1 minus p of t this is st over 2 now we can exponentiate both sides p of t times 1 minus p -nought is p -nought 1 -p of t times e to the s t over 2 now combining the p's together we can add p -nought p of t to both sides to get p of t uh sorry uh we can add we we can add p -nought p of t e to the s t over two, here p of t 1 minus p -naught plus p -nought e to the s t over 2 is equal to p -nought e to the s t over 2.
02:32
So then p of t is p -nought e to the s t over 2 divided by 1 minus p -0 plus p -0 e to the s t over 2.
02:44
And then we can divide by e to the s t over 2 to get p of t is p -nought divided by 1 minus p -0, e to the minus s t over 2 plus p -0.
03:05
So that gives our population as a function of time.
03:11
So if p -0 is not equal to 0, then the limit as t goes to infinity of p -t is going to be p -0 over 1 -p -0 times the limit as t goes to infinity.
03:26
E to the minus s, t over 2, because that's the only part that depends on t plus p -0, which is equal to p -naut over 1 minus p -naut times 0, because e to the minus s -t over 2 will go to zero as t goes to infinity, plus p -naught, which is p -naut, which is 1.
03:51
So the limit as t goes to infinity, p of t is 1.
03:59
So why does this make sense biologically? or biologically, if a causes faster reproduction, then the number of a carriers will increase faster than the number of little a carriers, such that the number of a carriers over the total number, the number of a's over the number of a's plus the number of a's is going to be dominated by the number of a's, which is one, because the number of a's is going to be much smaller than the number of, the number of little a's is going to be much smaller than the number of big a's...
