Problem 3 (20 pts) The functional response F(n) of a predator is the reproductiv

Question

Problem 3 (20 pts) The functional response F(n) of a predator is the reproductive rate of each predator individual as a function of prey density, n. A simple example would be F(n) = An, for a constant of proportionality A. One of your colleagues hypothesizes a more complicated-looking functional response F(n) = An n 1 2An n2 1 (2)

(a) Showing clearly what results from class for the addition and ratios of limits you are using, evaluate the limit limn0 F(n) where F(n) is as given in Equation (2). (

b) Again showing clearly what results from class for the addition and ratios of limits you are using, evaluate the limit limn F(n) where F(n) is as given in Equation (2).

(c) Something strange seems to be happening at n = 1, which seems like it should be a perfectly reasonable value for prey density. Remembering the trick from class that ( + 1)( 1) = 2 1 for any number , give the two terms in Equation (2) a common denominator so that you can combine their two numerators.

(d) Using your result from Part (c) or otherwise, evaluate limn1 F(n)

Thanks!!

Explanation / Answer

a>
F(n) = An/(n-1) - 2An/(n^2-1)
                          

lim n--> 0 F(n)
=> lim n--> 0 An/(n-1) - 2An/(n^2-1)
as n--> 0 the limit for the first term tends to 0 and the limit for the secons term tends to 0 as well.
hence lim n-->0 {An/(n-1) - 2An/(n^2-1)} is 0 -0 = 0

b>
lim n--> infinity F(n)
=> lim n--> infinity {An/(n-1) - 2An/(n^2-1)}
as n--> infinity the limit for the first term tends to infinity/infinity and the limit for the secons term tends to
infinity/infinity as well.
=> we have got an indeterminate form
so we'll have to simplify the F(n) in order to find the limit
Using L'Hospitals rule
that is differentiating the denominator and the numerator individually till the time we dond loose the indeterminent form.
lim n--> infinity {An/(n-1) - 2An/(n^2-1)}
lim n--> infinity {A/1 - 2A/(2n-1)} => A - 2A/infinity = A - 0 =A
hence the limit of F(n) is = A as n --> infinity

c>
at n=1
F(n)= infinity - infinity
so we'll have to do some simplification on F(n)
=> F(n) =An(n+1)/[(n-1)(n+1)] - 2An/(n^2-1)
=> F(n) =An(n+1)/(n^2-1) - 2An/(n^^2-1)
=> F(n) = (An^2-An)/(n^2-1) = An/(n+1)

d> Lim n --> 1 [F(n) from part c]
=> lim n --> 1 An/(n+1) = A/2
hence the limit for F(n) from part c when n--> 1 is A/2

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