Please show all work The figure shows a useful way to view the relationships amo

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Please show all work

The figure shows a useful way to view the relationships among these variables. Specifically, ecologists assume that the number of species S depeods direelly on A and z through a function of the form S = Az. Such a rclation&hip; makes sense: Wc expect the number of spccies to increase as the abundance increases and as The fraction of available niche space increases. We set that A and z arc also subiequently related to the disturbance level E. Wc must make some plausible a&sumptions; about these relationships. First, as The level of disturbance E incrcascs we expect the abundance A to decrease. For this reason. wc assume that this relationship is described by the linear function A(E)= Amax - aE., where Amax > 0 and a > 0 are given constants. Next, as the level of disturbance increases, the fraction of available niche space also increases. We assume that this relationship is described by the linear function z(E)= bE, where > 0 is a given constant. Given these assumptions. the goal is to leam about the relationship S-f(E) between S and E. Specifically we aim to extract the intermcdiaie disturbance principal from the modcl. Use the relationship S = Az and the Chain Rule to show that dS/dE = -azs/A +bs In A. Argue that when E is small (E 0). theo A Amax and z 0. Thercfore ds/dE>0. lnterpret this result that at email levels of disturbance. divenity increases. Argue that when E is large, then A is small and positive, and z = 1. Therefore ds/sE

Explanation / Answer

1)

A(E) = A(max) - aE

z(E) = bE

Now, S = A^(z)

=> S = [A(max) - aE ]^ (bE)

=> ln S = bE * ln [A(max) - aE ]

=> 1/S * dS/dE = b * ln [A(max) - aE ] + bE * (1/ (A(max) - aE ) ) * (-a)
=> dS/dE = -aZS/A +bS ln A

2)

A = A(max) - aE

So when E is small i.e when is nearly 0, A = A(max)

Z = bE

So when E is small i.e nearly equal to 0 , Z is also nearly 0.

Now, dS/dE = -aZS/A +bS ln A

When E is nearly 0 , Z is nearly 0. Hence -aZS/A is nearly 0.

S= A^(z) = A^(0) = 1

So , dS/dE = b Ln [ A(max) ] ,,, this is greater than 0 since both b and A(max) are greater than 0.

3)

A = A(max) - aE

So when E is large, A = A(max) - ( a large value )
Hence A is small and positive.. [ Since we are removing a very large value from the maximum value ]

Z = bE

So when E is large , Z which is a fraction, will tend to 1.

Now, dS/dE = -aZS/A +bS ln A

When E is large , A is very small. Z =1

S= A^(z) = A^(1) = A

So, dS/dE = -a + bA ln A

Now, Ln of a ver small positive number tends to ( - infinity ) .

So dS/dE is negative.

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