IB MATH METHODS SL - FISH ESCAPEMENT

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TJ Dhaliwal
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IB MATH METHODS SL

FISH ESCAPEMENT

Portfolio Assignment #3                                                        Type III

The Department of Fisheries and Oceans (DFO) sets targets for each river system for escapement (the number of fish that escape being caught by the ocean fishery and return to spawn). The salmon escapement in a river is currently estimated to be 2 million but has been declining by 8% per year. (That is, each year there are 8% fewer fish than the precious year.)

a) i) If this decline continues, predict the population (of the escapement) in 8 years.

The initial population of 2 million declines each year by 8%. Therefore we can write the population after 1 year as following:

Let PO be the initial population of 2 million.

Let PT be the population after T years.

P1 = PO – (PO x 0.08)

P1 = PO (1 – 0.08)

P1 = PO (0.92)

This will give us the answer for the population after 1 year, with an annual decline of 8%. Then to calculate the new population after 2 years, we calculate take the population after 1 year, while using the annual decline of 8% each year.

P2 = P1 (0.92)

P2 = PO (0.92)(0.92)

This is beginning to show a pattern for the population after a set amount of years, after an annual decline of 8% each year. Therefore, observing this pattern, we can utilize the following formula to generate the population after 8 years with the continuing decline:

PT = PO (0.92)T

Therefore, the population after T years (PT) is going to be a geometric series modelled by this preceding equation. Hence, to calculate the population after 8 years of decline, we use this formula and the numbers given:

P8 = 2 000 000 (0.92)8 

P8 = 1 026 437.746

P8 = 1.0264 million

Thus, the fish population after 8 years of decline is 1.0264 million.

ii)  Rewrite your equation from part (i) using a base of “e”.

      Using this new equation, what is the rate of decline?

      Why is this different from the rate of decline given originally as 8%?

Continuing with the equation, PT = PO (0.92)T, as used in part (i), I rewrote the geometric ratio, 0.92, in exponential form, or in base “e”:

Join now!

Let K be the new rate of decline.

0.92 = eK

ln(0.92) = ln (eK)

*ln(0.92) = **K ln(e)

K = ln(0.92)

* Using the laws of Logarithms, we can bring down the exponent of e and multiply by ln(e), as well as using ln on the other side of the equation.

** The figure “ln(e)” is equal to 1, therefore K multiplied by ln(e) is equal to K.

Using the new rate of decline, we can generate this new equation:

PT = PO (eln(0.92))T

The new rate of decline is ln(0.92), which is equal to ...

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