Fish harvesting A fishery manager knows that her fish...

Key Concepts

Exponential Growth

Exponential growth refers to a process where a quantity increases by a fixed percentage over equal time intervals. In population dynamics, this type of growth is modeled by a constant rate multiplying the current population, leading to progressively larger increases as the population grows.

Difference Equations

Difference equations, or recurrence relations, describe how a quantity changes from one time period to the next. They are particularly useful in modeling discrete processes such as population dynamics where changes occur at set intervals, and they can incorporate both multiplicative growth factors and additive or subtractive terms.

Infinite Geometric Series

An infinite geometric series is the sum of the terms of a geometric progression that continues indefinitely. This series converges when the common ratio is less than one in absolute value, and its sum can be calculated using a closed-form expression. This concept is useful for analyzing the cumulative effect of constant terms (like fixed removals) in a long-term process.

Steady-State Analysis

Steady-state analysis is the procedure of determining the long-term behavior of a dynamic system. In the context of population dynamics, it involves finding a value that the system approaches as time goes to infinity, where the effects of growth and reduction balance each other, resulting in no net change in the population.

Transcript

00:02 Hi, so today we're going to be looking at how much fish we will have after n months.

00:06 So we have a fish population that increases at a rate of 1 .5 % per month, and the amount of fish that is harvested from the group of fish is 120, with an initial amount of fish of 4 ,000.

00:21 So the first thing we're going to do is we're going to create an equation using our fn value, which is equal to our fish population after n months.

00:32 So fn is equal to the, how much the population was the month before, so which that would be is fn minus 1 plus the increase in population, or the rate of increase times the amount that there was the month before.

00:54 So 0 .015 times fn minus 1, and then minus 120 because, each month there are 120 fish harvest.

01:07 From here we can simplify this and make this equal to 1 .015 times fn minus 1 .20.

01:21 And from here we can actually expand this equation and we're going to substitute in for fn minus 1.

01:31 Using the exact same formula, we're able to form for our f of n value.

01:35 We can substitute in for fn minus 1, but this time we're going to have fn minus 2.

01:43 I'm going to show you how we're going to do that.

01:44 So this is all going to be equal to 1 .015 times 1 .015 f minus 2, minus 120, and then all that is going to be minus 120.

02:05 From here we can multiply out, or look like cross -sart are 1 .015, so we're going to have 1 .015 square times fn minus 2 minus 1 .015 times 120 minus 120.

02:27 And up here, though you can see that this is what we substitute.

02:31 In for our f minus one.

02:35 I just want to point that out.

02:37 From here we can further simplify our equation below and we can have a sorry color um 1 .015 square times fn minus 2 and that would be minus 120 times 1 .015.

02:57 It is important to keep this in this format for later on we'll need it when we'll just try to form more of a series type of format.

03:07 And then from here we can further expand by substituting in for our fn minus 2 values.

03:15 Doing the exact same thing we just did in the last couple steps.

03:20 So we're going to have here it's going to be 1 .015 squared times 1 .015 times fn minus 3.

03:32 Minus 120 minus 120 times 1 plus 1 .015.

03:42 Now from here, so we'll multiply through, so we're going to have 1 .015 cubed times fn minus 3 minus 120 times 120 times 1 .011 squared minus 120 times 1 plus 1 .015.

04:15 So we're going to have 1 .015 cube times fn minus 3 and then minus 120 times 1 plus 1 .015 plus 1 .015 squared.

04:35 From here we're going to skip some of our expansions.

04:39 I'm just going to end up with a 1 .015 now up to the n because we're going to say this is after n times.

04:48 So you've expanded it all the way through for as many times down as we can.

04:55 So this will be times n and this will be after the zero because as you can see this is decreasing always.

05:02 So we're going to be at the initial there.

05:04 Then we have this minus 120 times one plus 1 .015 plus 1 .0152 and then we're going to skip here as well.

05:20 I'm going to have 1 .015 to the n minus 1.

05:25 You can see above this is always, this value is always going to be one less than our n.

05:32 That is really because we have a 1 here to begin with.

05:36 So it took a little before we started our this sequence with it.

05:45 So from here what we're going to do is take equation from the chapter which is s n is equal to a times 1 minus r to be n over 1 minus r and our a value in this case is going to be negative 120...

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