Graphing a Fish Population Word Problem [Solved!]

X

a small lake on the island contains twoo species of fish: hake and redfish. the hake are predators that eat the redfish. the fish population in the lkae varies periodically with period 180 days. the number of hake varies btw. 500 and 1500, and the redfish varies from 1000 and 3000. the hake reach their maximum population 30 days after the redfish have reached their max. pop. in the cycle.

a.)sketch a graph that shows two complete periods of the population cycle for these species of fish. assume that t=0 corresponds to a time when the redfish population is at a max.
b.) find cosine functions of the form y=a cos k(t-b) + c that model the hake and redfish populations in the lake.

Relevant page

<a href="/analytic-trigonometry/analytic-trigo-intro.php">Analytic Trigonometry</a>

What I've done so far

my teacher CANNOT teach she gave us a huge project on graphing a word problem and didn't tell us how to do t and i have no idea what exactly to start with

i drew a graph and labeled it but i don't know where to go from there

X

Hi Stephanie

Actually, this is a good problem ^^

To start with, I don't know whether you found this page on IntMath:

<a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a>

Hopefully it will get you started.

Another page that will help (with the "+c" part of your project) is the first example on 

<a href="/trigonometric-graphs/6-composite-trigonometric-graphs.php">6. Composite Trigonometric Graphs</a>

Can you sketch the 2 graphs okay?

You then need to find amplitude, period and phase shift for each one.

The "+c" bit is quite straightforward.

Have a go at it and I can let you know if you are on the right track.

And concerning your teacher... actually, the best teachers are the ones that give you space to figure things out for yourself, so that you actually learn how to learn  ^^

I don't know whether these suggestions might help reduce your frustration:

<a href="http://www.intmath.com/blog/ten-ways-to-survive-the-math-blues">Ten Ways to Survive the Math Blues</a>

Regards

X

thanks for responding so soon. you have a point about my teacher its just annoying when we ask for help she says one thing and "changes her mind" the next day lol ok 

thanks much for the help i did look at most of your pages dealing with this problem ^^

i loved your smiley face by the way haha i can sketch the graphs and labeled them and everything it's just blank from there :p i will go back and look at your pages again and will definitely be back to ask you for help

                                       thank you,
                                     stephanie

X

I'm back!

Well, I tried to do the redfish one first, because that seemed easiest. This is what I got:

[graph]310,250;-40,365;-90,3050,90,1000;2000 + 1000cos((6.28 x)/180)[/graph]

Is it right?

X

It's good so far. 

What formula did you use? It needs to be in the format `y=a cos k(t-b) + c `

X

Because amplitude is `1000` and `b= (2 pi)/b = (2 pi)/180 = pi/90`, and the whole thing is moved up by `2000`, I got: 

`a = 1000`, `b = 0` and `c = 2000`.

So the equation is like this, right?

`y= 1000 cos (pi/90 t) + 2000`

X

Perfect! Now, how about the hake one?

X

The hake one peaks 30 days after the redfish.

So since phase shift `=-c/b`, I guess we put

`-c/b = 30`

`-c/(pi/90) = 30`

`-c = pi/90 xx 30`

`c = -pi/3`

But now it's confusing because your a, b, c are different to the a,b, and c in the question!

X

Yes, we need to be careful when the same letters are used for different variables.

You are on the right track. Can you complete that equation for the hake and graph it?

X

I got 

`a = 500`, `k=pi/90` and `c=-pi/3`

So the equation is:

`y=500 cos(pi / 90 t -pi/3) + 1000`

Here's my graph. I think it looks right!

[graph]310,250;-45,365;-50,1520,90,500;,500 cos(3.14 / 90 t -3.14/3) + 1000[/graph]

X

Great! Your function's almost right. It's not quite in the format `y=a cos k(t-b) + c`

X

Oh, I see.

`a=500`, `k=pi/90`, `-kb=-pi/3` so 

gives `-(pi/90)b=-pi/3` 

gives `b = 30`

`c = 1000`

So it must be

`y=500 cos(pi/90(t-30)) + 1000`

Here are the 2 graphs.

[graph]310,250;-50,365;-40,3170,90,500;1000 cos( 3.14/90(x)) +2000,500 cos( 3.14/90(x-30)) + 1000[/graph]

Looks good to me!

X

Yes, well done, Stephanie.   Good job!