# Need Help With Sinusoidal Graphs

• Jan 30th 2011, 02:59 PM
TacticalPro
Need Help With Sinusoidal Graphs
Hi all, I was sick and missed a day of pre-cal, and i missed an important lecture on sinusoidal graphs and word problems. I have a worksheet that has two problems, and I'm going to put the first one on so someone can help me and hopefully that will give me a jumpstart for doing the second. here it is:

fox population problem
Naturalists find that the populations of some kinds of predatory animal's vary periodically. Assume that the population T of foxes in a certain forest varies sinusoidally with time. Records started being kept when time t = 0 years. a minimum number, 200 foxes, occurred when t = 2.9 years. the next maximum, 800 foxes, occurred at t = 5.1 years.

1. sketch a graph of this sinusoid
2. write an equation expressing the number of foxes as a function of time, t.
3. predict the population when t = 7.
4. foxes are declared to be an endangered species when their population drops below 300. between what two nonnegative values of t were foxes first endangered?

all help/tips appreciated
• Jan 30th 2011, 03:51 PM
pickslides
Use $\displaystyle N = A\sin B(T-C)+D$

From your min,max values D= 500, A=300, do you know how?

For (T,N) use (2.9,200) , (5.1,800) to find B and C. Then you should be able to sktech it.
• Jan 30th 2011, 04:06 PM
Prove It
Have you at least sketched the graph yet?
• Jan 31st 2011, 05:55 AM
HallsofIvy
Look at the "basic" y= sin(x). That has period $\displaystyle 2\pi$, has a maximum of 1, at $\displaystyle \pi/2$, then goes back to 0 at $\displaystyle \pi$, goes to a minimum of -1 at $\displaystyle 3\pi/2$ and back to 0 at $\displaystyle 2\pi$. Notice that the distance between the max and min, $\displaystyle \pi/2$ to $\displaystyle 3\pi/2$ is $\displaystyle \pi$, half the period.

You are told that the minimum, 200, occured at t= 2.9 and the maximum, 800, at t= 5.1. The difference between those two times is 5.1- 2.9= 2.2 years so the period must be 2(2.2)= 4.4 years. In order to make $\displaystyle x= 2\pi$ when $\displaystyle t= 4.4$ we need $\displaystyle x= \frac{2\pi}{4.4}t$. But then when t= 5.1, $\displaystyle x= \frac{2\pi}{4.4}5.1= \fra{10.2}{4.4}\pi= 2.32\pi$ (approximately). But we want it to be equal to $\displaystyle \pi/2= .5\pi$ so we need to add $\displaystyle .5- 2.32= 2.68\pi$: we want $\displaystyle sin(\frac{10.2}{4.4}t+ 2.68\pi)$. The midpoint between 200 and 800 is (200+ 800)/2= 500. Since sine itself is centered on y= 0 and goes from -1 to 1, we want to multiply sine by (800- 500)= (500- 200)= 300 and then add 500.

Of course, Pickslides method works too. Do it both ways and see if you get the same thing.
• Jan 31st 2011, 03:36 PM
TacticalPro
thanks for the info i was able to piece it together and work it out! thanks