# Finding values between two mins or two maxes

• Sep 25th 2010, 10:41 PM
kelp
Finding values between two mins or two maxes
Hello,
I am trying to find the value between two maxes/two mins. I just need to see how long a wavelength is.
I have the equation. It is a cos function: 2 + 2.5cos(2pi*60*t).
So, In order to find all maxes and mins, I will derive the above function, and I get:
-sin(2pi*60*t)

I need to set this to zero to find the x values of the max and mins. This is where the problem comes in. I have forgotten how to solve for this:
0 = -sin(2pi*60*t)

I do the inverse of sin of both sides and get this:
0 = -(2pi*60*t)
What is the inverse sin of 0? It's a large range correct?
The next max or min can be found by doing:
pi = -(2pi*60*t) correct?

I tried that, and I get 0 and 1/30, which is def. not the interval between a max and a min. I was going to find the time between a max and a min and multiply by two to find the length of the wavelength. When I calculated this eq. into my graphing calculator, 1/15 is def. not the distance of a wavelength.

Thanks.
• Sep 25th 2010, 11:26 PM
Educated
The derivative is wrong.

$\displaystyle \frac{d}{dt}(2 + 2.5cos(2 \pi \cdot 60\cdot t)) \ne -sin(2pi\cdot 60\cdot t)$

Since the t variable is inside the cosine, you would have to use the chain rule to differentiate it.

Chain rule: $\displaystyle \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$
• Sep 26th 2010, 02:50 AM
HallsofIvy
You don't really need to know the derivative of "max" or "min" to find the wavelength.

You know (I hope!) that cos(x) has period (wavelength) $\displaystyle 2\pi$- that is, one wave starts at x= 0 and ends at $\displaystyle x= 2\pi$. That means that, for this function the wave starts where $\displaystyle 2\pi (60t)= 0$, or t= 0, and ends where $\displaystyle 2\pi (60t)= 2\pi$ or $\displaystyle t= \frac{1}{60}$. This function has a period (wavelength) of $\displaystyle \frac{1}{60}$.
• Sep 26th 2010, 11:44 AM
kelp
When I graph the function, I am getting x=~ 6 for the length of the wavelength and not 1/60.
• Sep 26th 2010, 07:03 PM
Educated
How are you graphing it? By hand? Graphical calculator? Computer program?

Here's a graph by Wolfram|Alpha of the equation: Wolfram|Alpha, plot[2 + 2.5cos(2pi*60*t)]

Are you graphing it in degrees or radians?