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eli123
find the period and the amplitude of the function y=6cospix
You can figure out the period from one time cos(pi.x) = 1 to the next time. So cos(pi.x) = 1 when x = 0, as cos(pi.0) = cos(0) = 1. What is the next value of x for which cos(pi.x) = 1?
in other words: cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 see a pattern?
As for the amplitude, that is the maximum value of this function. y = f(x) = 6 cos(pi.x). What is it's maximum value? Hint: it occurs when cos(pi.x) has its maximum value.
Yep so basically like james said, the amplitude is = to the maxium valuse that we can make 6cos(pi*x) Since the 6 is a constant the only thing we can change is the x. We need to change the x to something that will make "cos(pi*x)" as big as possible. if we plug in all the possible values around the unit circle you will find that cos never gets bigger than 1
6 is the amplitude, yes. What's the period.
@eyust707, you've got this one. Thanks.
i dont understand the period
cos(a x) any thing in front of x , in this case divide period= 2pi/a so cos(2 x) period = 2pi/2 = pi
The period of a function f is the smallest number T for which f(x + T) = f(x). For the function f(x) = 6 cos(pi.x), it is therefore the smallest number T such that 6cos(pi(x+T)) = 6cos(pi.x) that is cos(pi.(x+T)) = cos(pi.x) Now if the pi wasn't there, draw the function g(x) = cos(x). For what value of T is it the case that g(x + T) = g(x)? i.e., cos(x + T) = cos(x)?
As eyust noted above, cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 So what is T?
i.e., what is the period for the function g(x) = cos(x)?
It's clear that the period of g(x) = cos(x) is T = 2pi. Now that being the case, what is the period of the function f(x) where f(x) = 6 cos(pi.x) ? I.e., for what value of T is it the case that f(x+T) = f(x) cos(pi(x+T)) = cos(pi.x) ?
For example, for x = 0, cos(pi(0+T) = cos(pi.0) i.e., cos(piT) = cos(0) = 1 i.e., cos(pi.T) = 1. What is the smallest such number T so that is the case?
ooh i think i understand now
lets say for y(x)=-2cos4x the period would be 2pi?
By definition, it is the smallest number T such that y(x+T) = y(x) i.e., -2 cos(4(x+T)) = - 2 cos(4x) i.e., cos(4x + 4T) = cos(4x) Now cos has period 2pi. Hence 4T = 2pi or T = pi/2. Therefore the period of y(x) is T = pi/2.
or in other words, as imran was just writing, if you have a function f1(x) = sin(ax) or f2(x) = cos(ax), as both sin and cos have period 2pi, it follows that the period of both f1 and f2 is 2pi/a.
For example, the period T of f1 is the smallest number T such that f1(x + T) = f1(x) i.e., sin(a(x+T)) = sin(ax) i.e., sin(ax + aT) = sin(ax) i.e., aT = 2pi, because the period of sin is 2pi i.e., T = 2pi/a
No, it's just new for you. Do it a few more times and it'll be easy for you.
My teacher never taught me this and I'm trying to do the homework using the book but I find it really complicated
It's like the first time you saw algebra. It seemed hard, but now you can solve equations like 2x + 4 = 6 In your sleep.
yes but that is very simple math lol
For me, the questions you're asking are also very simple.
but there was a time when they were new for me too. Just stick with it, and do a few more problems!
Former University Lecturer
for y(x)=-2cos4x the amplitude is 2, correct?
Yes, the amplitude of y(x) = -2cos(4x) is 2, correct.
how do i find the frequency?
What's the definition of frequency, f?
I'll tell you. If a function is periodic, i.e., oscillates, it has a period, T such that f(t+T) = f(t) The frequency is the number of complete oscillations per unit of time. For example if T = 1, then there would be one oscillation per unit of time, seconds say. I.e., f = 1. If T = 2, there would be 1/2 an oscillation per second. I.e., f = 1/2. If T = 1/2, there would 2 oscillations per second, f = 2. Given all that, what is the relation between T and f?
ooohh so the period of y(x)=-2cos4x is pi/2? because 2pi/4 is pi/2
So now, returning to my last post and your question on frequency. What is the relationship between period T and frequency f?
1? because 1/2 times 1 is 2
Yes, so f = 1/T. Or T = 1/f
Hence the higher the period, the lower the frequency. Makes sense because if the period is longer, there can be less complete oscillations in a second. Or the lower the period, the higher the frequency.
The lower the frequency, the higher the period. The higher the frequency, the lower the period.
so if the period is pi/2, f=1/(pi/2)?
Yes. If T = pi/2, then f = 2/pi.
Which means every unit of time there are 2/pi complete oscillations.
I have another problem :/ E(t)=110cos(120pit-pi/3)?
So you should know enough now to try and figure this out. First, what's the amplitude?
Yes, exactly. Now remember that the period of cos and sin is \( 2\pi \). I.e., \[ \cos(x + 2\pi) = \cos(x) \ \ \ \ \hbox{ and } \ \ \ \sin(x + 2\pi) = \sin(x) \]
So go back to first principles to find the period T of your new function E(t). It is the number T such that E(t+T) = E(t) i.e., \[ 110 \cos(120\pi(t+T)-\pi/3) = 110 \cos(120\pi t-\pi/3) \] i.e., \[ \cos(120\pi(t+T)-\pi/3) = \cos(120\pi t - \pi/3) \] i.e., \[ \cos(120\pi t - \pi/3 + 120\pi T) = \cos(120\pi t - \pi/3) \] i.e., \[ 120\pi T = 2\pi\] because the period of cos is \( 2\pi \) Hence T = ... what?
If 120πT=2π, then T = ...
pi/60pi=1/60=0.016666667
T = 1/60. Don't write the decimal expansion unless you really, really have to. Always messy.
Now that looked complicated, but you can always just read it off from the coefficient of t. Given E(t)=110cos(120pi.t-pi/3), the coefficient of t is 120pi. Hence the period is T = 2pi/120pi = 1/60.
yes i actually got it! aah thank you soooo much!
What is the frequency of E(t). What is the value of f for that function?
Yes, exactly. If the units of t are seconds, E(t) has 60 complete oscillations per second. If the units of t are days, then E(t) has 60 complete oscillations per day. Etc.
If the units of t are seconds, E(t) has a complete oscillation every 1/60 seconds.
Try and answer the first part of this problem: http://openstudy.com/#/updates/4ed01d1de4b04e045af4e631
i think its 8 but im not sure