anonymous
  • anonymous
Can someone help me with determining the equation for a trig function? I'm aware of how to find the amplitude and period, i'm just confused with this graph!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
MTALHAHASSAN2
  • MTALHAHASSAN2
wait how can we find the amplitute
anonymous
  • anonymous
it's usually the "a", I know how to determine it by looking at an equation

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jim_thompson5910
  • jim_thompson5910
\[\large \text{amplitude} = \frac{|m-n|}{2}\] m = largest y coordinate of a point on the curve n = smallest y coordinate of a point on the curve
jim_thompson5910
  • jim_thompson5910
|dw:1444532222391:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444532254763:dw|
anonymous
  • anonymous
in this case it would be 2+4/2
anonymous
  • anonymous
which would be 3..
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
the midline is found by averaging the two extreme values \[\Large D = \frac{m+n}{2}\] y = D is the equation of the midline |dw:1444532397834:dw|
anonymous
  • anonymous
the midline is -1
jim_thompson5910
  • jim_thompson5910
|dw:1444532428898:dw|
jim_thompson5910
  • jim_thompson5910
`the midline is -1` yes
anonymous
  • anonymous
so for the period since the points aren't actually on the radians how would I calculate where it starts and where it ends?
jim_thompson5910
  • jim_thompson5910
look at the peak that lies on the y axis this is when x = 0 the next peak over to the right is at x = pi so the period is |pi - 0| = |pi| = pi units
anonymous
  • anonymous
I'm a little confused..
anonymous
  • anonymous
when x=0 y=2, is that the peak that you're talking about?
jim_thompson5910
  • jim_thompson5910
see the attached image
1 Attachment
anonymous
  • anonymous
ok thanks! I got that. I'm just confused why we are subtracting the pi-0. I thought the formula for period was 2pi/k
jim_thompson5910
  • jim_thompson5910
you might be thinking of T = 2pi/B and that is useful when you have an equation of this form y = A*cos(Bx - C) + D
jim_thompson5910
  • jim_thompson5910
I subtracted the x values to find the distance between them
jim_thompson5910
  • jim_thompson5910
the absolute value is to ensure the distance is positive
anonymous
  • anonymous
so for this graph we don't need to use that form. We just find distance simply by finding out the x values from the y and subtract them from one another?
jim_thompson5910
  • jim_thompson5910
you can pick on any 2 neighboring peaks and do the same thing
jim_thompson5910
  • jim_thompson5910
we can't use that form because we don't have the equation
anonymous
  • anonymous
oh ok! That's where I was confused at :)
jim_thompson5910
  • jim_thompson5910
you might have learned how to pick apart y = A*cos(Bx - C) + D (eg find the amplitude to be |A|, etc) but now we're going in reverse: finding the key elements and using them to construct the equation
anonymous
  • anonymous
so our equation would be y=3sin(x)-1?
anonymous
  • anonymous
y=sin(pi)-1
jim_thompson5910
  • jim_thompson5910
what is y when x = 0 ?
anonymous
  • anonymous
2
jim_thompson5910
  • jim_thompson5910
but does plugging in x = 0 give y = 2 when you use y = 3*sin(x) - 1 ?
anonymous
  • anonymous
no
jim_thompson5910
  • jim_thompson5910
so that equation doesn't fit
jim_thompson5910
  • jim_thompson5910
let's try cos instead of sin
jim_thompson5910
  • jim_thompson5910
if x = 0, does y = 2 for y = 3*cos(x) - 1 ?
anonymous
  • anonymous
it does fit
jim_thompson5910
  • jim_thompson5910
yep, ok how about if x = pi/2, does y = -4 like it should?
anonymous
  • anonymous
we know that cos(x)=1
anonymous
  • anonymous
I don't think so
anonymous
  • anonymous
cos of pi/2 is 1
jim_thompson5910
  • jim_thompson5910
cos(pi/2) = 0 actually
anonymous
  • anonymous
and 3-1=2, and 2-1=1
jim_thompson5910
  • jim_thompson5910
recall that the period is pi units so T = pi Solve the equation T = 2pi/B for B to get... T = 2pi/B pi = 2pi/B B*pi = 2pi B = 2pi/pi B = 2 so the equation should be y = 3*cos(2x) - 1
jim_thompson5910
  • jim_thompson5910
now if we try x = pi/2, then y = 3*cos(2x) - 1 y = 3*cos(2*pi/2) - 1 y = 3*cos(pi) - 1 y = 3*(-1) - 1 y = -3 - 1 y = -4 and it fits
anonymous
  • anonymous
I'm a little lost. Why are we using t=2pi/3?
anonymous
  • anonymous
B*
jim_thompson5910
  • jim_thompson5910
The period is T = pi going from peak to peak is pi units across
jim_thompson5910
  • jim_thompson5910
plug T = pi into T = 2pi/B and solve for B to get B = 2
jim_thompson5910
  • jim_thompson5910
T = 2pi/B is the period formula If you know B, you can find the period T (or vice versa)
anonymous
  • anonymous
so we just knew the T, but in order to find b we had to plug in pi for t
jim_thompson5910
  • jim_thompson5910
exactly
anonymous
  • anonymous
and how did you know that y=-4?
anonymous
  • anonymous
just the point on the graph?
jim_thompson5910
  • jim_thompson5910
1 Attachment
jim_thompson5910
  • jim_thompson5910
Yes the point (pi/2, -4) is on the graph Notice how in my work above I replaced x with pi/2 in the equation y = 3*cos(2x) - 1 Then I evaluated/simplified to get y = -4. So that confirms the equation works for x = pi/2
anonymous
  • anonymous
oh so you just replaced it to check to see if the equation that we wrote was correct.. makes sense. :) Thank you lots!
jim_thompson5910
  • jim_thompson5910
yes I recommend you check x = pi as well. Make sure that the output is y = 2
anonymous
  • anonymous
Thanks again :) You're a life saver! I honestly had a lot of trouble with this problem
jim_thompson5910
  • jim_thompson5910
I'm glad it's making sense now

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