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Hi I'm having trouble with these graphs, I don't know how to get the AMP or the period.
http://img.photobucket.com/albums/v6...e/problem5.gif
has amplitude _________ and period ________ .
and
http://img.photobucket.com/albums/v6...e/problem6.gif
shows the graph of the function
f(x) = _________________
Your answer may be of the form f(x)= asin(bx+c) where a, b, and c are small integers or Pi
The amplitude is the value at the peaks. The period is the time between two peaks (or two valleys).Quote:
Originally Posted by killasnake
Re graph #1 :Quote:
Originally Posted by killasnake
Amplitude is maximum vertical displacement from the horizontal axis of the sinusoidal curve.
Here the horizontal axis coincides with the x-axis. Maximum vertical displacement from the x-axis is shown as 3 or -3. Therefore, the amplitude of the curve is 3.
Period is one cycle. Any two points on the horizontal axis that show the same characteristic define the period of the sinusoidal curve. I see that you are using, or you mentioned, a sine curve in the other graph, so let us use the sine curve for this graph here. For a sine curve, it is easier to base from the beginning and end of one cycle of the curve. For one period, the sine curve begins at the horizontal axis, goes up to maximum height, goes down to "minimum" height, and then returns to the horizontal axis again.
In this graph, the horizontal axis is the x-axis, so look for a point near the (0,0) that is on the x-axis, or where y=0, where the graph is about to rise up. We see one at x = about, 0.3.
Then find the next one further to the right. We see it at x = about, 2.3.
So between x=about 0.3 to x=about 2.03, there is one cyle or one period.
The distance between this two points is (2.03 -0.3) = 2.
Therefore, the period of the graph is 2.
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Re graph #2:
" Your answer may be of the form f(x)= asin(bx+c) where a, b, and c are small integers or Pi. "
Same explanation as above.
So,
Amplitude = 2.
"a" = 2 --------------***
Period.
Nearest to (0,0), one point where y=0 is at x = about, -0.65.
Next one to the right is at x = about 1.3.
So, period is about 1.3 -(-0.65) = 1.95.
Period is (2pi)/b
So, about 1.95 = (2pi)/b
b = (2pi)/(about 1.95) = about pi.
Since it is mentioned that b is integer or pi, then b = pi --------***
Horizontal shift c :
Reason why I said start looking for y=0 nearest to the origin (0,0) is because that point defines the horizontal shift of the sinusoidal curve.
Nearest is found at x = about negative 0.65. But c must be an integer or pi.
Integer is no way, so c must be pi or "some" pi.
Phase or horizontal shift = c/b
0.65 = c/pi
c = 0.65pi
We express the 0.65 into pi-terms,
0.65 = n*pi
n = (0.65)/pi = (0.65)/(3.1416) = 0.2 = 1/5
Hence, c = (pi/5)(pi) = (pi^2)/5 -----------***
Therefore, curve #2 is the graph of
y = 2sin[(pi*x) +(pi^2)/5] ------answer.
"c" is positive because the "start" of the sine curve is to the left of the origin.
Check,
y = 0 at "start" of sine curve,
0 = 2sin[(pi*x) +(pi^2)/5]
0 = sin[(pi*x) +(pi^2)/5]
pi*x +(pi^2)/5 = arcsin(0) = 0
pi*x = -(pi^2)/5
x = -pi/5 = -0.6283 -----okay.