How do I find the curved length of one complete cycle of the graph of y=cos(x)? I have no clue where to begin. :/ How do I find the curved length of one complete cycle of the graph of y=cos(x)? I have no clue where to begin. :/ @Mathematics
iw as thinking it was 2pi, but thats not it is it .... thats the period of cos(x)
\[ds=\sqrt{dx^2+dy^2}\] is the basic set up for the integral of a curve
hmmm, i think you just integrate that .... from 0 to 2pi
\[\int_{0}^{2pi}\sqrt{1+sin^2(x)}\ dx\] maybe
well if you graph it, you would see the length = 2pi
the circumference of the unti circle is 2pi; but i dont think the length of the cos(x) curve is 2pi
you are talking about one cycle of the graph or a period i believe
1 period of that graph is 2 pi
unless you change the length of the perion by going y=2cosx or something like that it would be 2 pi
from x=0 to x=2pi
well i was talk in calc this semester that the length would = 1 period which would be 2 pi
a complete cycle equals 1 complete positive and negative
the length of the interval from x=0 to x=2pi is 2pi we are looking for the length of curve on that interval
which is a period. guess he would have to look at the book and see which they are looking for
the length of the curve would have to be greater than 2pi; since 2pi is the linear distance from the origin; and the curve, by nature of being other than linear; would be greater
period is not the length of the curve, but is the time that it takes for the curve to complete one cycle
amistre has the right formula for arc length at the top.
and i have the right answer :)
|dw:1321412075493:dw|
right right :)
amistre always gets credit first :(
no wait, isnt what you have just the area under the curve?
no
AH! i see it, my bad.
\[sin^2 + cos^2 = 1\] \[sin^2 = 1-cos^2\]
stupid minus sign
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