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Mathematics 19 Online
OpenStudy (anonymous):

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

OpenStudy (amistre64):

iw as thinking it was 2pi, but thats not it is it .... thats the period of cos(x)

OpenStudy (amistre64):

\[ds=\sqrt{dx^2+dy^2}\] is the basic set up for the integral of a curve

OpenStudy (amistre64):

hmmm, i think you just integrate that .... from 0 to 2pi

OpenStudy (amistre64):

\[\int_{0}^{2pi}\sqrt{1+sin^2(x)}\ dx\] maybe

OpenStudy (anonymous):

well if you graph it, you would see the length = 2pi

OpenStudy (amistre64):

the circumference of the unti circle is 2pi; but i dont think the length of the cos(x) curve is 2pi

OpenStudy (anonymous):

you are talking about one cycle of the graph or a period i believe

OpenStudy (anonymous):

1 period of that graph is 2 pi

OpenStudy (anonymous):

unless you change the length of the perion by going y=2cosx or something like that it would be 2 pi

myininaya (myininaya):

from x=0 to x=2pi

OpenStudy (anonymous):

well i was talk in calc this semester that the length would = 1 period which would be 2 pi

OpenStudy (anonymous):

a complete cycle equals 1 complete positive and negative

myininaya (myininaya):

the length of the interval from x=0 to x=2pi is 2pi we are looking for the length of curve on that interval

OpenStudy (anonymous):

which is a period. guess he would have to look at the book and see which they are looking for

OpenStudy (amistre64):

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

OpenStudy (amistre64):

period is not the length of the curve, but is the time that it takes for the curve to complete one cycle

OpenStudy (anonymous):

amistre has the right formula for arc length at the top.

myininaya (myininaya):

and i have the right answer :)

OpenStudy (amistre64):

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OpenStudy (anonymous):

right right :)

myininaya (myininaya):

amistre always gets credit first :(

OpenStudy (anonymous):

no wait, isnt what you have just the area under the curve?

myininaya (myininaya):

no

OpenStudy (anonymous):

AH! i see it, my bad.

OpenStudy (amistre64):

\[sin^2 + cos^2 = 1\] \[sin^2 = 1-cos^2\]

myininaya (myininaya):

stupid minus sign

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