anonymous
  • anonymous
Could anyone please help me learn how to determine the period of a polar or trigonometric parametric function? What I mean is, how do I determine the point in which the curve start tracing itself again over? Up to what point do I trace the whole curve? Our professor and math tutors suggest using the Calculor Window and increment the Tetha Max, but that is rather inaccurate if we need the exact bounds to find the Surface Area of a Curve. Thank you for your help.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i don't think there is a general method for that. It will depend on your functions. Basicly it will be epressed in terms of sin and cos and their combinations. So you will have to transfor the expression in some way to see it
amistre64
  • amistre64
you can tell by seeing if there is a "slope" in the arguement if its linear
amistre64
  • amistre64
\[a\ trig(mx+b)\] type thing; the m there would determine the perodicity based on the trig function itself

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amistre64
  • amistre64
for sin and cos; 2pi/m for tan ; pi/m
anonymous
  • anonymous
I see.
anonymous
  • anonymous
For example, we had x = e^sin(tetha) and y = e^cos(tetha) but we had no clue on how to determine the bounds of the integral.
anonymous
  • anonymous
sin cos are bouth periodic, with same period so... 2pi
anonymous
  • anonymous
AS myko states, I was kind of looking for a general rule, but it does seem to be something that is treated in a case by case basis.
amistre64
  • amistre64
yes, trig tends to be very case by case determined :/
anonymous
  • anonymous
But then we had x = cos3(tetha) and y = cos(tetha) and the period was no longer 0 - 2pi.
anonymous
  • anonymous
thats right, couse look: tg =sin/cos also involves sin and cos, like your exponencial, but has period pi, half the sin, cos
anonymous
  • anonymous
so, in the case of the last one, x = cos 3(tetha) and y = cos (tetha); I was thining the period would be 2pi, how would I determine the right one? Thank you
anonymous
  • anonymous
it would be 2pi i guess, couse you would choose the longest of bouth cos3x, cosx
anonymous
  • anonymous
the solutions manual says 0 - pi :s
anonymous
  • anonymous
hmm... cos3x =cos(2x+x)=cos2xcosx-sin2xsinx = cos^3x-sin^2xcosx -2sin^2xcosx=cos^3x -3sin^2xcosx maybe you can see it from here....
phi
  • phi
Here is how I would do it. There may be simpler ways?
anonymous
  • anonymous
@phi this is amazing!!!!!!!!! (exclamation points intended)

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