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anonymous

  • one year ago

What is the period? \( \huge y = 10 \sin(\frac{2\pi}{365}(x-50)) \) Is the period \( \huge frac{2\pi}{\frac{2\pi}{365}} \) which would be 365??

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  1. anonymous
    • one year ago
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    What is the period? \( \huge y = 10 \sin(\frac{2\pi}{365}(x-50)) \) Is the period \( \huge \frac{2\pi}{\frac{2\pi}{365}} \) which would be 365??

  2. anonymous
    • one year ago
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    @phi

  3. phi
    • one year ago
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    if you manipulate the coeff of x into the form \[ \frac{2\pi}{T} x \] then T is the period

  4. phi
    • one year ago
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    you have 2pi/365 the 365 matches T so yes, the period is 365

  5. anonymous
    • one year ago
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    Ok, so 365 is T so does the 50 compress the period?

  6. anonymous
    • one year ago
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    Or is shifts it to the left

  7. phi
    • one year ago
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    no the -50 * 2pi/365 is a phase shift (moves the curve to the right)

  8. anonymous
    • one year ago
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    Ok thanks so we have T = 365 with a phase shift to the right

  9. phi
    • one year ago
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    the red curve is the sin without the -50 term the black curve is yours.

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  10. anonymous
    • one year ago
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    Thank you!

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