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arcticf0x

  • 2 years ago

Show that every point on the curve y=b sin (x/a), where the curve meets the axes of x, is a point of inflextion. Do i have to plainly show that x=0 is a POI or its asking for something else?

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  1. amistre64
    • 2 years ago
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    inflection tends to be a found by a second derivative; and tested for cavage

  2. arcticf0x
    • 2 years ago
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    yeah i know that, but the language isnt very clear up there.

  3. amistre64
    • 2 years ago
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    looks like: -b sin(x/a)/a^2 is the second derivative to me

  4. arcticf0x
    • 2 years ago
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    yeah

  5. amistre64
    • 2 years ago
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    since b=0 is trivial; it looks to be what x/a = any mutiple of pi

  6. arcticf0x
    • 2 years ago
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    sorry didnt get that.

  7. amistre64
    • 2 years ago
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    how to "show" it? i aint got a clue

  8. amistre64
    • 2 years ago
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    sin(n*pi) = 0 so when x/a = n*pi we are at zero

  9. arcticf0x
    • 2 years ago
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    yeah thats where i came, so is that the answer fianlly, x=0

  10. amistre64
    • 2 years ago
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    x = a n*pi would seem to be the answer to me; but since n is an arbitrary integer then a*n would have to be an interger as well.

  11. amistre64
    • 2 years ago
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    other than that; I got no idea what the "answer" might entail ....

  12. arcticf0x
    • 2 years ago
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    umm, it says, where the curve meets the axes of x, so y=0 at that point, this only gives x=0.

  13. phi
    • 2 years ago
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    inflection point is where the second derivative changes sign. we know that y=b sin (x/a) crosses the x-axis when x/a is some multiple of 2pi the 2nd derivative at these same points is -b sin(x/a)/a^2 = 0 also, if we go a dx distance below x/a and a little distance dx above x/a, the 2nd derivative changes sign. So we are at an inflection point.

  14. arcticf0x
    • 2 years ago
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    That seems clear, understood, thank you phi and amistre for your time :)

  15. phi
    • 2 years ago
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    *crosses the x-axis when x/a is some multiple of pi

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