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a child puts beads on one spoke of a bicycle wheel. the tire has a diameter of 2ft. if the child rides so that the tire makes one full rotation every 15 sec, and the beads begin in the horizontal outward position, find an equation that models the position of the beads at time t.

  • Trig -

    we know we want

    y = a sin(bx+c) + d

    diameter is 2, so radius=1, and we have

    y = a sin(bx+c) + 1

    Starting at t=0, the bead starts going downward, so

    y = -sin(bx+c)+1

    If we make it so the axle is at (0,1) at t=0, we need to shift the graph to the right by 1, since the bead is on the rim

    y = -sin(b(x-1)) + 1

    The period is 15, so b = 2π/15

    y = -sin(2π/15 (x-1))
    = -sin(2π/15x - 2π/15) + 1

    Naturally, if we took the bead starting at (0,0), we'd have

    y = -sin(2π/15 x)

    but the axle-centered graph shifts things to the right and up.

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