Calculus

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Find the arc length of the curve described by the parametric equation over the given interval:
x=t^(2) + 1
y=2t - 3
0<t<1

  • Calculus -

    x = t^2 + 1
    t^2 = x-1
    t = √(x-1)

    y = 2t - 3
    t = (y+3)/2

    (y+3)/2 = √(x-1)
    y = 2√(x-1) - 3

    when t=0, x = 1, y=-3
    when t = 1, x = 2, y = -1
    so we want the length of the curve of
    y = 2√(x-1) - 3 from the point (1,-3) to (2, -1)

    If I recall, the length of a line segment
    = ∫√( 1 + (dy/dx)^2 ) dx from 1 to 2

    now dy/dx = 1/√(x-1)
    so (dy/dx)^2 = 1/(x-1)

    ∫√( 1 + (dy/dx)^2 ) dx from 1 to 2
    = ∫√( 1 + 1/(x-1) ) dx from 1 to 2
    = ∫√( 1 + (dy/dx)^2 ) dx from 1 to 2
    = ∫ ( (x+2)/(x+1) )^(1/2) dx from 1 to 2

    now running into a brick wall, can't seem to integrate that.

  • Calculus -

    ds = √(dx^2 + dy^2)

    s = ∫[0,1] √[(2t)^2+(2)^2] dt
    = ∫[0,1] √(4t^2+4) dt
    = 2∫[0,1] √(t^2+1) dt
    = t√(t^2+1) + arcsinh(t) [0,1]
    = √2+arcsinh(1)

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