Calculus 1

posted by TayB

Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola y = (x − 5)^2 that is closest to the origin.

  1. Nardos

    Two cars A and B are moving along a straight road in the same direction withe velocity of 25km/h and 40km/h, respectively. Find the velocity of car B relatively to car A?

  2. TayB

    Nardos, did you post to the wrong question by accident. Because what you posted does not even relate to my question

  3. Reiny

    let the point of contact be (a,b), then
    b = (a-5)^2 = a^2 - 10a + 25

    sketch the tangent and the normal
    y' = 2(x-5) = 2x - 10
    so the slope of the tangent at (a,b)
    = 2a - 10
    and the slope of the NORMAL at (a,b)
    = -1/(2a - 10) or 1/(10-2a)

    but using the old grade 9 way of doing slope ...
    slope of normal = (b-0)/(a-0) = b/a

    thus:
    1/(10-2a) = b/a
    a = 10b - 2ab
    subbing in b
    a = 10(a^2 - 10a + 25) - 2a(a^2 - 10a + 25)
    a = 10a^2 -100a + 250 - 2a^3 + 20a^2 - 50a
    2a^3 - 30a^2 + 151a - 250 = 0

    using x instead of a,
    f(x) = 2x^3 - 30x^2 + 151x - 250
    f ' (x) = 6x^2 - 60x + 151
    by Newtons' Method, the simplified expression is

    newx = (4x^3 - 30x^2 + 250)/(6x^2 - 60x + 151)
    according to my sketch x = 3.5 might be a good starting point

    newx = 3.765227...
    next newx = 3.765227..
    I think we got it after two iterations

    carry on, you have the x, now find the y

  4. Anonymous

    Reiny,
    Aren't we supposed to use distance formula for this though?

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