Calculus 1

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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.

  • Calculus 1 -

    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?

  • Calculus 1 -

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

  • TayB's question - Calculus 1 -

    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

  • Calculus 1 -

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

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