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Math Help - Check my calculations ^_^

  1. #1
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    Check my calculations ^_^

    I keep trying this question over and over again, and keep getting the wrong answer. I was wondering if somebody would be nice enough to check my calculations, and see where I went wrong (or if Iím going about it the wrong way!)

    Find the minimum distance from the parabola y=x^2 to the point (5,0).

    D = √[(x2-x1)^2 + (y2-y1)^2]

    I substituted in points (5,0) and (x, x^2)

    D = √[(x-5)^2 + x^2
    D = √2x^2 - 10x + 25

    d/dx = (4x-10)/ [blahblah not important]
    0 = 4x-10
    4x=10
    x=10/4

    Substituting this into the D formula doesn't give the right answer ... the answr is apparently 4.06

    Thanks!
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  2. #2
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    Quote Originally Posted by freswood
    I keep trying this question over and over again, and keep getting the wrong answer. I was wondering if somebody would be nice enough to check my calculations, and see where I went wrong (or if Iím going about it the wrong way!)

    Find the minimum distance from the parabola y=x^2 to the point (5,0).

    D = √[(x2-x1)^2 + (y2-y1)^2]

    I substituted in points (5,0) and (x, x^2)

    D = √[(x-5)^2 + x^2
    The above should be:

    <br />
D =\sqrt{ (x-5)^2 + x^4}<br />

    Also it is easier to minimise D^2

    RonL
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  3. #3
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    Oops silly me Even with that in mind, I still can't seem to solve it.

    d/dx = (2x^3 + x - 5) / [blah]

    so how do you solve:

    0 = 2x^3 + x - 5
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by freswood
    Oops silly me Even with that in mind, I still can't seem to solve it.

    d/dx = (2x^3 + x - 5) / [blah]

    so how do you solve:

    0 = 2x^3 + x - 5
    Unless your a dab hand with the cubic formula you solve it numericaly,
    the (real) solution is x \approx 1.235 , which you now
    need to substitute back into the distance formula to find the distance
    correspomding to this x.

    RonL
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  5. #5
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    Quote Originally Posted by freswood
    Oops silly me Even with that in mind, I still can't seem to solve it....
    so how do you solve:
    0 = 2x^3 + x - 5
    Hi, freswood,

    I can only think of 2 ways:

    1. Use an iteration (for instance Newtons Method)

    2. Use Cardanos Formula to solve a reduced cubic equation.

    Never mind which method you use, there is only one real solution:

    x=\sqrt[3]{\frac{5}{4}+\sqrt{\frac{677}{432}}}+\sqrt[3]{\frac{5}{4}-\sqrt{\frac{677}{432}}} \approx  1.23477

    Best wishes

    EB
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  6. #6
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    Thanks for that guys! It's pretty bizarre though - we haven't learnt that method of solving... after the holidays I'll go and ask my teacher how we're supposed to do it.
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