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Math Help - Calc 1 problem

  1. #1
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    Calc 1 problem

    hey, can anybody help solve this HW question? thanks!
    Write a function that models the distance D from a point on the line y = 3 x - 6 to the point (0,0) (as a function of x).
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  2. #2
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    Quote Originally Posted by smoti10 View Post
    hey, can anybody help solve this HW question? thanks!
    Write a function that models the distance D from a point on the line y = 3 x - 6 to the point (0,0) (as a function of x).

    A point in the plane belongs to your line iff it is of the form (x,3x-6) . Well, now write down the formula for the distance from this generic

    point on the line to the origin (0,0).

    Tonio
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  3. #3
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    Each point on the line can be expressed as a coordinate in the form...

    (x, 3x - 6).

    Why? A coordinate is of the form (x,y) and y = 3x - 6

    So what to do is think in terms of Pythagoras triangle... You have two sides, one of length x, one of length 3x-6 and one unknown side that you must find.

    Why? Well just think of the coordinate (x, 3x - 6) as being distance along the x-axis then distance up to get to the point. This will form a right angled triangle with side lengths x and 3x -6

    So, let's call this unknown side d, using the standard Pythagoras formula we get...

    d^2 = x^2 + (3x - 6)^2.

    Now you finish this off by expanding everything and you will be left with a formula of the form...

    d^2 = ...

    then just take square roots to get,

    d = \sqrt{\dots}

    which will be your answer.
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  4. #4
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    thanks for your help

    thank you very much for your help! now,once you got sqrt(10x^2 - 36x + 36) how can you find out where is the closest point to the point (0,0) on that line?
    Thanks again.
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by smoti10 View Post
    thank you very much for your help! now,once you got sqrt(10x^2 - 36x + 36) how can you find out where is the closest point to the point (0,0) on that line?
    Thanks again.
    Well you want to find the value of x that makes that function the smallest.

    This will happen at one of two places.

    A critical point,
    An end point of the range of x. (i.e. x = \pm \infty).

    Immediately you can see that that if x \to \pm \infty then f(x) \to \infty (if not I'll explain).

    So that leaves you to find critical points...

    I'll start you off...

    If f(x) = \sqrt{10x^2 - 36x + 36}

    => f'(x) = \frac{2(5x-9)}{\sqrt{10x^2 - 36x + 36}}.

    Solve for this to be 0 and what does that tell you..?
    Last edited by Deadstar; March 31st 2010 at 12:41 PM. Reason: Changed fixed point to critical point.
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  6. #6
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    Not "fixed point", critical point. A fixed point for a function, f(x), is a value of x such that f(x)= x.
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  7. #7
    Super Member Deadstar's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Not "fixed point", critical point. A fixed point for a function, f(x), is a value of x such that f(x)= x.
    Oops! Yes correct, I'm currently working away on my dynamical systems tutorials so probably just got a bit of cross subject banter lol.
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