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Math Help - Shortest distance possible...

  1. #1
    Member realintegerz's Avatar
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    Shortest distance possible...

    I see the graphs of these, using Geogebra, and I don't know how I can prove them..

    1) Point A is @ (0,2) and point B is on the graph of y=x^2 + 1, what should be the approximate location of B so that the distance from A to B is as short as possible?

    2) Point A is @ (3,1) and B is on a circle with center (1,2) and radius of 4. What is the closest point B should come to point A? Approximate to nearest thousandth.


    And I'm having difficulty with this problem as well, but it doesn't use the same ideology of the above two.

    A window is being built such that the bottom is a rectangle and the top is a semicircle. If there are 12 meter of framing material what should be the width of the window in order to let in the most light?


    Any help is appreciated...
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  2. #2
    Member realintegerz's Avatar
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    Ok, I got help [from outside MHF] for the last 2 problems but I still don't get the 1st one...

    can anyone please help?
    Last edited by mr fantastic; November 24th 2008 at 08:21 PM. Reason: Added [from outside MHF] to avoid confusion
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  3. #3
    o_O
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    Let d be the distance from (0,2) to the curve.

    Now if you remember your distance formula, ( d = \sqrt{(x-x_1)^2 + (y-y_1)^2}), we have this relationship: d^2 = (x-0)^2 + (y-2)^2 \ \iff \ d^2 = {\color{blue}x^2} + (y-2)^2

    Now, we can rearrange the equation of the curve to get: \color{blue}x^2 = y - 1

    So our relationship becomes: d^2 = {\color{blue}(y-1)} + (y-2)^2 \ \iff \ d^2 = y^2 - 3y + 3

    Differentiate both sides and solve for when the derivative d' is equal to 0 (as this will be the point where you have a minimum).

    Two things to note:
    (1) We are working in terms of y. So whatever your answer will be, it will give you the y-coordinate in which the distance is shortest from (0,2) to the curve. Plug this in to your equation of the curve to get the x-coordinates.

    (2) d is a function of y. So you must implicitly differentiate.
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  4. #4
    Member realintegerz's Avatar
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    Ok, so I solved for y as d was 0 on the left side

    I got y = 3/2 + sq. root ( - 3/4), assuming I ignored the negative root, y equaled 2.36

    and x is 1.16....


    I looked at a graph and the shortest distance is supposed to be 0.87, and the coordinates for B are (0.61, 1.38) to (0.79, 1.62)
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  5. #5
    Member realintegerz's Avatar
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    I got help earlier in the day about this problem, and my friend said that B is closest when it is perpendicular to the function
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  6. #6
    o_O
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    Are you sure you solved it correctly?

    You should get: y = \frac{3}{2} \ \Rightarrow \ x = \pm \frac{\sqrt{2}}{2} ..... i.e., \left(\pm \frac{\sqrt{2}}{2}, \frac{3}{2}\right)

    _______________________

    Yes, you can use the fact that the shortest distance from the point to a curve is when the line segment is perpendicular. This involves solving: {\color{red}\frac{y-2}{x-0}} = {\color{blue}-\frac{1}{y'}} \ \iff \ \frac{x^2 - 1}{x} = -\frac{1}{2x}

    The red represents the slope of the line segment from (0,2) to the a point on your curve. The blue represents the negative reciprical of the tangent at a given point. Whatever x values we get from solving this will be the values such that the line joining (0,2) to the curve is perpendicular to the curve. Note that x = 0 cannot be a solution to the equation but drawing a line segment from (0,2) to (0,1) fits the description above.

    Much shorter actually.
    Last edited by o_O; November 24th 2008 at 10:16 PM.
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