Results 1 to 4 of 4

Math Help - shortest distance

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    2

    shortest distance

    Determine the shortest distance from point A (-5, -2sqrt7) to the X-axis to point B ( 4, -sqrt7).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,915
    Thanks
    779
    Hello, strwbrry869!

    Determine the shortest distance from point A(\text{-}5,\:\text{-}2\sqrt{7}) to the x-axis to point B(4,\:\text{-}\sqrt7)
    There is a back-door approach to this problem.
    Code:
                        |             _
                        |         (4,√7)
                        |           o B'
                        |        *  :
        -5              |  P  *     :
      ---+--------------+--o--------+----
         :              *     *     :
         :           *  |        *  :
         :        *     |           o B_
         :     *        |         (4,-√7)
         :  *           |
       A o   _          |
      (-5,-2√7)         |

    Reflect point B over the x-axis to point B'.
    Draw line AB', intersecting the x-axis at P.

    P gives the shortest distance from A to P plus from P to B.
    . .
    (Do you see why?)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Using calculus, the problem gets VERY messy!


    The distance from A(-5,-2\sqrt{7}) to P(x,y) is:

    . d_1 \:=\:\sqrt{(x+5)^2 + (2\sqrt{7})^2} \:=\:\sqrt{x^2 + 10x + 53}


    The distance from B(4,-\sqrt{7}) to P(x,y) is:

    . . d_2 \;=\;\sqrt{(x-4)^2 + (\sqrt{7})^2} \:=\:\sqrt{x^2+10x + 23}


    The total distance is: . D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}


    And that is the function we must minimize . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    2
    thanks but im not good at math at all... do you know what the answer to that would be?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2009
    Posts
    2
    Continuing the Calculus way (the best I can):

    The distance formula, D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}, looks right. I think from there you would have to take the derivative of that and set it equal to 0 to find the minimum value.

    So:

    <br />
\frac{dD}{dx} = \frac{1}{2}(x^2+10x+53)^\frac{-1}{2}*(2x+10) + \frac{1}{2}(x^2+10x+23)^\frac{-1}{2}*(2x-8)<br />

    <br />
\frac{dD}{dx} = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}}<br />

    To find the minimum, we substitute 0 for \frac{dD}{dx}.

    <br />
0 = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}}<br />

    That is an extremely difficult equation to factor, so I just put it in my calculator and used the Equation Solver.
    I found that:

    <br />
x = 0.34110898362998....<br />

    So, I guess that that (0.3411, 0), would be your point of intersection on the x-axis.
    So you can now just find the distance between A and the point of intersection, and then the distance between B and the point of intersection, and add them together.

    Hope that helps a little. Seems like an extremely hard problem for an Algebra student. I'm in Calculus now, and that seems pretty difficult to me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Shortest distance
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 10th 2009, 10:37 PM
  2. Shortest distance possible...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 24th 2008, 11:02 PM
  3. xyz^2=5 shortest distance
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 27th 2008, 12:34 PM
  4. shortest distance
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 16th 2007, 10:10 AM
  5. shortest distance
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: December 18th 2006, 09:48 AM

Search Tags


/mathhelpforum @mathhelpforum