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Calculus problem involving finding coordinates

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Using calculus, find the coordinates of the point on the line y =-2x+5, which is closest to the origin, and the corresponding value of D

    2. Relevant equations

    y = -2x +5


    3. The attempt at a solution

    I know I need to find a line that is perpendicular to the line of the equation, but I'm not sure how to find the equation of this line so that I can make the equations equal to each other.

    Something like

    d(x) = x2+y2
    = x2 + (-2x+5)2
    = x2 + 4x2+25
    = 5x2 + 25

    am I on the right track?
     
  2. jcsd
  3. Aug 13, 2013 #2
    Assuming that [tex](-2x + 5)^{2} = 4x^{2} + 25[/tex] is terrible.
    I'm not good at calculus but this problem is quite easy, even though a mistake like this one can ruin your answer.
    Be careful that
    [tex](-2x + 5)^{2} = (-2x + 5)(-2x + 5) =\ ...[/tex]
     
  4. Aug 13, 2013 #3

    LCKurtz

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    To add to besulzbach's reply, note that what you are calculating is the distance squared from the origin to a point on the line. (Not that there's anything wrong with that.) :smile:
     
  5. Aug 13, 2013 #4
    If the line y1 = -2x + 5 has a slope of -2, then a line perpendicular will have a slope of 1/2.

    Since you are trying to find the points closest to the origin, the equation of the perpendicular line is y2 = (1/2)x

    This should be enough...
     
  6. Aug 13, 2013 #5

    LCKurtz

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    ...although whether it qualifies as a method "using calculus" is uncertain.
     
  7. Aug 13, 2013 #6
    True... he mentioned wanting to find a line perpendicular, so I went with that.
     
    Last edited: Aug 13, 2013
  8. Aug 13, 2013 #7
    Ah that's embarrassing! I forgot to expand the problem, I been away from math for to long haha. thanks for the replies guys!

    So
    is it 4x2 - 10x -10x + 25
    = 4x2-20x+25


    if f(x)= 4x2-20x+25
    then f'(x) = 8x -20
    and f''(x) = 8

    if 8x - 20 = 0
    8x= 20
    x= 20/8
    x=2.5

    y = -2x + 5
    y = -2(2.5) + 5
    y = 0

    So coordinates are (2.5, 0)
    How does this look
     
  9. Aug 13, 2013 #8
    Using Calculus:

    f(x) = x2 + y2, where y = -2x + 5

    so, f(x) = 5x2 - 20x + 25;
    therefore, f'(x) = 10x - 20.

    If f'(x) = 0 when f(x) is minimum, then

    0 = 10x - 20, so x = 2.

    Then, y = y(x) = y(2) = -2(2) + 5 = 1.

    The coordinate is (2, 1).


    NOTE: This agrees with the other "non-Calculus" method I used earlier.
     
    Last edited: Aug 13, 2013
  10. Aug 13, 2013 #9

    LCKurtz

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    You left out an ##x^2## from the other term.
     
  11. Aug 14, 2013 #10
    [tex]f(x) = 5x^{2} - 20x + 25 [/tex]
    Is NOT the function for the distance, it is the formula to find the square of the distance!
    Use that [itex] f(x) [/itex] to make a function [itex] d(x) [/itex] that gives you the distance for any input value of [itex] x [/itex].
    [tex] d(x) = \sqrt{5x^{2} - 20x + 25} [/tex]
    Any point [itex](x, (-2x + 5))[/itex] will be [itex]\sqrt{5x^{2} - 20x + 25} [/itex] from the origin. Now you can calculate D as you wanted.
     
  12. Aug 23, 2013 #11
    thanks for your help guys
    So the next question asks me to make a conjecture and prove it with algebra

    So I know that I need to do the same thing as with the first part but using

    y = ax + b

    f(x)=x2+y2

    f(x)=x2+(ax+b)(ax+b)

    f(x)=x2+a2 x2+2axb+b2

    then do I differentiate this?

    I'm not sure how to make a conjecture based on this though
     
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