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Minimum distance of a point on a line from the origin?

  1. May 7, 2012 #1
    Consider the graph of $${x^2 + 2 y^2 = 1}$$.

    What is the minimum distance of a point on the graph to the origin?

    When I calculated the point which has minimum distance from origin it came out to be (1,0) implying the min distance to be 1. The answer is 1/2. Help?
     
  2. jcsd
  3. May 7, 2012 #2
    Hi !
    Show your calcul in whole details. So, it will be possible to locate the mistake.
     
  4. May 7, 2012 #3
    Okay.
    We're given x^2+2*y^2=1.
    so x^2=1-2y^2

    now using distance formula
    d^2=x^2+y^2
    since x^2=1-2y^2, substituting it in the distance formula we get:
    d^2=1-2y^2+y^2=1-y^2;
    differentiating and then setting the eq to 0 we get;
    0=-4y
    or y=0. now x^2=1-2y^2=1
    so x=+-1
    so point having min distance form origin is (+-1,0)

    using the distance formula now
    d^2=x^2+y^2
    d=sqrt(1+0)=1
     
  5. May 7, 2012 #4
    Your computation is correct, but there is a snag :
    If the derivative of a function is 0, then the function is minimum or maximum.
    The function 1-y² is decreassing, so for y=1 we have the maximum of d², not the minimum.
    (1-y²) is the smallest for the heighest value of y². Since x²+2y²=1, the highest value is y²=1/2. Finally, the smallest d=sqrt(1/2) is obtained at (x=0, y=sqrt(1/2))

    Less confusing, d² = x²+y² = x²+(1-x²)/2 = (1/2)+(x²/2)
    The function of x is increassing. The derivative leads to x=0 and the function is minimum at the point (x=0, y=sqrt(1/2)).
     
  6. May 7, 2012 #5
    Okay I get the second part but still don't get the first part.

    Let's talk about the second part first. You defined x in terms of y and then solved. I defined y in terms of x. Since your answer is smaller then mine so it's logical that your correct. Now does htat mean that everytime I have to do each question twice i.e. first x interms of y and then y interms of x. And if both your and mine methods are correct should'nt both answers be same.

    Now could you explain the first part in greater detail. I could'nt understand the following line:

    The function 1-y² is decreassing, so for y=1 we have the maximum of d², not the minimum.
    If y=1; d^2=1-(1)^2=0
    So how is that the max of d^2?
     
  7. May 8, 2012 #6
    Think of this : no point exists on the curve x²+2y²=1 with y=1.
    So you cannot write : d²=1-(1)²=0. The distance between the origin and a non-existing point of the curve is a nonsense.
    Just draw the curve and see what is the range of y for really existing points.
     
    Last edited: May 8, 2012
  8. May 8, 2012 #7
    Ok I get it now. Thanks.
     
  9. May 9, 2012 #8

    mathwonk

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    that is obviously an ellipse, with top and bottom at (0,±1/sqrt(2)) and right hand endpoint at (±1,0).

    Thus there are 4 critical points for the distance function, namely the two points furthest away and the two points nearest the origin.
     
  10. May 9, 2012 #9
    Hi hivesaeed4

    Your calculations are correct, but you missed one detail.

    When looking for max/min of a function on an interval, you need to compare the values at the critical points (i.e. the points where derivative is 0 or does not exist) and at the endpoints.

    In your computations you forget to check endpoints [itex]y=\pm 1/\sqrt2[/itex], and each endpoint give you [itex]d^2=1/2<1[/itex]. So the minimal distance is [itex] 1/\sqrt2[/itex].
     
  11. May 9, 2012 #10
    Okay. But I've noticed in mathwonk's post that 4 critical points are taken into account. Does that mean the error I made was checking only two endpoints, instead of all four for the min/max values of the function?

    Does it also mean that for a given function whose min and/or max values are to be calculated, we should first find it's possible critical points and then use distance formula for each critical point?
     
  12. May 9, 2012 #11

    mathwonk

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    heres another point of view. the function x^2 + 2y^2 has gradient (2x, 4y). Look for points of the ellipse where the radius vector is parallel to this gradient.

    I.e. look for solutions (x,y) of x^2 + 2y^2, where (x,y) is parallel to (2x,4y). But these are never parallel unless either x or y = 0. there are 4 such points.
     
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