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Forming DE

  1. Nov 27, 2006 #1
    I need help. Im not getting the solution to this:

    A curve passes through the point (1,1) and has the property that the normal at any point on the curve is equidistant from the x-axis and the origin. Form the differential equation for the curve, find the curve and state which quadrant(s) the curve lies in.

    Here is what I did,
    The slope of the normal at any point would be -1/(dy/dx) and the equation of the normal at the point would be y=mx+c where m= -1/(dy/dx). If this is the normal at (1,1), then 1=m+c, => c=1-m

    Therefore, the equation of the line comes out to be y=mx+1-m. Rearranging this, y-mx-1+m=0. The distance of this line from the origin is :
    |-1+m| / ((1+m*m)^1/2) which is equal to 1 (distance from the x-axis as this is the normal at (1,1) ).

    After this, you substitute m= -1/(dy/dx) and you get your differential equation... only you dont... what am I doing wrong?
     
  2. jcsd
  3. Nov 27, 2006 #2

    HallsofIvy

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    I find the statement "the normal at any point on the curve is equidistant from the x-axis and the origin" confusing. The normal at that point is a line. I know what is meant by the distance from a line to a point but what is the "distance" between two lines in a plane? At the end, you appear to be using the distance from the point (1,1) to the x-axis as the "distance between the normal and the x-axis". That certainly isn't the way I would interpret the phrase.
     
  4. Nov 27, 2006 #3
    If that isnt it, then how would you look at the problem? What could a possible solution be?
     
  5. Nov 27, 2006 #4
    [tex] y - mx - 1 + m = 0 \Rightarrow y -(-\frac{1}{\frac{dy}{dx}})x-1-\frac{1}{\frac{dy}{dx}} = 0 [/tex]

    [tex] y +\frac{1}{\frac{dy}{dx}}(x-1) - 1 = 0 [/tex]

    [tex] \frac{1}{\frac{dy}{dx}}(x-1) = 1- y [/tex]


    [tex] \frac{1}{\frac{dy}{dx}} = \frac{1-y}{x-1} [/tex]

    [tex] \frac{dy}{dx} = \frac{x-1}{1-y} [/tex]
     
    Last edited: Nov 27, 2006
  6. Nov 28, 2006 #5
    Where do you use the condition that the perpendicular distance of the normal at a point is equal to its y co-ordinate? The answer given is d(x*x-y*y)=(2xy)dy. Is there something Im missing?
     
    Last edited: Nov 28, 2006
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