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ODE Problem, am I stupid?

  1. Feb 29, 2012 #1
    I have been reading Ordinary Differential Equations (Pollard) from Dover.
    The chapter I am in, is called Problems Leading to Differential Equations of The First Order - Geometric Problems.

    Problem :

    Find the family of curves with the property that the area of the region bounded by the x axis , the tangent line drawn at a point P(x,y) of a curve of the family and the projection of the tangent line on the x axis has a constante value A.

    image hosting png

    In the solution, they say the equation of the tangent line is y / (x - a) = y'

    They then solve, for a:

    a = x - (y/y')

    Afterwards, they obtain the distance QR = y/y'

    Therefore they have the area of the triangle. They integrate, bla blabla.

    Now, when I first looked this, it seemed pretty simple and straighforward. I understood every step. It was an elementary problem.

    But, today I gave it a second look, and now I just don't agree with the solution.
    Well, my question is y = mx + b;
    but m = y'.

    so, y = y' x + b.
    I don't agree with this since y defines the equation of the tangent line BUT y' defines the derivative of THE CURVE. therefore in my viewing, when they, in the solution, reach to QR = y/y', and then integrate they are mixing a fuction and a derivative of a diferent fuction.

    So, where is my reasoning wrong?
    Perhaps I should sleep more. ;D

    Thanks for all the explanations!
  2. jcsd
  3. Feb 29, 2012 #2
    The slope of a line tangent to a function at a point is the same as the value of the derivative of the function at that point, by definition; this also means that the derivative of the tangent line at a point is the same as the derivative of the function at that point, so [itex]y'_{line} = y'_{curve}[/itex].

    Since the line given by [itex]y = mx + b[/itex] is defined to be the tangent line to the curve, that means that [itex]m[/itex] must be equal to the [itex]y'[/itex] of the curve it is tangent to in order to statisfy that condition, which again, happens to also be the [itex]y'[/itex] of the line itself..
  4. Mar 1, 2012 #3


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    Well, it should be y= m(x- a)+ b.

    so y= y'(a)(x- a)+ b

    One definition of "derivative" (at a given point) is "slope of the tangent line" (at that point).
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