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ODE Problem, am I stupid? |
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| Feb29-12, 06:14 PM | #1 |
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ODE Problem, am I stupid?
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! |
| Feb29-12, 07:34 PM | #2 |
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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.. |
| Mar1-12, 07:25 AM | #3 |
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