# ODE Problem, am I stupid?

by c.teixeira
Tags: stupid
 P: 42 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!
P: 102
 Quote by c.teixeira 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!
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 $y'_{line} = y'_{curve}$.

Since the line given by $y = mx + b$ is defined to be the tangent line to the curve, that means that $m$ must be equal to the $y'$ of the curve it is tangent to in order to statisfy that condition, which again, happens to also be the $y'$ of the line itself..
Math
Emeritus
Thanks
PF Gold
P: 39,568
 Quote by c.teixeira 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. 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;
Well, it should be y= m(x- a)+ b.

 but m = y'. so, y = y' x + b.
so y= y'(a)(x- a)+ 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!
One definition of "derivative" (at a given point) is "slope of the tangent line" (at that point).

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