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ODE Problem, am I stupid?

by c.teixeira
Tags: stupid
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c.teixeira
#1
Feb29-12, 06:14 PM
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.


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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!
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Jasso
#2
Feb29-12, 07:34 PM
P: 102
Quote Quote by c.teixeira View Post
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 [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..
HallsofIvy
#3
Mar1-12, 07:25 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Quote Quote by c.teixeira View Post
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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