Solving ODE Problems: Understanding Tangent Lines and Integrating Functions

[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.

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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!

 

[Jasso]

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..

 

[HallsofIvy]

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).