1. ## Differential Equations.

Example 4: Find the curves that are perpendicular to the parabolas in Example 3.

It is said:

$\displaystyle \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y}$

My question is about $\displaystyle \frac{-x}{2y}$ and why it is so.

The equations for the parabolas in Example 3 are:

$\displaystyle x^2$
$\displaystyle 2x^2$
$\displaystyle -x^2$
$\displaystyle 0x^2$
$\displaystyle y=-2y^2$
$\displaystyle 100x^2$

I read through Example 3 and I understand to the extent that I follow the arithmetic and I see how the equations for the parabolas were derived.

The $\displaystyle \frac{-x}{2y}$ in $\displaystyle \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y}$ I am having some trouble understanding why that is or what is really being said. I don't think I've really been through this topic of a curve being perpendicular to another curve before.

Thank You...

2. ## Re: Differential Equations.

Hey sepoto.

I think you should keep in mind that two lines are perpendicular in two dimensions if m1*m2 = -1 where m1 is the gradient of one line and m2 is the gradient of the other. They must satisfy this relationship for the perpendicular attribute to hold.

3. ## Re: Differential Equations.

The family of parabolas you given are of the form $\displaystyle y= ax^2$, varying as a varies. $\displaystyle y'= 2ax$ so that at any point where a curve Y(x) is perpendicular to that, we must have $\displaystyle Y'= -1/(2ax)$. Since $\displaystyle y= ax^2$, $\displaystyle x/y= x/ax^2= 1/ax$ and so $\displaystyle Y'= -1/(2ax)= -x/2y$