Formation of Ordinary Differential Equation by Elimination/Examples/x^2 + y^2 equals a^2
Examples of Formation of Ordinary Differential Equation by Elimination
Consider the equation:
- $(1): \quad x^2 + y^2 = a^2$
This can be expressed as the ordinary differential equation:
- $\dfrac {\d y} {\d x} = -\dfrac x y$
which demonstrates that the radius of a circle where it meets the circle is perpendicular to the tangent at that point.
Proof
Differentiating with respect to $x$:
| \(\ds 2 x + 2 y \dfrac {\d y} {\d x}\) | \(=\) | \(\ds 0\) | Power Rule for Derivatives | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\d y} {\d x}\) | \(=\) | \(\ds -\dfrac x y\) | rearranging |
From Equation of Circle center Origin, $(1)$ is the equation of a circle whose center is at the origin.
The straight line $R$ from the origin to the point $\tuple {x, y}$ on the circle has slope $\dfrac y x$ by definition.
From Slope of Orthogonal Curves, a straight line $T$ through $\tuple {x, y}$ such that $\dfrac {\d y} {\d x} = -\dfrac x y$ is perpendicular to $R$.
But $T$ is by definition the tangent to the circle at $\tuple {x, y}$.
Hence we have shown that the radius of a circle where it meets the circle is perpendicular to the tangent at that point.
$\blacksquare$
Sources
- 1952: H.T.H. Piaggio: An Elementary Treatise on Differential Equations and their Applications (revised ed.) ... (previous) ... (next): Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: Examples for solution: $(5)$