Orthogonal Trajectories/Examples/Circles Tangent to Y Axis/Proof 2
Theorem
Consider the one-parameter family of curves:
- $(1): \quad x^2 + y^2 = 2 c x$
which describes the loci of circles tangent to the $y$-axis at the origin.
Its family of orthogonal trajectories is given by the equation:
- $x^2 + y^2 = 2 c y$
which describes the loci of circles tangent to the $x$-axis at the origin.
Proof
We use the technique of formation of ordinary differential equation by elimination.
Expressing $(1)$ in polar coordinates, we have:
- $(2): \quad r = 2 c \cos \theta$
Differentiating $(1)$ with respect to $\theta$ gives:
- $(3): \quad \dfrac {\d r} {\d \theta} = -2 c \sin \theta$
Eliminating $c$ from $(2)$ and $(3)$:
- $r \dfrac {\d \theta} {\d r} = -\dfrac {\cos \theta} {\sin \theta}$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
- $r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {\cos \theta}$
Using the technique of Solution to Separable Differential Equation:
- $\ds \int \frac {\d r} r = \int \dfrac {\cos \theta} {\sin \theta} \rd \theta$
which by Primitive of Reciprocal and various others gives:
- $\ln r = \map \ln {\sin \theta} + \ln 2 c$
or:
- $r = 2 c \sin \theta$
This can be expressed in Cartesian coordinates as:
- $x^2 + y^2 = 2 c y$
Hence the result.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories