Trajectories

Isogonal Trajectories

When two curves intersect in a plane, the angle between them is defined to be the angle made by their respective tangents drawn at their point of intersection.

In the above figure, $\alpha$ is the positive angle from the curve $c_1$ with tangent line $L_1$ to the curve $c_2$ with tangent line $L_2$; $\beta$ is the positive angle from the curve $c_2$ to the curve $c_1$. If we call $m_1$ the slope of $L_1$ and $m_2$ the slope of $L_2$, then by a formula in analytic geometry:

$$\begin{array}{cc}\tan \alpha =\frac{m_2-m_1}{1+m_1{m}_2};\tan \beta =\frac{m_1-m_2}{1+m_1{m}_2}.& \text{(14.11)}\end{array}$$

Definition 14.12 A curve which cuts every member of a given 1-parameter family of curves in the same angle is called an isogonal trajectory of the family.

If we call ${y_1}^{\prime }$ the slope of a curve of a given 1-parameter family, $y^{\prime }$ the slope of an isogonal trajectory of the family, and $\alpha$ their angle of intersection measured from the tangent line with slope ${y_1}^{\prime }$ to the tangent line with slope ${y_1}^{\prime }$, then by (14.11):

$$\begin{array}{cc}\tan \alpha =\frac{{y_1}^{\prime }-y^{\prime }}{1+y^{\prime }{y_1}^{\prime }}& \text{(14.13)}\end{array}$$

Orthogonal Trajectories

Definition 14.2 A curve which cuts every member of a given 1-parameter family of curves in a ${90}^{\circ }$ angle is called an orthogonal trajectory of the family.

Let ${y_1}^{\prime }$ be the slope of a give nfamily and let $y^{\prime }$ be the slope of an orthogonal family. Then, by a theorem in analytic geometry:

$$\begin{array}{cc}{y_1}^{\prime }{y}^{\prime }=-1,y^{\prime }=-\frac{1}{{y_1}^{\prime }}& \text{(14.21)}\end{array}$$

Orthogonal Trajectories in Polar Coordinates

In the above figure, call $P(r,\theta )$ the point of intersection in polar coordinates of two curves $c_1,c_2$, which are orthogonal trajectories of each other. Call ${\varphi }_1$ and ${\varphi }_2$ the respective angle the tangent to each curve $c_1$ and $c_2$ makes with the radius vector $r$ (measured from the radius vector counterclockwise to the tangent). Since the two tangents are orthogonal, it is evient form the figure that:

$${\varphi }_1={\varphi }_2+\frac{\pi }{2}$$

Therefore

$$\begin{array}{cc}\tan {\varphi }_1=\tan {\varphi }_2+\frac{\pi }{2}=-\frac{1}{\tan {\varphi }_2}& \text{(14.31)}\end{array}$$

As remarked previously in Example 13.3, in polar coordinates:

$$\begin{array}{cc}\tan {\varphi }_2=r\frac{d\theta }{dr}& \text{(14.32)}\end{array}$$

Therefore (14.31) becomes:

$$\begin{array}{cc}\tan {\varphi }_1=-\frac{dr}{rd\theta }& \text{(14.33)}\end{array}$$

Comparing (14.32) with (14.33) we see that if two curves are orthogonal, then $r\frac{d\theta }{dr}$ of one is the negative reciprocal of $r\frac{d\theta }{dr}$ of the other. Conversely, if one of two curves satisfies (14.32) and the other satisfies (14.33), then the curves are orthogonal.

Hence, to find an orthogonal family of a given family, we proceed as follows:

1. Calculate $r\frac{d\theta }{dr}$ of the given family.
2. Replace $r\frac{d\theta }{dr}$ by its negative reciprocal $-\frac{dr}{rd\theta }$
3. The family of solutions of this new resulting differential equation is orthogonal to the given family.

Some cool examples

Orthogonal Families of Hyperbolae

Apollonian Circles

The Cover of the Book