Chapter 3 · Section 3.1

3.1 Orthogonal and Oblique Trajectories

Draw the equipotential lines of an electrostatic field, and then draw the electric field lines. The two sets of curves cross each other at right angles everywhere. The same picture appears in heat conduction (isotherms and heat-flow lines) and in ideal two-dimensional flow (lines of constant potential and streamlines). In each case one family of curves is known, and the other family is wanted.

Finding the second family is a neat application of Chapter 2. The given family defines a first-order differential equation. A small change to that equation describes the curves that cut the family at a right angle, or at any fixed angle. Solving the new equation gives the trajectories.

A family of curves and its differential equation

A one-parameter family of curves is a set of curves described by one equation with one free constant, such as

F (x, y, c) = 0 or F (x, y) = c .

Through a typical point $(x, y)$ passes exactly one member of the family. That member has a definite slope at the point. So the family assigns a slope to every point, and this is exactly what a first-order equation $y^{'} = f (x, y)$ does.

To find that equation:

differentiate the family equation implicitly with respect to $x$ ;
eliminate the constant $c$ between the original equation and the differentiated one;
solve for $y^{'}$ .

Step 2 is the one students forget. The slope must be written in terms of the point $(x, y)$ alone. A formula that still contains $c$ does not tell you the slope at a point unless you also know which curve you are on.

When the family is written as $F (x, y) = c$ , the constant disappears by itself on differentiation:

F_{x} + F_{y} y^{'} = 0 ⟹ y^{'} = - \frac{F _{x}}{F _{y}} .

Orthogonal trajectories

Definition 3.1.1 (Orthogonal trajectory)

Let a one-parameter family of curves be given. A curve that meets every member of the family it crosses at a right angle is called an orthogonal trajectory of the family. "At a right angle" means that the tangent lines of the two curves at the crossing point are perpendicular.

Two lines with slopes $m_{1}$ and $m_{2}$ (neither vertical nor horizontal) are perpendicular exactly when $m_{1} m_{2} = - 1$ . So if a member of the family has slope $f (x, y)$ at the point $(x, y)$ , an orthogonal trajectory through that point must have slope $- 1/ f (x, y)$ . This holds at every point, which gives a new differential equation.

Figure 3.1.1. At the crossing point the family has slope

f (x, y)

and the trajectory has slope

- 1/ f (x, y)

. The two dashed tangent lines are perpendicular.

Theorem 3.1.1 (Equation of the orthogonal trajectories)

Suppose the family satisfies $y^{'} = f (x, y)$ in a region $R$ where $f$ is continuous and $f (x, y) \neq = 0$ . Then, in $R$ , the orthogonal trajectories of the family are the solution curves of

y^{'} = - \frac{1}{f ( x , y )} .

If the family is given as $F (x, y) = c$ , with $F$ continuously differentiable, the orthogonal trajectories satisfy the differential-form equation

F_{y} d x - F_{x} d y = 0.

The second form follows from the first: $f = - F_{x} / F_{y}$ , so $- 1/ f = F_{y} / F_{x}$ , i.e. $F_{x} d y = F_{y} d x$ . It has a nice meaning. The vector $(F_{x}, F_{y}) = \nabla F$ is perpendicular to the level curves of $F$ , so the orthogonal trajectories are the curves that always point along the gradient. In electrostatics, with $F = V$ the potential, they are the field lines of $E = - \nabla V$ . The differential form also handles points where the trajectory is vertical ( $F_{x} = 0$ ) without any special treatment.

Watch the excluded points

Where $f (x, y) = 0$ the family is horizontal, so the trajectory must be vertical. Vertical lines $x = const$ are not graphs $y (x)$ , and solving $y^{'} = - 1/ f$ can miss them. Always ask at the end: is some vertical line (or horizontal line, where $f$ is infinite) also a trajectory?

Method in a nutshell

Write the differential equation $y^{'} = f (x, y)$ of the given family. Eliminate $c$ .
Replace $f (x, y)$ by $- 1/ f (x, y)$ . (Or, for $F (x, y) = c$ , write $F_{y} d x - F_{x} d y = 0$ .)
Solve the new equation with a method from Chapter 2 (often it is separable, homogeneous or exact). Its one-parameter family of solutions is the family of orthogonal trajectories.
Check for vertical or horizontal lines that were lost along the way.

Worked examples

Example 3.1.1

Find the orthogonal trajectories of the hyperbolas $x y = c$ .

Solution

Equation of the family. Here $F (x, y) = x y$ , so $F_{x} = y$ and $F_{y} = x$ . Differentiating $x y = c$ gives $y + x y^{'} = 0$ , i.e. $y^{'} = - \frac{y}{x}$ .
Replace the slope. The trajectories satisfy $y^{'} = \frac{x}{y} .$
Solve. The equation is separable: $y d y = x d x$ . Integrating, $\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + \frac{k}{2} ⟹ x^{2} - y^{2} = k .$
Lost lines? On the $x$ -axis $f = 0$ , so a trajectory must be vertical there; the hyperbolas $x^{2} - y^{2} = k$ with $k > 0$ do cross the $x$ -axis vertically. Nothing is missing. (The pair of lines $y = \pm x$ is the member $k = 0$ .)

The trajectories are again hyperbolas, rotated by $4 5^{\circ}$ (Figure 3.1.2). In fluid mechanics, these two families are the streamlines and equipotentials of the flow that hits a flat wall head-on near a stagnation point.

Figure 3.1.2. The hyperbolas

x y = c

(solid) and their orthogonal trajectories

x^{2} - y^{2} = k

(dashed). Every crossing is a right angle.

You can explore this with the tool below. Type a family $F (x, y) = c$ (for example x^2 + 4*y^2 or y/x^3) and it draws both families.

Example 3.1.2

Find the orthogonal trajectories of the exponential curves $y = c e^{- 2 x}$ .

Solution

Differentiate. $y^{'} = - 2 c e^{- 2 x}$ .
Eliminate $c$ . From the family, $c e^{- 2 x} = y$ . Hence $y^{'} = - 2 y .$ Notice that $c$ is gone: the slope at $(x, y)$ is $- 2 y$ , whatever curve we are on.
Replace the slope. The trajectories satisfy $y^{'} = \frac{1}{2 y}$ .
Solve. Separate: $2 y d y = d x$ , so $y^{2} = x + k .$
Lost lines? The member $c = 0$ is the $x$ -axis, where $f = 0$ . The parabolas $y^{2} = x + k$ have vertical tangents exactly on the $x$ -axis, so they do cross it at right angles. No extra curve is needed.

The trajectories are parabolas opening to the right, all with the same shape and shifted along the $x$ -axis.

Example 3.1.3

Find the orthogonal trajectories of the family of ellipses $x^{2} + 2 y^{2} = c x$ .

Solution

Completing the square, $(x - \frac{c}{2})^{2} + 2 y^{2} = \frac{c ^{2}}{4}$ . These are ellipses centred on the $x$ -axis, all passing through the origin. Here the constant does not vanish on differentiation, so we must eliminate it.

Differentiate. $2 x + 4 y y^{'} = c$ .
Eliminate $c$ . From the family (for $x \neq = 0$ ), $c = \frac{x ^{2} + 2 y ^{2}}{x}$ . Substitute: $4 y y^{'} = \frac{x ^{2} + 2 y ^{2}}{x} - 2 x = \frac{2 y ^{2} - x ^{2}}{x},$ so $y^{'} = \frac{2 y ^{2} - x ^{2}}{4 x y}$ .
Replace the slope. $y^{'} = \frac{4 x y}{x ^{2} - 2 y ^{2}} .$ The right side is a function of $y / x$ , so the equation is homogeneous (Section 2.2).
Substitute $y = vx$ . Then $y^{'} = v + x v^{'}$ and $v + x v^{'} x v^{'} = \frac{4 v}{1 - 2 v ^{2}} = \frac{4 v}{1 - 2 v ^{2}} - v = \frac{3 v + 2 v ^{3}}{1 - 2 v ^{2}} .$
Separate. $\frac{1 - 2 v ^{2}}{v ( 3 + 2 v ^{2} )} d v = \frac{d x}{x} .$ Partial fractions give $\frac{1 - 2 v ^{2}}{v ( 3 + 2 v ^{2} )} = \frac{1}{3} \cdot \frac{1}{v} - \frac{8}{3} \cdot \frac{v}{3 + 2 v ^{2}} .$
Integrate. $\frac{1}{3} ln ∣ v ∣ - \frac{2}{3} ln (3 + 2 v^{2}) = ln ∣ x ∣ + C .$ Multiply by 3 and exponentiate: $\frac{v}{( 3 + 2 v ^{2} ) ^{2}} = K x^{3} .$
Back to $x, y$ . With $v = y / x$ , the left side is $\frac{y / x}{( 3 x ^{2} + 2 y ^{2} ) ^{2} / x ^{4}} = \frac{x ^{3} y}{( 3 x ^{2} + 2 y ^{2} ) ^{2}}$ . Cancel $x^{3}$ : $y = K (3 x^{2} + 2 y^{2})^{2}, or (3 x^{2} + 2 y^{2})^{2} = k y .$
Lost lines? Dividing by $v$ removed $v = 0$ , i.e. the $x$ -axis $y = 0$ . It is indeed a trajectory: the ellipses cross the $x$ -axis vertically. (It corresponds to $K = 0$ in $y = K (3 x^{2} + 2 y^{2})^{2}$ .)

Figure 3.1.3. The ellipses

x^{2} + 2 y^{2} = c x

(solid) and their orthogonal trajectories

(3 x^{2} + 2 y^{2})^{2} = k y

(dashed). Both families pass through the origin: the ellipses arrive tangent to the

y

-axis, the trajectories tangent to the

x

-axis.

Orthogonal trajectories in polar coordinates

Some families are much simpler in polar coordinates $(r, θ)$ : spirals, roses, circles through the origin. For a curve $r = r (θ)$ , the useful angle is $ψ$ , the angle between the radius vector and the tangent line. It satisfies

tan ψ = \frac{r}{d r / d θ} .

Let two curves pass through the same point, with derivatives $r_{1}^{'}$ and $r_{2}^{'}$ there. They are perpendicular when their angles $ψ$ differ by $9 0^{\circ}$ , i.e. when $tan ψ_{1} tan ψ_{2} = - 1$ :

\frac{r}{r _{1}^{'}} \cdot \frac{r}{r _{2}^{'}} = - 1 ⟺ r_{1}^{'} r_{2}^{'} = - r^{2} .

(You can confirm this by writing the tangent vectors $(r^{'} cos θ - r sin θ, r^{'} sin θ + r cos θ)$ and setting their dot product to zero; it equals $r_{1}^{'} r_{2}^{'} + r^{2}$ .) So the polar version of the recipe is:

Polar coordinates

Find the equation of the family in the form $\frac{d r}{d θ} = g (r, θ)$ , with the constant eliminated.
Replace $g (r, θ)$ by $- \frac{r ^{2}}{g ( r , θ )}$ .
Solve the new equation for $r (θ)$ .

Example 3.1.4

Find the orthogonal trajectories of the logarithmic spirals $r = c e^{2 θ}$ .

Solution

Equation of the family. $\frac{d r}{d θ} = 2 c e^{2 θ} = 2 r$ .
Replace. The trajectories satisfy $\frac{d r}{d θ} = - \frac{r ^{2}}{2 r} = - \frac{r}{2} .$
Solve. Separate: $\frac{d r}{r} = - \frac{d θ}{2}$ , so $ln r = - \frac{θ}{2} + C$ and $r = k e^{- θ /2} .$

The trajectories are again logarithmic spirals, winding the other way (Figure 3.1.4). For the family, $tan ψ = r / (2 r) = \frac{1}{2}$ : every spiral crosses every ray from the origin at the same angle. This constant-angle property is why logarithmic spirals appear in frequency-independent spiral antennas.

Figure 3.1.4. Logarithmic spirals

r = c e^{2 θ}

(solid) and their orthogonal trajectories

r = k e^{- θ /2}

(dashed).

Oblique trajectories

Now let the crossing angle be any fixed angle $α$ with $0 < α < 9 0^{\circ}$ .

Definition 3.1.2 (Oblique trajectory)

A curve that meets every member of a one-parameter family it crosses at the same angle $α$ , $0 < α < 9 0^{\circ}$ , is called an oblique trajectory of the family (at angle $α$ ).

Let us derive the slope step by step.

Angle of the family's tangent. A curve of the family with slope $f$ at $(x, y)$ has a tangent line inclined at the angle $β = arctan f$ to the $x$ -axis.
Angle of the trajectory's tangent. The trajectory crosses at angle $α$ , so its tangent is inclined at $β + α$ (turning counter-clockwise from the family's tangent).
Slope of the trajectory. By the addition formula for the tangent, $tan (β + α) = \frac{tan β + tan α}{1 - tan β tan α} = \frac{f + tan α}{1 - f tan α} .$

Theorem 3.1.2 (Equation of oblique trajectories)

If the family satisfies $y^{'} = f (x, y)$ , its oblique trajectories at angle $α$ satisfy

y^{'} = \frac{f ( x , y ) + tan α}{1 - f ( x , y ) tan α}

at the points where $1 - f tan α \neq = 0$ .

Turning clockwise instead gives a second family, with $tan α$ replaced by $- tan α$ . Both families cross the given curves at the angle $α$ ; they differ in the sense of rotation. As $α \to 9 0^{\circ}$ , divide numerator and denominator by $tan α$ : the slope tends to $- 1/ f$ , and we recover Theorem 3.1.1.

Method in a nutshell

Find $y^{'} = f (x, y)$ for the family, with $c$ eliminated.
Replace $f$ by $\frac{f + tan α}{1 - f tan α}$ .
Solve the new equation.

Example 3.1.5

Find the oblique trajectories that cross the curves $y = c e^{x}$ at $4 5^{\circ}$ .

Solution

Equation of the family. $y^{'} = c e^{x} = y$ , so $f (x, y) = y$ .
Replace. With $tan 4 5^{\circ} = 1$ : $y^{'} = \frac{y + 1}{1 - y} .$
Separate. For $y \neq = - 1$ : $\frac{1 - y}{1 + y} d y = d x .$
Simplify the integrand. Divide: $\frac{1 - y}{1 + y} = - 1 + \frac{2}{1 + y}$ .
Integrate. $- y + 2 ln ∣1 + y ∣ = x - k ⟹ x + y - 2 ln ∣1 + y ∣ = k .$
The lost solution. The constant function $y = - 1$ also solves $y^{'} = (1 + y) / (1 - y)$ . It is a trajectory: where it meets the family ( $y = - 1$ ), the family has slope $- 1$ , and a horizontal line meets a line of slope $- 1$ at $4 5^{\circ}$ .

The trajectories are vertical where $y = 1$ (denominator zero). This is not a problem: the curve simply turns back there, as Figure 3.1.5 shows.

Figure 3.1.5. The curves

y = c e^{x}

(solid) and their

4 5^{\circ}

trajectories

x + y - 2 ln (1 + y) = k

(dashed, drawn for

y > - 1

). The line

y = - 1

is also a trajectory.

Engineering remark

In a two-dimensional potential problem, the potential $V (x, y)$ and the "stream function" $Ψ (x, y)$ whose level curves are the field lines are often the real and imaginary parts of one analytic function. For example, $x^{2} - y^{2}$ and $2 x y$ come from $z^{2}$ , with $z = x + i y$ . Exercise 7 shows the same for $z^{3}$ . You will meet this idea again in courses on complex analysis and electromagnetics.

Exercises

In Exercises 1–6, find the orthogonal trajectories of the given family.

$y = \frac{c}{x ^{3}}$
Answer
$y^{'} = - 3 y / x$ ; trajectories $3 y^{2} - x^{2} = k$ .
$x^{2} + 4 y^{2} = c$
Answer
$y = k x^{4}$ (and the line $x = 0$ ).
$y^{3} = c x^{2}$
Answer
$y^{'} = \frac{2 y}{3 x}$ ; trajectories $3 x^{2} + 2 y^{2} = k$ .
$y = sin x + c$
Answer
$y = - ln ∣ sec x + tan x ∣ + k$ .
$y = c (x^{2} + 1)$
Answer
$y^{'} = \frac{2 x y}{x ^{2} + 1}$ ; trajectories $y^{2} = - \frac{x ^{2}}{2} - ln ∣ x ∣ + k$ .
$x^{2} + x y + y^{2} = c$
Answer
The trajectory equation $y^{'} = \frac{x + 2 y}{2 x + y}$ is homogeneous; $x + y = k (x - y)^{3}$ (and the line $y = x$ ).
The equipotential lines of a two-dimensional field are $x^{3} - 3 x y^{2} = c$ . Find the field lines.
Answer
The equation $- 6 x y d x - (3 x^{2} - 3 y^{2}) d y = 0$ is exact; field lines $3 x^{2} y - y^{3} = k$ .
Find the orthogonal trajectories of the four-petal roses $r = c sin 2 θ$ .
Answer
$d r / d θ = 2 r cot 2 θ$ for the family; trajectories $r^{4} = k cos 2 θ$ .
Find the orthogonal trajectories of the spirals $r = c θ$ , $θ > 0$ .
Answer
$d r / d θ = r / θ$ for the family; trajectories $r = k e^{- θ^{2} /2}$ .
Find the oblique trajectories that cross the parabolas $y = x^{2} + c$ at $4 5^{\circ}$ .
Answer
$y^{'} = \frac{2 x + 1}{1 - 2 x}$ ; $y = - x - ln ∣1 - 2 x ∣ + k$ .
Show that every logarithmic spiral $r = k e^{a θ}$ ( $a > 0$ ) crosses every ray $θ = const$ at the same angle, and find that angle.
Answer
$tan ψ = r / (d r / d θ) = 1/ a$ , so $ψ = arctan (1/ a)$ at every point.
The lines $y = 2 x + c$ all have slope $2$ . Find the oblique trajectories at the angle $α$ with $tan α = 2$ , for both senses of rotation.
Answer
Counter-clockwise: $y^{'} = \frac{2 + 2}{1 - 4} = - \frac{4}{3}$ , the lines $4 x + 3 y = k$ . Clockwise: $y^{'} = \frac{2 - 2}{1 + 4} = 0$ , the horizontal lines $y = k$ .