2.5: Orthogonal Trajectories of Curves and Pursuit Curves

Learning Objectives

• Understand and solve differential equations to find orthogonal trajectories of a given family of curves.
• Understand and derive the pursuit curve equations for a moving body following a target, and determine conditions for interception.

There are many problems from geometry which have lead to the study of differential equations. One such problem is the construction of orthogonal trajectories. Give a a family of curves, \(y_{1}(x ; a)\), we seek another family of curves \(y_{2}(x ; c)\) such that the second family of curves are perpendicular to the given family. This means that the tangents of two intersecting curves at the point of intersection are perpendicular to each other. The slopes of the tangent lines are given by the derivatives \(y_{1}^{\prime}(x)\) and \(y_{2}^{\prime}(x)\). We recall from elementary geometry that the slopes of two perpendicular lines are related by

\[y_{2}^{\prime}(x)=-\dfrac{1}{y_{1}^{\prime}(x)} \nonumber \]

Example \(\PageIndex{1}\)

Find a family of orthogonal trajectories to the family of parabolas \(y_{1}(x ; a)=a x^{2}\).

Solution

We note that the new collection of curves has to satisfy the equation

\[y_{2}^{\prime}(x)=-\dfrac{1}{y_{1}^{\prime}(x)}=-\dfrac{1}{2 a x} \nonumber \]

Before solving for \(y_{2}(x)\), we need to eliminate the parameter \(a\). From the given function, we have that \(a=\dfrac{y}{x^{2}}\). Inserting this into the equation, we have

\[y^{\prime}(x)=-\dfrac{1}{2 a x}= -\dfrac{1}{2 \left(\dfrac{y}{x^{2}}\right) x}= -\dfrac{x}{2 y}\nonumber \]

Thus, to find \(y_{2}(x)\), we have to solve the differential equation

\[2 y y^{\prime}+x=0\nonumber \]

Noting that \(\left(y^{2}\right)^{\prime}=2 y y^{\prime}\) and \(\left(\dfrac{1}{2} x^{2}\right)^{\prime}=x\), this (exact) equation can be written as

\[\dfrac{d}{d x}\left(y^{2}+\dfrac{1}{2} x^{2}\right)=0\nonumber \]

Integrating, we find the family of solutions,

\[y^{2}+\dfrac{1}{2} x^{2}=k \nonumber \]

In Figure \(\PageIndex{1}\) we plot both families of orthogonal curves.

Figure \(\PageIndex{1}\): The family of blue parabolas \(y=a x^{2}\) are orthogonal trajectories of the family of black ellipses \(y^{2}+\dfrac{1}{2} x^{2}=k\). Multiple parabolas and ellipses are drawn on the xy-plane where \(-5 \leq x \leq 5\) and \(-5 \leq y \leq 5\) (CC BY-NC-SA 3.0, Russell Herman via A First Course in Differential equations).

Another application that is interesting is to find the path that a body traces out as it moves towards a fixed point or another moving body. Such curves are know as pursuit curves. These could model aircraft or submarines following targets, or predators following prey. We demonstrate this with an example.

Example \(\PageIndex{2}\)

A hawk at point \((x, y)\) sees a sparrow traveling at speed \(v\) along a straight line. The hawk flies towards the sparrow at constant speed \(w\) but always in a direction along the line of sight between their positions. If the hawk starts out at the point \((a, 0)\) at \(t=0\) when the sparrow is at \((0,0)\), then what is the path the hawk needs to follow? Will the hawk catch the sparrow? The situation is shown in Figure \(\PageIndex{2}\). We pick the path of the sparrow to be along the \(y\)-axis. Therefore, the sparrow is at position \((0, v t)\).

Figure \(\PageIndex{2}\): A hawk at point \((x, y)\) sees a sparrow at point \((0, v t)\) and always follows the straight line between these points. (CC BY-NC-SA 3.0; Russell Herman via A First Course in Differential equations).

Solution

First we need the equation of the line of sight between the points \((x, y)\) and \((0, v t)\). Considering that the slope of the line is the same as the slope of the tangent to the path, \(y=y(x)\), we have

\[y^{\prime}=\dfrac{y-v t}{x} \nonumber \]

The hawk is moving at a constant speed, w. Since the speed is related to the time through the distance the hawk travels. We need to find the arclength of the path between \((a, 0)\) and \((x, y)\). This is given by

\[L=\int d s=\int_{x}^{a} \sqrt{1+\left[y^{\prime}(x)\right]^{2}} d x . \nonumber \]

The distance is related to the speed, \(w\), and the time, \(t\), by \(L=w t\). Eliminating the time using \(y^{\prime}=\dfrac{y-v t}{x}\), we have

\[\int_{x}^{a} \sqrt{1+\left[y^{\prime}(x)\right]^{2}} d x=\dfrac{w}{v}\left(y-x y^{\prime}\right) \nonumber \]

Furthermore, we can differentiate this result with respect to \(x\) to get rid of the integral,

\[\sqrt{1+\left[y^{\prime}(x)\right]^{2}}=\dfrac{w}{v} x y^{\prime \prime} \nonumber \]

Even though this is a second order differential equation for \(y(x)\), it is a first order separable equation in the speed function \(z(x)=y^{\prime}(x)\). Namely,

\[\dfrac{w}{v} x z^{\prime}=\sqrt{1+z^{2}} \nonumber \]

Separating variables, we find

\[\dfrac{w}{v} \int \dfrac{d z}{\sqrt{1+z^{2}}}=\int \dfrac{d x}{x} \nonumber \]

The integrals can be computed using standard methods from calculus. We can easily integrate the right hand side,

\[\int \dfrac{d x}{x}=\ln |x|+c_{1} . \nonumber \]

The left hand side takes a little extra work using trigonometric substitution method. Let \(z=\tan \theta \). Then \(d z=\sec ^{2} \theta d \theta\). The methods proceeds as follows:

\[ \begin{aligned} \int \dfrac{d z}{\sqrt{1+z^{2}}}&=\int \dfrac{\sec ^{2} \theta}{\sqrt{1+\tan ^{2} \theta}} d \theta \\[4pt]\ &=\int \sec \theta d \theta \\[4pt] &=\ln (\tan \theta+\sec \theta)+c_{2} \\[4pt] &=\ln \left(z+\sqrt{1+z^{2}}\right)+c_{2} \end{aligned}\label{1.44} \]

Putting these together, we have for \(x>0\),

\[\ln \left(z+\sqrt{1+z^{2}}\right)=\dfrac{v}{w} \ln x+C \nonumber \]

Using the initial condition \(z=y^{\prime}=0\) and \(x=a\) at \(t=0\),

\[0=\dfrac{v}{w} \ln a+C \nonumber \]

or \(C=-\dfrac{v}{w} \ln a\).

Using this value for \(c\), we find

\[ \begin{aligned} \ln \left(z+\sqrt{1+z^{2}}\right) &=\dfrac{v}{w} \ln x-\dfrac{v}{w} \ln a \\[4pt] \ln \left(z+\sqrt{1+z^{2}}\right) &=\dfrac{v}{w} \ln \dfrac{x}{a} \\[4pt] \ln \left(z+\sqrt{1+z^{2}}\right) &=\ln \left(\dfrac{x}{a}\right)^{\dfrac{v}{w}} \\[4pt] z+\sqrt{1+z^{2}} &=\left(\dfrac{x}{a}\right)^{\dfrac{v}{w}} \end{aligned} \nonumber \]

We can solve for \(z=y^{\prime}\), to find

\[y^{\prime}=\dfrac{1}{2}\left[\left(\dfrac{x}{a}\right)^{\dfrac{v}{w}}-\left(\dfrac{x}{a}\right)^{-\dfrac{v}{w}}\right] \nonumber \]

Integrating,

\[y(x)=\dfrac{a}{2}\left[\dfrac{\left(\dfrac{x}{a}\right)^{1+\dfrac{v}{w}}}{1+\dfrac{v}{w}}-\dfrac{\left(\dfrac{x}{a}\right)^{1-\dfrac{v}{w}}}{1-\dfrac{v}{w}}\right]+k \nonumber \]

The integration constant, \(k\), can be found knowing \(y(a)=0 \). This gives

\[ \begin{aligned} 0 &=\dfrac{a}{2}\left[\dfrac{1}{1+\dfrac{v}{w}}-\dfrac{1}{1-\dfrac{v}{w}}\right]+k \\[4pt] k &=\dfrac{a}{2}\left[\dfrac{1}{1-\dfrac{v}{w}}-\dfrac{1}{1+\dfrac{v}{w}}\right] \\[4pt] &=\dfrac{a v w}{w^{2}-v^{2}} \end{aligned}\label{1.46} \]

The full solution for the path is given by

\[y(x)=\dfrac{a}{2}\left[\dfrac{\left(\dfrac{x}{a}\right)^{1+\dfrac{v}{w}}}{1+\dfrac{v}{w}}-\dfrac{\left(\dfrac{x}{a}\right)^{1-\dfrac{v}{w}}}{1-\dfrac{v}{w}}\right]+\dfrac{a v w}{w^{2}-v^{2}} \nonumber \]

Can the hawk catch the sparrow? This would happen if there is a time when \(y(0)=v t\). Inserting \(x=0\) into the solution, we have \(y(0)=\dfrac{a v w}{w^{2}-v^{2}}=v t \). This is possible if \(w>v \).