Preface

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This section presents important applications to geometry; namely, it discusses deriving orthogonal trajectories based on differential equations.

Orthogonal Trajectories

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The word orthogonal comes from the Greek ορθη (right) and γωνια (angle); the word trajectory comes from the Latin trajectus (cut across). Therefore, a curve that cuts across certain others at right angles is called an orthogonal trajectory of those others.

Now we extend this definition to the family of curves that form the general solution of the differential equation. Suppose that the differential equation \( y' = f(x,y) \) has the family of solutions in implicit form \( F(x,y) = C , \) where C is an arbitrary constant. We refer to the set of curves on the plane as the family of orthogonal trajectories is they form the general solution to the differential equation \( y' = - 1/f(x,y) . \) For example, a family of exponential curves b^x and b^-x/(lnb)² are orthogonal.

or Hence, two families of solutions are orthogonal if they are solutions of the differential equations:

\[ \frac{{\text d}y}{{\text d}x} = f(x,y) \qquad\mbox{and}\qquad \frac{{\text d}y}{{\text d}x} = \frac{-1}{f(x,y)} \]

or

\[ M(x,y)\,{\text d}x + N(x,y)\,{\text d}y =0 \qquad\mbox{and}\qquad N(x,y)\,{\text d}x - M(x,y)\,{\text d}y =0 . \]

Example: Orthogonal trajectories Example: Orthogonal trajectories of a separable equation

 

Consider a curve whose equation is expressed in polar coordinates (r,θ). In calculus it is shown that the angle ψ, measured positive in the counterclockwise direction from the radius vector to the tangent line at a point, is given by If two curves are orthogonal at some point having angles ψ₁ and ψ₂, then ψ₂ = ψ₁ + π/2. So \( \tan\psi_2 = - \cot\psi_1 = -1/\tan\psi_1 . \) Hence, if two curves are to cut one another at right angles, then at the point of intersection the value of the product \( \displaystyle r\,\frac{{\text d}\theta}{{\text d}r} \) for one curve must be the negative reciprocal of the value of that product for the other curve.

When polar coordinates are used, suppose a family of curves forms the general solution to the differential equation \( P(r,\theta )\,{\text d}r + Q(r,\theta )\,{\text d}\theta = 0 . \) Then \( \displaystyle r\,\frac{{\text d}\theta}{{\text d}r} = - \frac{r\,P}{Q} . \) Hence, the family of the orthogonal trajectories of the solutions must be solutions of the equations

\[ \frac{{\text d}\theta}{{\text d}r} = f(r,\theta ) \qquad\mbox{and}\qquad \frac{{\text d}\theta}{{\text d}r} = \frac{-1}{r^2 f(r,\theta )} \]

or

\[ P(r,\theta )\,{\text d}r + Q(r,\theta )\,{\text d}\theta =0 \qquad\mbox{and}\qquad Q(r,\theta )\,{\text d}r - r^2 P(r,\theta )\,{\text d}\theta =0 . \]

Example: Cardioids

 

If we are given a family of curves that satisfies the differential equation \( y' = f(x,y) \) and we want to find a family of curves that intersects this family at a constantt angle θ, we have to solve the differential equation

The above formula is based on the trigonometric identity: where m₁ = tanφ and m₁ = tanθ. Example: Oblique trajectories

 

 

Example: Temperature isotherms

 

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