Orthogonal trajectories

Orthogonal trajectories are lines intersecting a given family of curves at right angles. If a $y_{1}'$ ${\ displaystyle y_ {1} '}$ ${\ displaystyle y_ {1} '}$ Is the angular coefficient of the tangent to the orthogonal trajectory, and $y_{2}'$ ${\ displaystyle y_ {2} '}$ ${\ displaystyle y_ {2} '}$ Is the angular coefficient of the tangent to the curve of this family, then $y_{1}'$ ${\ displaystyle y_ {1} '}$ ${\ displaystyle y_ {1} '}$ and $y_{2}'$ ${\ displaystyle y_ {2} '}$ ${\ displaystyle y_ {2} '}$ must at each point satisfy the condition of orthogonality :

y_{1}'=-{1 \over y_{2}'}

{\ displaystyle y_ {1} '= - {1 \ over y_ {2}'}}

{\ displaystyle y_ {1} '= - {1 \ over y_ {2}'}}

Suppose we have a family of curves $g(x,y)=C$ ${\ displaystyle g (x, y) = C}$ ${\ displaystyle g (x, y) = C}$ where $C$ ${\ displaystyle C}$ $C$ Is a constant. Then orthogonal trajectories can be found by solving a system of differential equations :

\nabla f(x,y)\cdot \nabla g(x,y)=0

{\ displaystyle \ nabla f (x, y) \ cdot \ nabla g (x, y) = 0}

{\ displaystyle \ nabla f (x, y) \ cdot \ nabla g (x, y) = 0}

Using the definition of a gradient , you can write:

\nabla f(x,y)=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right)

{\ displaystyle \ nabla f (x, y) = \ left ({\ frac {\ partial f} {\ partial x}}, {\ frac {\ partial f} {\ partial y}} \ right)}

{\ displaystyle \ nabla f (x, y) = \ left ({\ frac {\ partial f} {\ partial x}}, {\ frac {\ partial f} {\ partial y}} \ right)}

In this way:

\nabla f(x,y)\cdot \nabla g(x,y)=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right)\cdot \left({\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right)={\frac {\partial f}{\partial x}}\cdot {\frac {\partial g}{\partial x}}+{\frac {\partial f}{\partial y}}\cdot {\frac {\partial g}{\partial y}}=0

{\ displaystyle \ nabla f (x, y) \ cdot \ nabla g (x, y) = \ left ({\ frac {\ partial f} {\ partial x}}, {\ frac {\ partial f} {\ partial y}} \ right) \ cdot \ left ({\ frac {\ partial g} {\ partial x}}, {\ frac {\ partial g} {\ partial y}} \ right) = {\ frac {\ partial f} {\ partial x}} \ cdot {\ frac {\ partial g} {\ partial x}} + {\ frac {\ partial f} {\ partial y}} \ cdot {\ frac {\ partial g} {\ partial y}} = 0}

{\ displaystyle \ nabla f (x, y) \ cdot \ nabla g (x, y) = \ left ({\ frac {\ partial f} {\ partial x}}, {\ frac {\ partial f} {\ partial y}} \ right) \ cdot \ left ({\ frac {\ partial g} {\ partial x}}, {\ frac {\ partial g} {\ partial y}} \ right) = {\ frac {\ partial f} {\ partial x}} \ cdot {\ frac {\ partial g} {\ partial x}} + {\ frac {\ partial f} {\ partial y}} \ cdot {\ frac {\ partial g} {\ partial y}} = 0}

Examples

Suppose we have a family of straight lines passing through the origin given by the equation $y=kx$ ${\ displaystyle y = kx}$ . Differentiating this equation in a variable $x$ ${\ displaystyle x}$ we get:

y'=k=const

{\ displaystyle y '= k = const}

Exclude the parameter $k$ ${\ displaystyle k}$ from the system:

~\mathrm {{\begin{cases}y=kx\\y'=k\end{cases}}\Rightarrow y'={\frac {y}{x}}}

{\ displaystyle ~ \ mathrm {{\ begin {cases} y = kx \\ y '= k \ end {cases}} \ Rightarrow y' = {\ frac {y} {x}}}}

Replace $y^{'}$ ${\ displaystyle y '}$ on $\left(-{\frac {1}{y'}}\right)$ ${\ displaystyle \ left (- {\ frac {1} {y '}} \ right)}$ :

-{\frac {1}{y'}}={\frac {y}{x}}\Rightarrow y'=-{\frac {x}{y}}\Rightarrow {\frac {dy}{dx}}=-{\frac {x}{y}}

{\ displaystyle - {\ frac {1} {y '}} = {\ frac {y} {x}} \ Rightarrow y' = - {\ frac {x} {y}} \ Rightarrow {\ frac {dy} {dx}} = - {\ frac {x} {y}}}

We got a typical differential equation with separable variables. Integrating, we get:

ydy=-xdx\Rightarrow \int {ydy}=-\int {xdx}\Rightarrow {\frac {x^{2}}{2}}+{\frac {y^{2}}{2}}=C

{\ displaystyle ydy = -xdx \ Rightarrow \ int {ydy} = - \ int {xdx} \ Rightarrow {\ frac {x ^ {2}} {2}} + {\ frac {y ^ {2}} {2 }} = C}

This equation is nothing but the equation of a circle of radius ${\sqrt {2C}}$ ${\ displaystyle {\ sqrt {2C}}}$ . Really:

R^{2}=2C\Rightarrow x^{2}+y^{2}=R^{2}

{\ displaystyle R ^ {2} = 2C \ Rightarrow x ^ {2} + y ^ {2} = R ^ {2}}

Literature

Elsgolts L. E. Differential equations and calculus of variations. M .: Nauka, 1969. (p. 23, Example 8)

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