Orthogonal Trajectories
Solved Problems
Example 3.
Find the orthogonal trajectories of the family of curves given by the power function \(y = C{x^4}.\)
Solution.
\(1)\) Determine the differential equation for the given family of power curves:
\[y = C{x^4},\;\; \Rightarrow y' = 4C{x^3}.\]
By solving the system of two equations and eliminating \(C,\) we get:
\[C = \frac{y}{{{x^4}}},\;\; \Rightarrow y' = 4 \cdot \frac{y}{{{x^4}}} \cdot {x^3} = \frac{{4y}}{x}.\]
\(2)\) Replacing \(y'\) with \(\left( { - \frac{1}{{y'}}} \right)\) gives:
\[ - \frac{1}{{y'}} = \frac{{4y}}{x},\;\; \Rightarrow y' = - \frac{x}{{4y}}.\]
The last expression is the differential equation of the orthogonal trajectories.
\(3)\) By integrating we can find the algebraic equation of the orthogonal trajectories:
\[y' = - \frac{x}{{4y}},\;\; \Rightarrow
\frac{{dy}}{{dx}} = - \frac{x}{{4y}},\;\; \Rightarrow
4ydy = - xdx,\;\; \Rightarrow
4\int {ydy} = - \int {xdx} ,\;\; \Rightarrow
4 \cdot \frac{{{y^2}}}{2} = - \frac{{{x^2}}}{2} + C,\;\; \Rightarrow
4{y^2} + {x^2} = 2C.\]
Divide both sides by \(2C:\)
\[\frac{{4{y^2}}}{{2C}} + \frac{{{x^2}}}{{2C}} = \frac{{2C}}{{2C}},\;\; \Rightarrow
\frac{{{y^2}}}{{\frac{C}{2}}} + \frac{{{x^2}}}{{2C}} = 1,\;\; \Rightarrow
\frac{{{y^2}}}{{{{\left( {\sqrt {\frac{C}{2}} } \right)}^2}}} + \frac{{{x^2}}}{{{{\left( {\sqrt {2C} } \right)}^2}}} = 1.\]
We obtain the equation of the family of ellipses, which are the orthogonal trajectories for the given family of power curves \(y = C{x^4}.\) The ratio of the lengths of semiaxes for these ellipses is
\[\frac{{\sqrt {2C} }}{{\sqrt {\frac{C}{2}} }} = \frac{{\sqrt 2 }}{{\sqrt {\frac{1}{2}} }} = {\left( {\sqrt 2 } \right)^2} = 2.\]
Schematically the graphs of the families of curves are shown in Figure \(4.\)
Figure 4.
Example 4.
Determine the orthogonal trajectories of the family of sinusoids \(y = C\sin x.\)
Solution.
\(1)\) Differentiating the given equation with respect to \(x\) gives:
\[y' = C\cos x.\]
By substituting \(C = \frac{y}{{\sin x}}\) we find the differential equation of the given sinusoidal curves:
\[y' = \frac{y}{{\sin x}}\cos x = y\cot x.\]
\(2)\) Replace \(y'\) with \(\left( { - \frac{1}{{y'}}} \right)\) to write the differential equation of the orthogonal curves:
\[- \frac{1}{{y'}} = y\cot x,\;\; \Rightarrow y' = - \frac{1}{{y\cot x}} = - \frac{{\tan x}}{y}.\]
\(3)\) Now we can integrate this differential equation:
\[y' = - \frac{{\tan x}}{y},\;\; \Rightarrow
\frac{{dy}}{{dx}} = - \frac{{\tan x}}{y},\;\; \Rightarrow
ydy = - \tan xdx,\;\; \Rightarrow
\int {ydy} = - \int {\tan xdx} ,\;\; \Rightarrow
\frac{{{y^2}}}{2} = \ln \left| {\cos x} \right| + \ln C,\;\; \Rightarrow
\frac{{{y^2}}}{2} = \ln \left( {C\left| {\cos x} \right|} \right).\]
It follows from here that
\[C\left| {\cos x} \right| = \exp \left( {\frac{{{y^2}}}{2}} \right),\;\; \Rightarrow
\cos x = \pm \frac{1}{C}\exp \left( {\frac{{{y^2}}}{2}} \right).\]
By denoting \({C_1} = \pm \frac{1}{C},\) we obtain the final implicit equation of the orthogonal trajectories:
\[\cos x = {C_1}\exp \left( {\frac{{{y^2}}}{2}} \right).\]
