9.3 Orthogonal Trajectories

Example 1: Sketching Orthogonal Trajectories

Since orthogonal trajectories are simply families of curves that intersect at right angles at every intersection, we just need to be able to plot implicit families of curves to be able to show graphical representations of orthogonal tranjectories.

I like to define the family as a function of the constant, then use a loop to get a list of members of the family of curves.

Clear[x,y,k,c]

family1[k_]=y==k*(x^2)
list1=Table[family1[k],{k,-2,2,1/4}]
p1=ContourPlot[Evaluate[list1],{x,-4,4},{y,-4,4},ContourStyle->Blue]

Here is the second family of curves, notice I use a different constant (I didn’t have to).

family2[c_]=((x^2)/2)+(y^2)==c
list2=Table[family2[c],{c,-2,5,1/2}]
p2=ContourPlot[Evaluate[list2],{x,-4,4},{y,-4,4},ContourStyle->Red]

And here we show the overlay of the two families of curves.

Show[p1,p2,AspectRatio->1]

If we zoom in we can visually verify the orthogonal intersections.

Show[p1,p2,AspectRatio->1,PlotRange->{{1,1.5},{1,1.5}}]

Example 2

Clear[x,y,k,c]

family1[k_]=y==k*Exp[-x]
list1=Table[family1[k],{k,-2,2,1/4}];
p1=ContourPlot[Evaluate[list1],{x,-4,4},{y,-4,4},ContourStyle->Blue];

family2[c_]=x==((y^2)/2)+c
list2=Table[family2[c],{c,-2,5,1/2}];
p2=ContourPlot[Evaluate[list2],{x,-4,4},{y,-4,4},ContourStyle->Red];

Show[p1,p2,AspectRatio->1]

Example 3

Clear[x,y,k,c]

family1[k_]=(y^2)==2 Log[Abs[x]]+2 x+k
list1=Table[family1[k],{k,-8,10,1}];
p1=ContourPlot[Evaluate[list1],{x,-4,4},{y,-4,4},ContourStyle->Blue];

family2[c_]=-x+Log[Abs[x+1]]==Log[Abs[y]]+c
list2=Table[family2[c],{c,-2,4,1/2}];
p2=ContourPlot[Evaluate[list2],{x,-4,4},{y,-4,4},ContourStyle->Red];

Show[p1,p2,AspectRatio->1]

Example 4

Clear[x,y,k,c]

family1[k_]=(x^2)+(y^2)==2k*x
list1=Table[family1[k],{k,-8,10,1/2}];
p1=ContourPlot[Evaluate[list1],{x,-4,4},{y,-4,4},ContourStyle->Blue];

family2[c_]=(y/((x^2)+(y^2)))==c
list2=Table[family2[c],{c,-2,2,1/4}];
p2=ContourPlot[Evaluate[list2],{x,-4,4},{y,-4,4},ContourStyle->Red];

Show[p1,p2,AspectRatio->1]