1. Orthogonal Trajectories ELECTRICAL ENGINEERING
MAHATMA GANDHI INSTITUTE OF TECHNICAL EDUCATION &RESEACH CENTRE, NAVSARI
Orthogonal Trajectories
BE 3RD SEM
ELECTRICAL ENGINEERING
DIV:-A

2. GUIDED BY: FACULTY OF MATHS DEPT. MGITER NAVSARI
REPRESENTED BY:- KANSARA JUGAL J KIKANI PRADEEP c LAD JAY B LAD JAY R LADUMOR SHAILESH S
GUIDED BY:
FACULTY OF MATHS DEPT.
MGITER
NAVSARI

3. Definition of orthogonal trajectories
Two families of curves are such that each curve in either family is(whenever they intersect) to every curve in the other family. Each family of curves is orthogonal trajectories of the other. In case the two families are identical,they we say that the family is self-orthogonal.
Orthogonal trajectories has important applications in the eld of physics For example, the equipotential lines and the streamlines in an irratational 2D flow are orthogonal.

4. ORTHOGONAL CURVES
Two curves (graphs) are said to be orthogonal at a point if their tangents lines are perpendicular at that point. That is, the slopes of the two tangent lines are negative reciprocals of each other.

5. ORTHOGONAL TRAJECTORIES
When all the curves of one family of curves
G(x, y, c1) = 0 intersect orthogonally all the curves of
another family H(x, y, c2) = 0, then the families are
said to be orthogonal trajectories of each other.
NOTE: Orthogonal trajectories occur naturally in the construction of meteorological maps and in the study of electricity and magnetism.

6. FINDING ORTHOGONAL TRAJECTORIES
To find the orthogonal trajectory of a given family of curves
Find the differential equation that describes the family.
2. The differential equation of the second, and orthogonal, family is then
3. Solve the equation in Step 2.

7. An orthogonal trajectory of
a family of curves is a curve
that intersects each curve
of the family orthogonally—
that is, at right angles.

8. Each member of the family y = mx
of straight lines through the origin
is an orthogonal trajectory of the
family x2 + y2 = r2 of concentric
circles with center the origin.
We say that the two families are
orthogonal trajectories of each other.

9. Orthogonal Trajections
Let
f(x,y,a)= (1)
represent a family of curves,one curve for each value of the
parameters a.
Differentiating, we get
f/ x+ f/ y dy/dx= (2)

10. Eliminating a between (1) and (2),we get a differential
equation of the first order
(x,y,dy/dx)=0
of which (1) is the general solution.
Now we want a family of curves
cutting every member of (1) at right
angle at all points of intersection.
At a point of intersection of the two curves,
x,y are the same but the slope of the second
curve is negative reciprocal of the
slope of the first curve.

11. As such differential equation
Of the family of orthogonal trajectories is
(x,y,-1/ dy/dx)=0
Integrating, we get g(x,y,b)=0
which give the orthogonal trajectories of the family (1)
Let the original family be y = mx, when m is a parameter then
dy/dx = m
and eliminating m, we get the differential equation of this
Concurrent family of straight lines as y/x=dy/dx
To get the orthogonal trajectories,
we replace dy/dx by-1/ dy/dx to get
y/x=-1/ dy/dx

12. Integrating
x2+y2=a2
which gives the orthogonal trajectories as concentric
circle. Find the orthogonal trajectories of the family
of confocal conics
x2/a2++y2/b2+=1
where  is a parameter. Differentiating,we get
x2/a2++y2/b2+ dy/dx=0
Eliminating  ,we get
(x-y)(x+ y)= (a2+b2) ; =dy/dx
To get the orthogonal trajectories, we replace  by -1/  to get
(-x/;-y)(x- y/)=-1/ (a2-b2 ) or (x-y)(x+ y)= (a2+b2)

13. As such the family of confocal conics is self-orthogonal,
i.e. for every conic of the family,there is another
with same focii which cuts it at right angles.
One family consists of confocal
ellipses and the other consists of
confocal hyperbolas with the same focii.
In polar coordinates after getting
the differential equation of the family
of curves,we have to replace r d/dr by
-1/ r d/dr and then integrate the
resulting differential equation.

14. Orthogonal trajectories in polar cordinates
Orthogonal trajectories in polar cordinates

15. Then if the original family is
r=2a cos
with a0 as a parameter, its differential equation
is obtained by eliminating a and
dr/d=-2asin 
to get
r dr/d=-cot 
Replacing r dr/d by -( r dr/d)-1 , we get
r dr/d=tan 
Integrating we get r=2bsin 

16. The orthogonal trajectories are shown
The circles of both families pass through the origin, but while
the centre of one family lie on x-axis, the centres of the
orthogonal family lie on y-axis.

17. ORTHOGONAL TRAJECTORIES IN PHYSICS
Orthogonal trajectories occur in various branches of physics.
In an electrostatic field, the lines of force are orthogonal to the lines of constant potential.
The streamlines in aerodynamics are orthogonal trajectories of the velocity-equipotential curves.

18. Example Find the orthogonal trajectories
of family of straight lines through
the origin.
Solution :
The family of straight lines through the origin is given by y = mx
The ODE for this family is xy’-y = 0
The ODE for the orthogonal family is x + yy’ = 0
Integrating we find X^2 + y^2 = C;
which are family of circles with centre at the
Origin.