Orthogonal Trajectories ELECTRICAL ENGINEERING MAHATMA GANDHI INSTITUTE OF TECHNICAL EDUCATION &RESEACH CENTRE, NAVSARI Orthogonal Trajectories BE 3RD SEM ELECTRICAL ENGINEERING DIV:-A
GUIDED BY: FACULTY OF MATHS DEPT. MGITER NAVSARI REPRESENTED BY:- KANSARA JUGAL J. 130330109031 KIKANI PRADEEP c. 130330109032 LAD JAY B. 130330109033 LAD JAY R. 130330109034 LADUMOR SHAILESH S. 130330109035 GUIDED BY: FACULTY OF MATHS DEPT. MGITER NAVSARI
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.
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.
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.
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.
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally— that is, at right angles.
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.
Orthogonal Trajections Let f(x,y,a)=0----------------(1) represent a family of curves,one curve for each value of the parameters a. Differentiating, we get f/ x+ f/ y dy/dx=0--------(2)
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.
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
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)
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.
Orthogonal trajectories in polar cordinates Orthogonal trajectories in polar cordinates
Then if the original family is r=2a cos with a0 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
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.
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.
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.