One Parameter Family of Curves

ONE PARAMETER

FAMILY OF CURVES

One Parameter Family of Curves; we will learn

how to use DEs for finding curves that intersect

given curves at right angle, a task which arises

rather often in applications.

If for each fixed real value of c the equation

F(x,y,c) = 0 (1)

represents a curve in the xy-plane and if for

variable c, it represent infinitely many curves

then totality of these curves is called a One

Parameter Family of Curves, and c is called the

Parameter of the family. For example

F(x,y,c) = x + y + c = 0 represents a family of

0),,( 222 cyxcyxF

01dx

dy

parallel straight lines correspond to precisely one value of the parameter c. The DE of this eq is

And eq

represents a family of concentric circles of radius c with centre at the origin. The DE of this eq is

022 dx

dyyx

y

x

dx

dy

2cxy cxdx

dy2

The DE of the family of parabola is

From eq we have

Put this value of c in DE

Or differentiate

which is same as above

2x

yc

x

y

dx

dy 2

2x

yc

x

y

dx

dy

x

y

x

y 220 32

/

The general solution of a first order DE involves a parameter c and thus represents a family of curves. This yields a possibility for representing many one-parameter families of curves by first order DEs.

ORTHOGONAL TRAJECTORIES

Orthogonal Trajectories In many engineering applications, a family of curves is given and it is required to find another family whose curves intersect each of the given curves at right angles. Then the curves of two families are said to be mutually orthogonal, and they form orthogonal net, and the curves of the family to be obtained are called Orthogonal Trajectories of the given curves.The DEs may be used for finding curves that intersect given curves at right angle. The new curves are then called Orthogonal Trajectories of the given curves (and conversely). Example, in electrostatics the equipotential lines of concentric circles(cylinders in space

appearing as circles in cross section) and their orthogonal trajectories are straight lines(plain in space through the axis of cylinders).

It is important to note that the angle of intersection of two curves is defined to be the angle between the tangents of the curves at point of intersection

Equipotential lines and lines of electric force(dashed) between two concentric cylinders, are orthogonal trajectories of each other

Or Orthogonal TrajectoriesConsider a one-parameter family of curves in the xy-plane defined by F(x, y, c) = 0 (1)where c denotes the parameter. The problem is to find another one-parameter family of curves, called the orthogonal trajectories of the family of curves in eq (1) and given analytically by G(x, y, k) = 0 (2)such that every curve in this new family of eq (2) intersects at right angles every curve in the original family of eq (1).We first implicitly differentiate eq (1) with respect to x, then eliminate c between this derived equation and eq (1)

This gives an equation connecting x, y, and which we solve for to obtain a differential equation of the form (3)The orthogonal trajectories of eq (1) are the solutions of (4)For many families of curves, one cannot explicitly solve for dy/dx and obtain a differential equation of the form of eq (3).

/y/y

),(/ yxfy

),(

1/

yxfy

CARTTESIAN

COORDINATES

Consider a family of curves F(x,y,c) = 0 (1)Which is represented by a DE

(2)

corresponding orthogonal can be found as followFrom eq (2) it is observed that a curve of given family which passes through a pointHas the slope at that point . The of the orthogonal trajectory through at this point should be negative reciprocal of that is because this is the condition for tangents of two curves at to be perpendicular as we know that

),( yxFdx

dy

),( 00 yx),( 00 yxf

),( 00 yx

),,( 00 yxf ),,(/1 00 yxf

),( 00 yx

Consequently the DE of the orthogonal trajectory is (3)

and trajectories are obtained by solving this eq (3).Note Here orthogonal is another word for perpendicular.

),(

1

yxFdx

dy

2121

11)(

mmmm

OR If every member of a family of curves is a solution of M(x,y)dx + N(x,y)dy = 0 …..(1)While every member of second family of curves is a solution of related equation N(x,y)dx - M(x,y)dy = 0 ……………………(2)Then each curve of the one family is orthogonal to every curve of the other family. Each family is said to be a family of orthogonal trajectories of the other.As from eq (1), we have

And from eq (2) we have

Which shows eq (1) = -1/eq (2)

),(

),(

yxN

yxM

dx

dy

),(

),(

yxM

yxN

dx

dy

Find the eq of orthogonal trajectory of the curve

Soln: Differentiating the eq

Now by using eq

We get

),(

1

yxFdx

dy

2cyx

dx

dycy21

c

xyas

cxcydx

dy ....

2

1

2

1

cx

cxdx

dy2

211

Is the eq of the orthogonal trajectory

22 ....2

)(2y

xcas

y

xx

y

x

dx

dy

xdxydyy

x

dx

dy2

2

cyxcxy

2222

22

Q; In the electric field between two concentric cylinders as shown in figure, the equipotential lines( equal to curves of constant potential) are circles given by

Find the trajectories(the curves of electric force).Soln: Differentiating the eq

……………(1)

][),( 22 voltsconstyxyxU

022 dx

dyyx 0

dx

dyyx

y

x

dx

dy

Now by using eq

Eq (1) can be written as

Is the eq of the orthogonal trajectory

x

dx

y

dyx

y

dx

dy

xcy

cxy

lnln

),(

1

yxFdx

dy

Find the eq of orthogonal trajectory of the curve

Soln: Differentiating the eq

Now by using eq

We get

),(

1

yxFdx

dy

2)( cxy )(2 cx

dx

dy

ydx

dy

2

1

yxcasyyxxdx

dy ....2)(2

Taking square of both sides

is the eq of the orthogonal trajectory

cxycxy

334]3

2[2 2

32

3

dxdyy 2

)(34 2

3

xcy

23 )(916 xcy

Find the eq of orthogonal trajectory of the curve

cxxy 22