orthogonal trajectories in cartesian coordinates

ORTHOGONAL TRAJECTORIESIN CARTESIAN COORDINATES

Submitted by:

RAFAE ASHFAQ

08-EE-40

ORTHOGONAL TRAJECTORIESIN CARTESIAN COORDINATES

Submitted by:

SHEHZEB ASHRAF

08-TE-24

CONTENTS

• This presentation includes the following– INTRODUCTION TO ORTHOGONAL

TRAJECTORIES.– FINDING ORTHOGONAL TRAJECTORIES– APPLICATIONS– EXAMPLES

INTRODUCTION

ORTHOGONAL TRAJECTORIESIn dictionary

ORTHOGONAL: Derived from greek word “Ortogonios” meaning right angled. TRAJECTORY: Derived from the latin word “Trajectoria” meaning cut across.

So, orthogonal trajectories means cutting something across the right angle.

CARTESIAN COORDINATES locate a point in terms of its perpendicular distance from (two or three) mutually perpendicular axes

ORTHOGONAL TRAJECTORIESIN GEOMETRY

• In geometry, orthogonal means perpendicular and trajectories mean curves or surfaces cutting another family of curves or surfaces at a constant angle. As we are relating it with the word “orthogonal” so “ORTHOGONAL TRAJECTORIES” means curves or surfaces intersecting a family of curves or surfaces at right angle where the angle is defined as the angle between the tangents of the two curves.

ORTHOGONAL FAMILIES

Similarly two families of curves will be orthogonal if each curve in one family intersect the other family at right angle. In other words, two families

F(x, y, c) = 0 and G(x, y,K) = 0

are orthogonal families if each member

of one family is an orthogonal trajectory of

the other family.

FINDING

ORTHOGONAL TRAJECTORIES

METHODS TO FIND ORTHOGONAL TRAJECTORIES

There are two ways in which we can find the differential equations to obtain orthogonal trajectories.

Before going into the detail of these methods we should be familiar with some of the useful concepts which are related to these concepts.

DERIVATIVE AS THE SLOPE OF THE CURVE

• Let AB be the arc of the graph of f defined by the equation y=f(x).

• Let P(x,f(x)) and Q(x+δx,f(x+δx))

be two neighbouring points on

the arc AB. The line PQ is

secant of the curve and it

makes <α with the x-axis.

Drawing the ordinates of PM,

QN and the perpendicular

PR to NQ, we have

O

y

M NS x

A R

B

P(x,f(x))

Q(x+δx),f(x+δx))

and

RQ = NQ –NR =NQ –MP= f(x+δx) – f(x)

PR = MN= ON –OM = x+δx-x = δx

Thus, tan m <α =tan m <RPQ

Revolving the secant line

PQ about P towards P,

some of its successive

positions PQ1, PQ2 , PQ3 , . . .

are shown in the figure.

δx

f(x)δx)f(x

PR

RQ

O

y

MS x

Q

P

Q1

Q2

Q3

R3

T

Points Qi (i=1,2,3,…) are getting closer and closer to the point P and PRi i.e.,δ xi (i=1,2,3,…) are approaching to zero, that is,

Thus, the slope of the tangent line to the graph of f at (x,f(x)) is f’(x) or dy/dx.

XTPmtan(x)f'or

XTPmtanδx

f(x)δx)f(xlim

0δxasXTPmtanδx

f(x)δx)f(xor

0δxwhenXTPmtanαmtan

0δx

SLOPE OF A CURVE PERPENDICULAR TO ANOTHER CURVE

• Consider two curves C1 and C2 which intersect each other at an angle α

at an arbitrary point P.

In other words the

angle between their

tangents at P will be

equal to angle α.φ

C1

C2

Ψ

α

P

x

y

O

• To relate it to orthogonal trajectories, we consider that C2 is perpendicular to C1. So,

angle α =90° or π/2 rad

The angle between x-axis and the tangent of C1 at P is Φ and the angle between x-axis and the tangent of C2 at P is Ψ. So,

angle α = Ψ – Φ

According to the basic trigonometric identity

φΨ

φΨ

φ)(ψ

πα

tantan

tantan

tan

tantan

1

2

As α = π/2, so tan α will be infinite, but the numerator of the above equation is finite. So the above equation will be infinite if and only if

1+ tanΨ tanΦ = 0

Showing that )(1tanψ

1Φtan

• Now, if dy/dx is the slope of the curve C1 at P,

dy/dx = tanΦ

and if dyo/dx is the slope of its orthogonal trajectory at P,

dyo/dx = tanΨ

Thus from eq 1, we see that these two slopes are related by the expression

dxdy

1dx

dy

0

FINDING THE ORTHOGONAL TRAJECTORIES

An important application of first order DE’s is to find the orthogonal trajectories of a family of curves F1, given by the equation:

f(x, y,C) = 0

This can be done by two ways1. By using differential 2. By differentiating with respect to x.

BY USING THE DIFFERENTIAL

Suppose we have to find the orthogonal trajectories of a family of curves F1, given by the equation:

f(x, y,C) = 0 ….. (1)To find the orthogonal trajectories, we may derive the differential equations, whose solutions are described by these trajectories. For this, we proceed as follows

The differential of eq. 1 will be dy

y

fdx

x

fdf

• Solving for dy/dx, we have

If F2 is the family of curves perpendicular to F1,which

is given by

f(x0, y0,C0) = 0

then as we have proved earlier, its slope will be related to the slope of F1 as below,

)(2

δyδfδxδf

dx

dy

• Putting the value of dy/dx from eq. 2 in the above equation

or

dxdy

1dx

dy

0

0

δyδfδxδf

1dx

dy

)(3x

y

f

f

δxδfδyδf

dx

dy

• By integrating eq.3, we can get the required equation of the family F2 which is orthogonal to the family F1.

BY DIFFERENTIATING WITH RESPECT TO x

Suppose we have to find the orthogonal trajectories of a family of curves, given by the equation:

f(x, y,c) = 0 ….. (1)To find the orthogonal trajectories, we would have to find the differential equations, and by integrating them we shall find the equations described by these trajectories. For this, we proceed as follows,

Differentiating both sides with respect to x, we shall get the constant canceled, terms

• Containing y will give dy/dx. Solving for dy/dx we shall get a function in c and y variables with no constant. If constant is not eliminated yet, then by solving the equation.1 for c and then putting the value of c in dy/dx will give us the derivative or slope of the given family.

• Now, as we have proved, the family of curves passing perpendicularly through the given family of curves will have slope or derivative related to the given family will be

• By integrating this equation, we can get the required family of curves.

dxdy

1

dX

dY

EXAMPLE• Find the orthogonal trajectory of the

following:

x2 + y2 = c2

SOLUTION

We have the family

x2 + y2 = c2

Differentiating both sides with respect to x, we have

2x + 2y dy/dx = 0

The orthogonal family will have the slope

Re arranging and integrating both sides

y

x

dx

dy

x

y

dx

dy

yx

1dx

dy

lnylnclnx

dyy

1dx

x

1

which is the required

family of lines. These

two families can be

sketched as shown in

the figure.

cxy

APPLICATIONS

APPLICATIONS• Orthogonal trajectories has vast applications in

Physics, Engineering and many other fields. Some of them are– Heat conduction– Fluid dynamics/hydrodynamics – Electromagnetic theory– Electrostatics– Gravitation– Elasticity– Meteorology (Weather mapping)– Geometric optics

HEAT CONDUCTION

• Orthogonal trajectories arise in thermodynamics as well. In heat conduction, isotherms are orthogonal to the heat flow.

Fluid dynamics/hydrodynamics

• Orthogonal trajectories arise in fluid dynamics as well. In two dimensional fluid dynamics, orthogonal trajectories express the relationship between the curves followed by the fluid particles called streamlines, and the associated equipotential lines along which a function called the fluid potential is constant.

GRAVITATION

• Orthogonal trajectories also occur in gravitational study. If the original family represent the lines of force in a gravitational field, its orthogonal trajectories represent the equipotentials which are the curves along which gravitational potential is constant.

METEOROLOGY

• As another example, consider a weather map. The curves represent the isobars, which are curves connecting all cities that report the same barometric pressure at the weather bureau.

WORLD GLOBE

• Meridians and parallels on the world globe are orthogonal trajectories of each other.

ELASTICITY

• Orthogonal trajectories arise in elasticity as well. When stress is applied in a body, strain is produced. It is seen that the lines of stress are orthogonal to the lines of constant strain.

Geometric Optics• In geometric optics wave fronts are orthogonal

trajectories of the rays.Light rays

Wave fronts