ORTHOGONAL TRAJECTORIES
(Polar Coordinates)
Topic No 1.1.1.2.2
In this lecture, we are going to discuss about orthogonal trajectories of a given family of
curves in polar coordinates. This is one of the most familiar applications of ordinary differential
equations of first order.
Expected outcomes of this lecture are:
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understand the procedure to find orthogonal trajectories of the given family of
curves in polar coordinates
apply the procedure to find orthogonal trajectories of the given family of curves
in polar coordinates
Appreciate some of the applications of orthogonal trajectories.
To understand this concept we require the knowledge of
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1. How to formulate a first order differential equation.
2. How to solve a first order differential equation.
3. Some basic concepts of Geometry (angle between curves, slope etc.)
I organize my lecture as follows:
➢ Procedure to find orthogonal trajectories of given family of curves (polar coordinates)
➢ Implement the above procedure to solve the problems
➢ Finally, discuss some of the applications of orthogonal trajectories
First, Let us recollect the definition of orthogonal trajectories
Definition:
Two families of curves such that every member of either family cuts each member of
the other family at right angles are called orthogonal trajectories of each other.
First, we shall see the procedure for obtaining orthogonal trajectories of a given family
of curves when they are given in polar coordinates.
Consider the family of curves 𝑭(𝒓, 𝜽, 𝒌) = 𝟎, where ‘k’ is an arbitrary constant. Suppose
the associated differential equation may be found as
𝒅𝒓
𝒅𝜽
= 𝒇(𝒓, 𝜽), on eliminating the
arbitrary constant ‘k’.
For the given curve that passes through 𝑃(𝑟, 𝜃),
the slope at ‘P’ is
𝑑𝜃
𝑇𝑎𝑛∅ = 𝑟 𝑑𝑟
⃗⃗⃗⃗⃗ and the tangent Tc at the point 𝑃(𝑟, 𝜃) to the
(where ∅ is angle between the radius vector 𝑂𝑃
curve ‘C’)
Now the slope of the curve ‘D’ that is orthogonal to the given curve ‘C’ is
𝑡𝑎𝑛(∅0 ) = 𝑡𝑎𝑛(∅ + 900 ) = −𝑐𝑜𝑡(∅) = −
=−
1
𝑡𝑎𝑛∅
1 𝑑𝑟
𝑟 𝑑𝜃
Thus for getting differential equation of the orthogonal trajectory
𝑟
Or
𝑑𝜃
𝑑𝑟
𝑑𝑟
𝑑𝜃
is to be replaced by −
1 𝑑𝑟
𝑟 𝑑𝜃
is to be replaced by −𝑟 2
𝑑𝜃
𝑑𝑟
And we have to solve this new differential equation to get the equation of orthogonal
trajectories.
Working procedure to find the orthogonal trajectories of the family of curves in polar
coordinates:
(i) Suppose given family of curve is 𝑭(𝒓, 𝜽, 𝒄) = 𝟎 where ′𝑐′ is parameter.
(ii) Form the differential equation of the curve in the form 𝑓 (𝑟, 𝜃,
(iii) Replace
𝑑𝑟
𝑑𝜃
by −𝑟 2
𝑑𝜃
𝑑𝑟
𝑑𝑟
𝑑𝜃
) = 0 by eliminating ′𝑐′.
in the differential equation obtained in (ii) to get differential equation
of orthogonal trajectories.
(iv) Solve the differential equation of the orthogonal trajectories 𝑓 (𝑟, 𝜃, −𝑟 2
𝑑𝜃
𝑑𝑟
) = 0.
Problem: Find the orthogonal trajectories of the family of curves (cardioids) 𝑟 = 𝑎(1 + cos 𝜃),
where ′𝑎′ is a parameter.
Solution: Given family of curves 𝑟 = 𝑎(1 + cos 𝜃) where ′𝑎’ is parameter.
Applying logarithms on both sides, we get
log 𝑟 = log 𝑎 + log(1 + cos 𝜃)
differentiating with respect to ′𝜃 ′ we get,
1 𝑑𝑟
𝑟 𝑑𝜃
=
1
1+cos 𝜃
. (− sin 𝜃) ⟹
𝑑𝑟
𝑑𝜃
=−
𝑟 sin 𝜃
1+cos 𝜃
To get differential equation of orthogonal trajectories replace
−𝑟 2
⟹
𝑑𝜃
𝑑𝑟
=−
1+cos 𝜃
sin 𝜃
⟹∫
⟹∫
𝑟 sin 𝜃
1+cos 𝜃
1
𝑑𝜃 = 𝑑𝑟
𝑟
1+cos 𝜃
sin 2𝜃
2 cos2
𝜃
2
1
𝑑𝜃 = ∫ 𝑑𝑟
𝑟
𝜃
2
𝜃
2
2 sin cos
𝑑𝜃 = log 𝑟
𝜃
⟹ ∫ cot 𝑑𝜃 = log 𝑟
2
𝜃
⟹ 2 log sin + log 𝑐 = log 𝑟
2
𝑑𝑟
𝑑𝜃
by −𝑟 2
𝑑𝜃
𝑑𝑟
, then we get
𝜃
⟹ log sin2 + log 𝑐 = log 𝑟
2
𝜃
⟹ log c sin2 = log 𝑟
2
𝜃
⟹ c sin2 = 𝑟
2
⟹ 𝑐(
1−cos 𝜃
2
)=𝑟
⟹ 𝑟 = 𝑘 (1 − cos 𝜃) is the equation of orthogonal trajectories (here k=c/2) of given
family of curves 𝑟 = 𝑎(1 + cos 𝜃).
Problem: Find orthogonal trajectories of the family of curves 𝑟 𝑛 = 𝑎𝑛 cos 𝑛𝜃
Solution: Given family of curves is 𝑟 𝑛 = 𝑎𝑛 cos 𝑛𝜃 , ----- (1) where ‘a’ is a parameter.
Taking logarithms on both sides of (1),
𝑛. log(𝑟) = 𝑛. log(𝑎) + log (cos 𝑛𝜃) ----- (2)
Differentiating (2) w.r.to ′𝜃′ we get,
𝑛.
⇒
1 𝑑𝑟
r 𝑑𝜃
1 𝑑𝑟
r 𝑑𝜃
=0+
1
cos 𝑛𝜃
(− sin 𝑛θ). 𝑛
= − tan 𝑛𝜃 ----- (3), this is differential equation of given family
of curves.
𝑑𝑟
𝑑𝜃
To get the differential equation of orthogonal trajectories of (1), replace 𝑑𝜃 𝑏𝑦 − 𝑟 2 𝑑𝑟
in (3)
Thus we get,
⇒𝑟
𝑑𝜃
𝑑𝑟
1
r
(−𝑟 2
𝑑𝜃
𝑑𝑟
) = − tan 𝑛𝜃
= tan 𝑛𝜃 ----- (4), is the differential equation of orthogonal trajectories.
Let us solve this differential equation to get the equation of orthogonal trajectories.
(4) can be written as,
𝑑𝜃
tan 𝑛𝜃
=
1
𝑟
𝑑𝑟
1
⇒ ∫ cot 𝑛𝜃 𝑑𝜃 + log 𝑐 = ∫ 𝑑𝑟
𝑟
1
⇒ 𝑙𝑜𝑔 (sin 𝑛𝜃) + 𝑙𝑜𝑔𝑐 = 𝑙𝑜𝑔 (𝑟)
𝑛
⇒ 𝑙𝑜𝑔 (sin 𝑛𝜃) + 𝑛. 𝑙𝑜𝑔𝑐 = 𝑛. 𝑙𝑜𝑔 (𝑟)
⇒ 𝑙𝑜𝑔 (𝑐 𝑛 sin 𝑛𝜃) = 𝑙𝑜𝑔 (𝑟 𝑛 )
⇒ 𝑟 𝑛 = 𝑐 𝑛 sin 𝑛𝜃, is the equation of orthogonal trajectories of
𝑟 𝑛 = 𝑎𝑛 cos 𝑛𝜃
Problem: Find the orthogonal trajectories of the family of curves 𝑟 =
2𝑎
1+cos 𝜃
, where ′𝑎′
is a parameter.
Solution: Given family of curves 𝑟 =
2𝑎
where 𝑎 is parameter.
1+cos 𝜃
Applying logarithms on both sides, we get
log 𝑟 = log 2𝑎 − log(1 + cos 𝜃)
differentiate above equation with respect to ′𝜃′
1 𝑑𝑟
=
𝑟 𝑑𝜃
1
1+cos 𝜃
. sin 𝜃 ⟹
𝑑𝑟
𝑑𝜃
=
𝑟 sin 𝜃
1+cos 𝜃
To get the differential equation of orthogonal trajectories replace
−𝑟 2
⟹
𝑑𝜃
𝑑𝑟
𝑟 sin 𝜃
=
1+cos 𝜃
1+cos 𝜃
sin 𝜃
⟹∫
⟹∫
1
𝑑𝜃 = − 𝑑𝑟
𝑟
1+cos 𝜃
sin 2𝜃
1
𝑑𝜃 = − ∫ 𝑑𝑟
𝑟
2 cos2
𝜃
2
𝜃
2
𝜃
2
2 sin cos
𝑑𝜃 = − log 𝑟
𝜃
⟹ ∫ cot 𝑑𝜃 = − log 𝑟
2
𝜃
⟹ 2 log sin = − log 𝑟 + log 𝑐
2
𝜃
⟹ log sin2 = − log 𝑟 + log 𝑐
2
𝜃
𝑐
2
𝑟
⟹ log sin2 = log
𝜃
𝑐
2
𝑟
⟹ sin2 =
𝑑𝑟
𝑑𝜃
by −𝑟 2
𝑑𝜃
𝑑𝑟
to get,
⟹
1−cos 𝜃
=
2
⟹𝑟=
𝑐
𝑟
2𝑐
1−cos 𝜃
Problem: Find the orthogonal trajectories of the family of curves 𝑟 = 2𝑎(cos 𝜃 + sin 𝜃),
where ′𝑎′ is a parameter.
Solution: Given family of curves 𝑟 = 2𝑎(cos 𝜃 + sin 𝜃) where 𝑎 is parameter.
Applying logarithms on both sides, we get
log 𝑟 = log 2𝑎 + log(cos 𝜃 + sin 𝜃)
differentiate above equation with respect to ′𝜃′
1 𝑑𝑟
𝑟 𝑑𝜃
=
1
cos 𝜃+sin 𝜃
. (− sin 𝜃 + cos 𝜃) ⟹
𝑑𝑟
𝑑𝜃
=
(cos 𝜃−sin 𝜃)
cos 𝜃+sin 𝜃
To get differential equation of orthogonal trajectories replace
−𝑟
𝑑𝜃
𝑑𝑟
⟹−
⟹∫
=
𝑑𝑟
𝑑𝜃
by −𝑟 2
𝑑𝜃
𝑑𝑟
to get,
(cos 𝜃−sin 𝜃)
cos 𝜃+sin 𝜃
(cos 𝜃+sin 𝜃)
cos 𝜃−sin 𝜃
(− cos 𝜃−sin 𝜃)
cos 𝜃−sin 𝜃
1
𝑑𝜃 = 𝑑𝑟
𝑟
1
𝑑𝜃 = ∫ 𝑑𝑟
𝑟
⟹ log (cos 𝜃 − sin 𝜃) + log 𝑐 = log 𝑟
⟹ 𝑟 = 𝑐 (cos 𝜃 − sin 𝜃) is the equation of orthogonal trajectories of 𝑟 = 2𝑎(cos 𝜃 + sin 𝜃)
Exercise to students:
Problem: Find the orthogonal trajectories of the family of curves 𝑟 2 = 𝑎2 sin 2𝜃 , where ‘𝑎′ is
a parameter.
The concept of orthogonal trajectories is of wide use in applied mathematics especially in field
problems.
❖ In an electric field, curves of electric force are the orthogonal trajectories of
equipotential lines.
❖ In fluid flow problems the stream lines and equipotential lines are orthogonal
trajectories of each other. For instance in heat flow problems, isothermals and heat flow
lines are orthogonal trajectories of each other.
You can appreciate the engineering applications of orthogonal trajectories through the
following problems:
1. The electric lines of force of two opposite charges of the same strength at (-1, 0) and (1, 0)
are the circles through (-1, 0) and (1, 0) whose equation is given by 𝑥 2 + (𝑦 − 𝑐)2 = 1 + 𝑐 2 .
Show that the equipotential lines are the circles (𝑥 + 𝑑)2 + 𝑦 2 = 𝑑 2 − 1.
2. If the stream lines of the flow in the channel are 𝑥𝑦 = 𝑐 find equipotential lines.
3. If the isotherms in a body are 𝑇(𝑥, 𝑦) ≡ 2𝑥 2 + 𝑦 2 =constant. Find the curves along which
heat will flow.