maths

1. GEC BHAVNAGAR

2. OrthogonalTrajectories

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

4. OrthogonalTrajectories

5. METHODMETHOD
Given family of curve ( , , )F x y c o=
STEP 1 Find the differential equation for the given
family of curves, by differentiating Eq.1.
( , )
dy
f x y
dx
=
STEP 2 Find the differential equation of the Orthogonal
Trajectory
1
( , )
dy
dx f x y
= −
STEP 3
Eq.3.
Eq.2.
Eq.1.
Solution of Eq.3. will be the equation of the
family of orthogonal trajectories
( , , )G x y c o=
OrthogonalTrajectoriesChapter3

6. Orthogonal Curves (1)
=
=By differentiation we get: . Hence the family of parabola
in question satisfies the differential equation 2 .
2
dy
x
dx
dy
x
dx
2
Consider the family of parabola . Find the family of curves
which intersect the above family of parabola perpendicularly.
y x C= +
Example
Solution
Two curve intersect perpendicularly if the product of the slopes of
the tangents at the intersection point is -1.
The differential equation for the orthogonal
family of curves.
= −
1
2
dy
dx x
OrthogonalTrajectoriesChapter3

7. Orthogonal Curves (2)
= − ⇒ = − ⇒∫ ∫
1 1 1
2 2
dy
dy dx
dx x x
It remains to solve
2
Consider the family of parabola . Find the family of curves
which intersect the above family of parabola perpendicularly.
y x C= +
Example
Solution
(cont’d)
The figure on the right shows these
two orthogonal families of curves.
= −
1
2
dy
dx x
= − +
1
ln
2
y x C
OrthogonalTrajectoriesChapter3

8. Describe the orthogonal
trajectories for the family
of curves given by
xy = C
Solution
y2
– x2
= K
y2
– x2
= K

10. Find the orthogonal
trajectories of
the family of curves x = ky2
,
where k
is an arbitrary constant.
1
1 2 or =
2
dy dy
ky
dx dx ky
=
2
1 1
2 2
dy
xdx ky y
y
= =
Solution

11. 2
dy y
dx x
=
2dy x
dx y
= −
2
2
2
2
2
2
2
y dy xdx
y
x C
y
x C
= −
= − +
+ =
∫ ∫