03.1 orthogonal treajectories (1)
DESCRIPTION
mathsTRANSCRIPT
GEC BHAVNAGAR
O
rthog
onal
Tra
ject
orie
s
• An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally—that is, at right angles.
Ort
hogo
nal T
raje
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METHODMETHODGiven 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 ydx
STEP 2 Find the differential equation of the Orthogonal Trajectory
1( , )
dydx 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
Orth
ogon
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Chap
ter 3
Orthogonal Curves (1)
By differentiation we get: . Hence the family of parabola
in question satisfies the differential equation 2 .
2
dy xdx
dy xdx
2Consider the family of parabola . Find the family of curveswhich 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
2dydx xOr
thog
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Tra
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apte
r 3
Orthogonal Curves (2)
1 1 1
2 2dy dy dxdx x x
It remains to solve
2Consider the family of parabola . Find the family of curveswhich 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
2dydx x
1 ln2
y x C
Orth
ogon
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Chap
ter 3
Describe the orthogonal trajectories for the family of curves given by
xy = C
Solution
y2 – x2 = K
y2 – x2 = K
Find the orthogonal trajectories of the family of curves x = ky2, where k is an arbitrary constant.
11 2 or = 2
dy dykydx dx ky
2
1 12 2
dyxdx ky yy
Solution
2dy ydx x
2dy xdx y
22
22
2
2
2
y dy x dx
y x C
yx C