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Orthogonal trajectories

Date post: 14-Apr-2017
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Names of group members 1.Syed Umar Rasheed (14093122-002) 2.Adnan Aslam (14093122-003) 3.Bilal Amjad (14093122-004) 4.Inzmam Tahir (14093122-005)
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Page 1: Orthogonal trajectories

Names of group members

1. Syed Umar Rasheed (14093122-002)2. Adnan Aslam (14093122-003)3. Bilal Amjad (14093122-004)4. Inzmam Tahir (14093122-005)

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Orthogonal Trajectories in rectangular coordinate system

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Definition

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

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Suppose that we have a family of curves given by and another family of curves such that at any intersection of the curves of the first family with a curve of the second family, the tangents of the curves are perpendicular.

Orthogonal trajectories, therefore, are two families of curves that always intersect perpendicularly.

𝑓 (𝑥 , 𝑦 )=0

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Let us consider the example of the two families

If we draw the two families together on the same graph we get

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METHOD( , )dy f x y

dx

1( , )

dydx f x y

Find the differential equation for the given family of curves, by differentiating

Find the differential equation of the Orthogonal Trajectory

Separating variables and integrating the above differential equation we get the algebraic equation of the family of orthogonal trajectories.

Given family of curve 𝑓 (𝑥 , 𝑦 )=0

STEP 1

STEP 2

STEP 3

𝑑𝑥𝑑𝑦=− 𝑓 (𝑥 , 𝑦)or

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Orthogonal Curves2Consider 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.

𝑑𝑦𝑑𝑥 =− 1

2𝑥

By differentiation we get . Hence the family of parabola in question satisfies the differential equation .

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1 1 1

2 2dy dy dxdx x x

The figure on the right shows these two orthogonal families of curves.

1

2dydx x

1 ln2

y x C

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Example Find the orthogonal trajectories of all parabolas with vertices at the origin and foci on the x-axis𝒚𝟐=𝟒𝒂𝒙𝒚𝟐

𝒙 =𝟒𝒂

STEP 1: Differentiate Eq 1Eq 1

𝒙 𝒅𝒅𝒙 ( 𝒚𝟐)− 𝒚𝟐 𝒅

𝒅𝒙 (𝒙)

𝒙𝟐 =𝟎

𝟐 𝒙𝒚 𝒅𝒚𝒅𝒙 −𝒚

𝟐=𝟎

𝒅𝒚𝒅𝒙 =

𝒚𝟐 𝒙

STEP 2: Differential equation of the orthogonal trajectories𝒅𝒚𝒅𝒙 =−𝟐𝒙𝒚

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STEP 3: Solving the differential equation by method of separation of variables 𝒅𝒚𝒅𝒙 =−𝟐𝒙𝒚−𝒚𝒅𝒚=𝟐𝒙𝒅𝒙 ,integrating both sides𝒙𝟐+

𝒚𝟐

𝟐 =𝑪

𝟐 𝒙𝟐+𝒚𝟐=𝑪𝟐 𝒙𝟐+𝒚𝟐=𝒃𝟐

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Exercise 9.7

Find the orthogonal trajectories of each of the following.

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Applications• Equipotential Lines and Electric Fields

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• Electromagnetic waves consist of both electric and magnetic field waves. These waves oscillate in perpendicular planes with respect to each other.

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• The light rays of sun that passes through the orbits of the planets.

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References • www.slideshare.com• www.quora.com• www.mathcity.com• www.wikipedia.com


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