Ch. 3 Pg. 176

Differential equation of the orthogonal trajectory via equation (10).

Ch 3- Sec 2- Pro 8 Parts a , c Page 1 / 4

Show that each pair of equations (a - c) in polar coordinates describes a set of orthogonal trajectories:

Since

out of phase they are orthogonal. All this aside

The lines x = c and y = C are orthogonal lines.Alternatively

Now use equation (17) The assumed orthogonal trajectory of (10):

Ch. 3 Pg. 177

(22) is the differential equation of the orthogonal trajectory

Ch 3- Sec 2- Pro 8 Parts a , c

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Since (22) = (15) the pair isorthogonal.

( B )If the polar pair describes a set of orthogonal trajectories then

the differential equation of the orthogonal trajectory of equation (1) will equal the differential equation of equation (2)

(1,2)

Note: In equations ( 1) & (2) sine and cosine are out of phase by 90degrees. Therefore they are orthogonal.

Ch. 3 Pg. 178

Differential equation of orthogonal trajectory of equation (1)

Ch 3- Sec 2- Pro 8

Parts b , c Page 3 / 4

Now find the differential equation of equation (2):

Therefore, since the differential equation of eqn(2) = differentialequation of eqn (1). The polar coordinate pair describes a set of orthogonal trajectories.

( C )Show that the following pair of polar equations describes a set

of orthogonal trajectories.

(1 , 2)

First find the differential equation of the orthogonal trajectories of equation (1).

Ch. 3 Pg. 179

Differential equation of theorthogonal trajectory.

Ch 3- Sec 2- Pro 8 Parts c Page 4/ 4

Now if equation (2) is the orthogonal trajectory, its differential should equal equation (6).

Therefore eqn (6) =eqn (18). And the pair is an orthogonal pair.

Ch. 3 Pg. 180

Differential equation of orthogonal trajectory

Ch 3- Sec 2- Pro 9 Parts a , b Page 1 / 7

Let k be a fixed parameter that specifies the family and c varies from one curve to the next in a given family. ( A ) Show

that the orthogonal trajectories of are ellipses for k > 0,

and hyperbolas for k < 0 , with the equation .

Hint: assume x > 0 , y > 0 and drop the | abs | values. The extension

to other quadrants is made by symmetry. :Hint

Solution:

Ch. 3 Pg. 181

Ch 3- Sec 2- Pro 9 Parts a , b

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If k = 1 , equation (11) is an ellipse (Circle) centered at the origin.

As k grows or k > 0 the diagonal of the ellipse along the x - axis becomes less and less and does not grow proportional to the y - axis diameter. Therefore an ellipse results.

Note: is the standard equation of the ellipse.

If k < 0 then

Ch. 3 Pg. 182

Family of parabolas that pass through the origin

Ch 3- Sec 2- Pro 9 Parts a , b

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For

Here C limits how large x & y can be. I.e. not greater

than C .Therefore the hyperbola closes, to become an ellipse.

( B )Sketch the original family and the orthogonal trajectories for

k = { 2, 1, 0, -2 }.

Solution:

Ch. 3 Pg. 183

Family of straight lines passing through the origin.

Set of lines orthogonal to the y - axis

Ch 3- Sec 2- Pro 9

Parts b Page 4/ 7

See figure 4 below for

Ch. 3 Pg. 184

Ch 3- Sec 2- Pro 9

Parts b Page 5 / 7

For the orthogonal trajectories @ k = 0 and

we get the family of straight lines

orthogonal to the x - axis. See figure 5. (Orthogonal to figure 3)

Ch. 3 Pg. 185

Ch 3- Sec 2- Pro 9 Parts b

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A family of circles (ellipses) centered about the origin. Compare

with orthogonal curves in figure 2. See figure 6 below where c = 0 is a point at the origin.

At k = 2

See orthogonal curve in figure 1.See also figure 7 below:

Ch. 3 Pg. 186

Ch 3- Sec 2- Pro 9 Parts b

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Then @ k = -2

A family of hyperbolas

Ch. 3 Pg. 187

Differential of the original family.

Ch 3- Sec 2- Pro 10Page 1 / 2

If , show that the following families are orthogonal

trajectories of each other:

@ x > 0 and y > 0[ When k is an integer, then the ‘@’ condition can be replaced by

]

[ (8) is the differential of the orthogonal trajectory ]

Now let’s check the orthogonal trajectory.

Ch. 3 Pg. 188

Ch 3- Sec 2- Pro 10Page 2 / 2

[ Differential of orthogonal trajectory. (13) = (8) ]

Ch 3- Sec 2- Pro 11 Parts a , d Page 1 / 6

Find the orthogonal trajectories of , assuming

when necessary that x > 0 , y > 0 , or both. What happens if k = 1, 2, ...

Solution:

Then eliminate C

Ch. 3 Pg. 189

Differential of orthogonal trajectory

Differential of original family

Ch 3- Sec 2- Pro 11 Parts a , d

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An Exact differential of (10) yields the following:

Ch. 3 Pg. 190

NoOrthogonality

A family of lines passing through the point (1,0)

Ch 3- Sec 2- Pro 11 Parts b, d

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( B )If k = 0

( C )If k = 1 then

See figure C - 1.

And

See figure C-2 and the compilation figure C-3 below.

Ch. 3 Pg. 191

Ch 3- Sec 2- Pro 11 Parts c, d

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Ch. 3 Pg. 192

Ellipse not centered @ the origin

Ch 3- Sec 2- Pro 11

Parts c, d Page 5/ 6

Figure C - 3. Compilation of figure C-1 & C-2

( D ) Hint: For k = 2 solve (36) for its orthogonal trajectory.If k = 2 then equation (19) becomes

For k = 2 we must start the problem again. So

Ch. 3 Pg. 193

Ch 3- Sec 2- Pro 11 Parts d

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Then eliminate c from (42) & (40)

The orthogonal trajectory @ k = 2.

Ch. 3 Pg. 194

Ch 3- Sec 2- Pro 12 Page 1 / 3

This problem is harder ( the algebra ) than those previously done.

If a and b are constant and is a parameter, show that the family of curves

(1)

satisfies an equation, free of , that is unaltered when y’ is

replaced by (-1 / y’ ) . This means that the family is self - orthogonal.

Hint: Differentiate and use the result to eliminate from

the original equation (1). After clearing of fractions you will have

So,

Substitute (10) into (5).

Ch. 3 Pg. 195

Ch 3- Sec 2- Pro 12 Page 2 / 3