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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
Page 2 / 4
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
Page 2 / 7
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
Page 3/ 7
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
Page 6 / 7
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
Page 7/ 7
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
Page 2 / 6
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
Page 3 / 6
( 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
Page 4/ 6
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
Page 6/ 6
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