4. 2 The Polar Equation of the Conic Sections
7. Mathematical Modelling with Analytic Tools
7.1 Orthogonal Trajectories
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Consider the ellipses in the plane determined by the
x2C3 y2=c equation.
By changing the value of the c parameter located in the equation we can get a family of ellipses which is called a class of ellipses. Look for the equation of the class of curves that contains one free parameter in the plane of the ellipses and the arbitrary curve of which perpendicularly crosses any of the ellipses. In other words, the orthogonal trajectory of the class of ellipses with one parameter is to be determined.
Draw the ellipses and their orthogonal trajectories in the same coordinate system.
It is not easy to imagine two classes each of which contains infinite amount of curves and the elements of which perpendicularly crosses each other in pairs. This means that if we consider an arbitrary element of a class of curves then it is crossed perpendicularly by many infinite elements of the other class of curves. This is horrible but we do hope that by the end of the worksheet we will be more enthusiastic about this.
For the better understanding of orthogonal trajectories let’s see an easier task. Examine the class of lines determined by the y=mx equation. The elements of the class of lines are lines which go through the origin with different slopes. It is obvious that every concentric circle with a [képlet]
origin centre perpendicularly crosses any of these lines because it is known that the tangent of the circle is perpendicular to the line which connects the point of tangency with the centre of the circle. According to this, the orthogonal trajectory of the y=mx one-parameter class of lines is the one-parameter class of concentric circles that can be described by the [képlet] equation. Draw these in the same coordinate system to illustrate the two perpendicular classes of curves.
restart
e:=seq plot cos 1
10 m Pi t, sin 1
10 m Pi t,t=K5 ..5 ,m=K5 ..5 :
# egyenesek
k:=seq plot m cos t ,m sin t ,t= 0 ..2 Pi ,m= 0 ..5 : # körök plots[display]([e,k],scaling=constrained,title="Circles and their Ortogonal Trajectories");
(7.1.3) (7.1.3)
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(7.1.2) (7.1.2) (7.1.1) (7.1.1) K4 K2 0 2 4
K4 K2 2 4
Circles and their Ortogonal Trajectories
We put the graphs of the circles and lines in the e and k variables then had them displayed in the same coordinate system by the display procedure.
A question arises at this point: are there any other classes of curves with the same properties besides the circles? Similar uniqueness issues and the usage of the free parameters seem to refer to the fact that we are facing with differential equations.
We recommend the following method to solve the task. First, work out a differential equation the general solution of which is the initial class of ellipses. Then change the slope of the tangent to the slope of the perpendicular line in this differential equation and solve the equation generated. The class of curves, which serves as a general solution, provides the orthogonal trajectory of the class of ellipses. So let’s see it.
x2C3 y x 2=c2
x2C3 y x 2=c2 d
dx %
2 xC6 y x d
dx y x = 0
Notice that in the first command we used the y(x) instead of the y which denotes that the y is the function of the x. This technique is generally used when we work with differential equations. We wrote the equation of the ellipses and differentiated it by the x in the second command. As we can see, the system created the derivative of the function correctly. What would have been the
solution of the derivation if we had used the y in the first command? In this case, there would be 0 instead of the derivative of the y because the system would have considered the y as the c
constant.
The differential equation of the ellipses can be expressed from the equation above with a simple mathematical sorting of equation. We can express the derivative of the y from it.
d
dx y x =solve %, d dx y x
d
dx y x =K1 3
x y x
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(7.1.5) (7.1.5) (7.1.4) (7.1.4)
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According to this, the slope of the tangent of the ellipse is m=-x/3y at the [x,y] coordinate point and the slope of the perpendicular line is -1/m. Naturally the latter is the slope of the tangent of the perpendicular class of curves at the [x,y] point.
Now let’s write the differential equation of the perpendicular class of curves and solve it with the dsolve procedure. Its syntax is easy to remember because its parameters are similar to those of the solve procedure. The first parameter of the dsolve is the differential equation to be solved and its second parameter is the unknown y(x) function.
d
dx y x =K 1 rhs (7.1.3)
d
dx y x = 3 y x x dsolve (7.1.4),y x
y x =_C1 x3
We received the general solution of the differential equation. The solution is the one-parameter class of third-degree curves. This class of curves is the orthogonal trajectory of the class of ellipses mentioned in the task. We will get back to the illustration of the task later. Now let’s discuss the _C1 variable.
Maple is often forced to create new variables during runtime. Previously we became familiar with the _Z variable used in the RootOf notation. In our case it has to denote the constant multiplier appearing in the general solution of the differential equation. Since the system generates variable names that start with the underscore character (_) during runtime we should not start the name of our variables with this character because we can easily get confused. You are highly
recommended to avoid this situation.
The graphic display of the two classes of curves is ahead. Probably most of the readers would suggest that we should write procedures for the drawing of the ellipses and power functions. This would be an appropriate way but we know a better one. The DEtools package is the collection of such procedures. Leave the colon from the end of the following instruction to see the procedures of the DEtools.
with DEtools :
The DEplot procedure displays the solutions of the firs-order differential equation and the tangent field of the solutions in the same coordinate system. Obviously we should use this.
DEplot(diff(y(x), x) = -x/(3*y(x)), y(x), x = -2 .. 2, {[y(1)
=1/2], [y(1)=1], [y(1)=3/2], [y(1)=(-1)/2],[y(1)=-1], [y(1)=
(-3)/2]}, scaling=constrained, arrows=SMALL, thickness=2, linecolor=blue);
Warning, plot may be incomplete, the following errors(s) were issued:
cannot evaluate the solution further right of 1.3228757, probably a singularity
cannot evaluate the solution further left of -1.3228757, probably a singularity
Warning, plot may be incomplete, the following errors(s) were
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(7.1.9) (7.1.9)
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(7.1.10) (7.1.10)
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(7.1.7) (7.1.7)
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(7.1.6) (7.1.6)
(7.1.8) (7.1.8)
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issued:
cannot evaluate the solution further right of 1.3228757, probably a singularity
cannot evaluate the solution further left of -1.3228757, probably a singularity
x K2 K1 1 2
y x
K1.5 K1 K0.5 0.5 1 1.5
Well, although we should see a class of ellipses the graph is a bit crisscrossed. Why is the DEplot drawing such strange graphs? Maybe because the differential equation of (4) is singular. This means that the right side of the differential equation at the point of the x axis is meaningless. In this case the denominator is 0. The tangents of the ellipses at the common intersection with the x axis are parallel with the y axis thus their slopes do not exist in this case. Briefly, the y cannot be expressed as the function of the x at these points.
What is the solution in this case? Let’s approach the issue parametrically. Assume that both the x and y are the functions of the t variable.
x t 2C3 y t 2=c2
x t 2C3 y t 2=c2
We can get the differential equation of the class of curves by the derivation of the equation by the t. Notice that the singularity of the differential equation can be ceased if the derivative of the x is substituted with the y. In this case we can simplify the equation with the y.
de0:= d dt %
de0:= 2 x t d
dt x t C6 y t d
dt y t = 0 h1:= d
dt x t =y t
h1:= d
dt x t =y t subs h1,de0
2 x t y t C6 y t d
dt y t = 0 h2:= d
dt y t =solve %, d dt y t
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(7.1.10) (7.1.10)
(7.1.11) (7.1.11)
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(7.1.12) (7.1.12) h2:= d
dt y t =K1 3 x t
The h1 and h2 make a system of differential equation concerning the x(t) and y(t) unknown functions the solutions of which are the ellipses appearing in the task.
Use the DEplot procedure of the DEtools package to draw the solutions. But before this we have to give the initial values for which we would like to get the solutions.
e:= h1,h2
e:= d
dt x t =y t , d
dt y t =K1 3 x t k:= 0, 1, 1
2 , 0, 1, 1 , 0, 1, 3
2 , 0, 1, 2 , 0, 1, 5
2 , 0, 1, 3 k:= 0, 1, 1 , 0, 1, 2 , 0, 1, 3 , 0, 1, 1
2 , 0, 1, 3
2 , 0, 1, 5 2
R:=DEplot e, x t ,y t ,t= 0 ..15,k,stepsize= 0.1,scene= x,y ,x=K3 ..3,y=K2 ..2, title="Ellipses",scaling=constrained,thickness= 2,linecolor=blue,arrows=none : R
x
K3 K2 K1 0 1 2 3 y
K2 K1 1 2 Ellipses
The call sequence of the DEplot procedure is
DEplot diffegyenlet,változók,t_intervallum,kezdeti értékek, opciók , or
DEplot diffegyenlet,változók,t_intervallum,x_intervallum,y_intervallum, opciók . The differential equation can consist of a one, two or even more dimensional system or a single higher order differential equation. The variables contain the names of the unknown functions. The t_interval is the calculation interval of the independent variable. The initial values have to be determined by the type of the differential equation. In our case the set consists of three-element lists. The first, second and third elements of the list give the initial value of the t, the x and the y.
The x and y intervals of the representation are not needed to be given.
We can use the options of the plot. Maple uses the stepsize option for the numeric calculation of the integral curve. This option gives the step size of the independent variable. If it does not appear then the default is [képlet] in which case [a,b] is the interval of the t.
The scene=[x,y] or the scene=[x,y,t] option gives the plot domain of the solution. In the first case the system draws the curve in the x,y plane and t is the parameter. In the second case we get a 3-D
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(7.1.13) (7.1.13)
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curve. This option is useless in the case of a first-order equation. The thickness=n option determines the thickness of the line. The n can be 0,1,2 and 3. The default is 0.
Let’s get to know the procedures of the DEtools package better and draw the direction field of the ellipses with the help of the dfieldplot procedure.
m[1]:=dfieldplot([diff(x(t),t)=y(t), diff(y(t),t)=-1/3*x(t)], [x(t),y(t)], t=0..1, x=-3..3, y=-3..3, arrows=SLIM, scaling=
constrained):
m1
x K3 K2 1 2 3
y
K3 K2 K1 1 2 3
The parameters of the dfieldplot and the DEplot procedures are similar. The arrows option is responsible for the display of the arrows of the vector field. The following arrows can be given:
• thin
• slim
• thick
• line
• none
So far we have created two plot objects which we can display with the display procedure in the same coordinate system.
plots[display]({R,m[1]}, title = `Ellipszisek and fields`);
Notice how smoothly the ellipses osculate into the direction field. This last graph could have been created by the DEplot procedure if we had used the arrows option. However, in this case we could not make a difference between the line thickness of the ellipses and the arrows.
Let’s continue the illustration of the class of curves and deduce the parametric system of the differential equation of the perpendicular class of curves.
de1:=subs d
dt x t =K d
dt y t , d
dt y t = d
dt x t ,de0 de1:=K2 x t d
dt y t C6 y t d
dt x t = 0
As you can see we have received the differential equation of the orthogonal trajectory from the differential equation of the ellipse by having exchanged the coordinates of the tangent of the parametric plane curve and given an opposite sign to it. So we have executed the following substitutions:
(7.1.17) (7.1.17)
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(7.1.16) (7.1.16)
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(7.1.15) (7.1.15) (7.1.14) (7.1.14)
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d
dt y t = d
dt x t . Cease the singularity of the differential equation received.
h1:= d
dt y t =y t
h1:= d
dt y t =y t subs h1,de1
K2 x t y t C6 y t d
dt x t = 0 h2:=isolate %, d
dt x t
h2:= d
dt x t = 1 3 x t ortogonal:= h2,h1
ortogonal:= d
dt x t = 1
3 x t , d
dt y t =y t
We entered the h1 variable then substituted it into the differential equation of the orthogonal trajectory. Then we expressed the derivative of the x(t) from the equation and received the system of differential equation of the orthogonal trajectory provided by the [h2, h1] list.
First, let’s draw the vector field with the dfieldplot procedure.
m[2] := dfieldplot(ortogonal, [x(t), y(t)], t = 0 .. 1, x = -3 .. 3, y = -3 .. 3, arrows = SLIM, scaling = constrained, title = `Perpendicular Vector Field`):
m2
x K3 K2 1 2 3
y
K3 K2 K1 1 2 3
Perpendicular Vector Field
We can get an interesting graph if we display the two vector fields drawn with thin arrows in the same coordinate system. Notice that every arrow has its perpendicular counterpart.
plotsdisplay m2,m1
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x K3K2 1 2 3
y
K3 K2 K1 1 2 3
Perpendicular Vector Field
To draw the graphs of the solutions we have to give the initial values from which the solutions originate then display the perpendicular class of curves with the DEplot procedure.
k:= 0,K1, 1
8 , 0,K1, 1
4 , 0,K1, 1
2 , 0, 1, 1
2 , 0, 1, 1 , 0, 1, 2 , 0,K1, 1 , 0,K1, 2 , 0,K1, K1
8 , 0,K1, K1
4 , 0,K1, K1
2 , 0,K1,K1 , 0,K1,K2 , 0, 1, 1
8 , 0, 1, 1
4 , 0, 1, K1
8 , 0, 1, K1
4 , 0, 1, K1
2 , 0, 1,K1 , 0, 1,K2 :
Orto:=DEplot(ortogonal, [x(t),y(t)], -5..5, k, stepsize=.2, scene=[x, y], x=-3..3, y=-3..3, title= `Perpendicular Set of Curves`, scaling=constrained, thickness=2, linecolor=blue, arrows=none):
Orto
x K3 K2 1 2 3
y
K3 K2 K1 1 2 3
Perpendicular Set of Curves
The common display of the solutions and the vector field perfectly illustrates how smoothly the class of curves osculates into its direction field.
DEplot(ortogonal, [x(t),y(t)], -5..5, k, stepsize=.2, scene=
[x, y], x=-3..3, y=-3..3, title= "Ortogonális Trajectories and Vector Fields", scaling=constrained, thickness=2,
linecolor=blue);
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x K3K2 1 2 3
y
K3 K2 K1 1 2 3
Ortogonális Trajectories and Vector Fields
Let’s dot the i’s and cross the t’s. Display the perpendicular class of curves of the ellipses and orthogonal trajectories in the same coordinate system.
plots[display]({R, Orto}, title=`Ellipses with Perpendicular Curves`);
x K3 K2 1 2 3
y
K3 K2 K1 1 2 3
Ellipses with Perpendicular Curves
We have solved the task. We have become familiar with the perpendicular class of curves of the ellipses by having used the DEtools package. .
What Have You Learnt About Maple?
We can differentiate expressions and equations with the diff procedure. Its syntax is d
dx egyenlet, In the case of 1-D Math the input is diff(egyenlet,x).
The dsolve procedure is used for the solution of the differential equations. Its syntax is [képlet].
We get the solution in the form of an equality if there is an explicit or implicit solution to the y (x).
The DEplot procedure draws solutions that belong to certain initial values. The DEplot can be applied for an arbitrary system of differential equation. Its syntax is
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DEplot differenciálegyenletrendszer,változók,t=t0..t1,kezdeti értékek,x=a..b,y=c..d, további opciók .
The dfieldplot procedure gives the drawing of the direction field of the 1-D or 2-D differential equations. Its syntax is
dfieldplot f x,y ,g x,y , x,y ,a..b,x=c..d,y=e..f,arrows=a nyilak tipusa .
Exercises
1. Find the orthogonal trajectories of the following class of curves and draw them in the same coordinate system with the class of curves.
y=m x x2Cy2= 2 a x y=a x2; x2Cy2= 2 a y y2C2 a x= 0; x2Qy2=a2
x y=c; y=c eQx
2. Solve the parametric system of differential equation of the class of ellipses. Prove that the solution satisfies the x2C3 y2=c2 equation.
3. Solve the parametric system of differential equation called orthogonal of the orthogonal trajectories. (see (17)) Prove that the solution received satisfies the y=c x3 equation.
7.2 Catenaries
Consider a perfectly flexible string or a chain with uniformly distributed load which we suspend by their ends.
a) Determine the function which describes the equilibrium figure of the string if the string has to bear only its own weight.
b) Determine the arc length above the [0,x] interval of the string as the function of the x.
This issue perfectly illustrates the relationship between the continuous and discrete modelling. On the one hand, the string can be considered as a solid with continuously distributed load. On the other hand, it can be considered as a curve consisting of particles in which case the connecting parts can only affect each other through the adjacent elements. A small part of the string transfers the pulling force affecting along the tangent of the curve to the adjacent part which the string resists thus does not stretch. However, the parts are unable to resist the cross directional force, the twisting force and the thrust. The figure of the string changes when it is affected by these forces.
The equilibrium figure of the string can be seen below. This curve is called the catenary.
(7.2.3) (7.2.3) (7.2.2) (7.2.2)
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(7.2.5) (7.2.5)
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(7.2.4) (7.2.4)
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(7.2.1) (7.2.1)
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In the figure the deepest point of the sag was denoted with the A and it was entered on the y axis from a 0<a distance from the x axis. The coordinates of another, arbitrary P point of the string was denoted with [x,y] and the length of the AP string part was denoted with s. If the w is the weight of a unit long part of the string then the weight of the AP arc of the string is ws. Three different forces are applied on the AP arc:
• the ws weight force is applied vertically downwards on the centre of the AP arc of the string
• the H pulling force in the A point, which is transferred by the string parts because of the left side suspension, points towards the negative direction of the x axis
• the P point is the striking point of the T holding force and its direction is the tangent of the string which forms a phi angle with the x axis.
Let’s start the solution of the task by writing the assumption of the balance of force for the x and y directional components separately.
restart
T cos phi =H
T cos φ =H T sin phi =w s
T sin φ =w s
After we express the T from the (1) and substitute it to the (2) we can easily determine the slope of the curve we are looking for.
isolate T cos phi =H,T
T= H cos φ subs %,T sin phi =w s
H sin φ
cos φ =w s
% H
sin φ
cos φ = w s H
(7.2.9) meredekség:=studentpowsubs w
H = 1
a ,(7.2.5) meredekség:= sin φ
cos φ = s a For the sake of clarity, the a=H/w notation has been introduced.
So we have found a relation between the slope of the function we are looking for and the s arc length. But it is known that the slope, the substitution value of the differential and the arc length can be expressed by the well-known formula
0 x
1C v
vx y 2 dt
If we do these substitutions in the equalities received for the slope then we get the differential equation concerning the y=y x function.
ivhossz:=s=
We have received a first-order differential equation on the right side of which there is an integral.
To get rid of this, let’s differentiate both sides of the first-order differential equation by the x thus we get a second-order, nonlinear differential equation which is partial in the x and can be solved by the dsolve procedure.
d
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To solve the task we have to determine the value of the _C1 and _C2 free parameters appearing in the general solution of the differential equation.
The value of the derivative of the function is 0 at the x=0 because the tangent of the curve is parallel with the x axis in the A point.
d
Notice that the diff procedure disregarded the fact that it had received an inequality as a first operand. It did the derivation on both sides of the equality. After this we substituted zero into the place of the derivative of the x and the function and solved the equality for _C1.
Let’s substitute the _C1=0 value to the “solution” and simplify the expression.
megoldás:=subs C1,megoldás
megoldás:=y x =a cosh x
a C_C2
Without any constraints on the generalities we can assume that the y(0)=a will be true. This can be achieved by shifting the coordinate system. In this way we get the zero value for the _C2 which we can substitute into the solution and the get the function describing the catenary.
subs x= 0,y 0 =a,megoldás
> Let’s draw the catenary by the a=1 parameter value.
plot subs a= 1,rhs láncgörbe ,x=K2 ..2,y= 0 ..3,scaling=constrained
Here is the exponential syntax of the catenary for the sake of those who are interested. This syntax is provided by the exp option of the convert procedure.
lhs láncgörbe =convert rhs láncgörbe , exp y x =a 1
The solution of the b part of the task, that is, the determination of the arc length of the catenary is still ahead. Naturally it is as easy as pie. We have the formula with which we can calculate the arc length so we only have to substitute the indefinite y(t) located in it with the function that describes
The solution of the b part of the task, that is, the determination of the arc length of the catenary is still ahead. Naturally it is as easy as pie. We have the formula with which we can calculate the arc length so we only have to substitute the indefinite y(t) located in it with the function that describes