Revisiting Analytic Geometry (1)

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Analytic geometry is essentially the application of algebra to geometry, whereby we are able to reference any number of two dimensional curves, e.g. circle, ellipse, parabola, and hyperbola, in terms of algebraic equations.  It is first important to see how each of these curves is formed, by taking 2 dimensional cuts of conic sections.   These curves are then recognized as distinct equations.

Once we generate the curve F(x,y) = 0,  we can then study its peculiar geometric properties (for example those for a parabola differ from those for an ellipse) as well as find the equation of a curve once its geometric properties are known

 The basic conic sections (2nd degree curves)  are shown in the diagram above. As seen there, the circle is formed from a straight cut in one plane, the ellipse is formed from a cut at an oblique angle with the plane now at some angle – say Θ  - from the horizontal.  The parabola represents an even greater oblique cut and which lacks any end closure. I.e. it is open at one end.  The hyperbola carries this further, with the cut going through the conic section on one side only – so as to form two different curves facing each other.

   Probably the most straightforward equation and curve to begin examining is the circle. Let us consider one defined by the equation:

x ²    +    y ²  =  4

This is very simple to plot as a graph and one ends up with the figure shown below:

This is actually an equation of basic form: 

x ²    +    y ²  =      r ² 

where r denotes the radius. Since the square root of  4  =  r²   is 2, then we see from the graph the radius is 2.  But this simple example is deceiving because it may give the impression that all equations of the circle consist of three terms, and all circles are centered at the origin (0, 0). In fact, most interesting cases are circles which are off -center, and also of more complicated equations.


Consider, for example, the equation:

x ²    +    y ²   + 2x   - 4y  =  11

Is this a circle? Yes, it is, but we must work to get it into the correct form to analyze, or graph.
First, write the left side as:

x ²    +  2x  +  y ²   - 4y 

Now complete the square for the expression in x and y to get:

(x + 1) ²    + (y - 2) ²   =  x ²    +  2x  + 1  +    y²   - 4y    + 4

Re-configure the entire equation paying attention to terms added to both sides:

(x ²    +  2x  + 1)   +    (y²   - 4y    + 4)   = 


11 +   4   +   1 =   16


And if we now write in the most general form using coordinates (h, k) for the center:
(x - h)²    + (y - k ) ²   =  r ² 

We see at a glance, that h = -1 and k =2 with the radius r = the square root of 16 or 4.  We can check this by doing the graph:



Inspection of the above graph shows that indeed:   (h, k) = (-1, 2), the coordinates of the center and the radius is 4 units. There is also another, longer way to approach the problem using a general analytic equation, viz.


Ax ²  +    Bxy +   Cy ²   + Dx  + Ey  + F   = 0


This can be reduced to the earlier equation:

 (x - h) ²    + (y - k ) ²   =  r ²    

By completing the square, which is left as an exercise.

The general, analytical expression for the radius is then obtained:

  r ²     = (D ²   +    E ²      - 4AF) /  4 A ²    

We check to see that this works for the example problem:

x ²    +    y²   + 2x   - 4y  =  11

Rewritten as:

x ²    +    y²   + 2x   - 4y  -  11  = 0

So:  A = 1,  D = 2,   E = -4  and F = -11

Then:

r ²     = [2 ²   +    (-4) ²    - 4 (1)(-11))]  /  4  ²  

r ²     =  [4 + 16  + 44]/   4   =   64/  4   =   16

And since:  r ²     =  16 then r = 4  which  checks out.  The coordinate of the center will also be found to be:


h =  (-D/ 2A)   and k =  (-E/ 2A)

Again, checking this for the example:

h =   -2/ 2(1) =   -1   and k  =  -(-4)/ 2 =  4/2 = 2


so (h, k ) =   (-1, 2)

which again checks.


We conclude that the circle with center at some graphical coordinate (h, k) can be expressed:

(x - h)²    + (y - k ) ²   =  r² 

The parabola -specific equation:   x ²  =    4py 

Is obtained by using:

A = 1,  E = - 4p, and B = C = D = F = 0

Note that the terms:   Ax² ,      Bxy ,     Cy²

are the second degree or quadratic terms with  Bxy the "cross product" term.


Most problems to do with this term involve eliminating it using a rotation of Cartesian axes such that:



(x)
(y)  =

(cos Θ.   -    sin  Θ) (x')
(sin Θ.    +  cos Θ)  .(y')


Performing the matrix operation for rotation:

x = x' cos Θ   -    y' sin  Θ

y = x' sin Θ.   +  y'  cos Θ

Example: What if  Θ  =  45 degrees?

Then:  cos (45)  = sin (45)    =  1 / Ö 2

Then:  x =  x' / Ö 2   -   y' / Ö 2  =    (x'   -   y') / Ö 2

y  =   x' / Ö 2   +  y' / Ö 2  =   ( x'   +   y' ) / Ö 2  

Now, if  we apply the rotation of axes equation to the general analytic equation for 2nd degree curves what do we get? It can be shown we have:

A'x' ²  +    B' x' y' +   C' y' ²   + D' x'  + E'y'  + F'    = 0

These have the following relations to the coefficients:

A'  =   A cos² Θ  +   B cos Θ sin  Θ  + C sin ² Θ

B'    = B (cos² Θ  - sin ² Θ) + 2 (C - A) sin  Θ cos Θ  


C =   A sin ² Θ  -   B  sin  Θ cos Θ +   C cos² Θ    



D'   =  D cos Θ + E sin  Θ  

 E' =   - D  sin  Θ   + E'  cos Θ

 F'    = F     

The technique for getting rid of the cross product terms is pretty straightforward, given an angle or rotation  Θ  can always be found such that the new cross product term is eliminated.  How to do this? Simply set  B' = 0 in the 2nd equation in the set above and solve for the angle Θ. Of course, it helps to have at hand a couple of trigonometric identities:

cos² Θ  -    sin ² Θ    =    sin 2Θ 

And:

2  sin  Θ cos Θ      = sin 2Θ

So that:

B' =  B cos 2Θ  +   (C - A) sin 2Θ     

So B' will vanish if we choose:

cot 2Θ  =   (A  - C) / B

Example : Eliminate the cross product term in the equation:

x ²  +   xy  +  y²    =  3

And thereby identify the 2nd degree curve and graph it.


Solution:

The given equation has: A = B = C = 1

Choose Θ  according to:

 cot 2Θ  =   (A  - C)/ B  =   (1 - 1)/ 1 = 0/1 = 0

Then:  2Θ  =  90 deg     so: Θ  =  45 deg    

Then:  x  =   (x'   -   y')  / Ö 2       

And:   y = (x'   +   y' ) / Ö 2  
  
Now write:

3(x') ²  +  ( y') ²    =  6

Divide through by 6:

(x') ² / 2   +  ( y') ²  / 6  =  1

which is an ellipse with foci on the y' - axis.

Problems for Math Mavens:

1)Find the coordinates of the center of each for the following circles and the radius r. Sketch each of the circles

a)  x ²    +    y ²  - 2 y   =  3

b) 2x ²    +  2  y ²   + x  + y = 0

c) x ²    +    y²  + 2x  = 8

2) Use the general analytic equation to write  the specific equation for the ellipse  with:

A = 9,  B= 0, C = 4,  D= 36,  E =  -8 and F = 4

And sketch the resulting ellipse.

3) Determine the equation to which:

x²  +   xy  +  y²    =  1

reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve. 

Analytic Geometry Conclusion: Generalizing 2nd Degree Curves

                                                    

                                    Fig. 1.Conic sections generated in analytic geometry

While we have examined separately the circle, ellipse, parabola and hyperbola it should also be noted that all are second degree curves - what we call "conic sections" - which can be portrayed as special cases of a general analytic equation, viz.

Ax²  +    Bxy +   Cy²   + Dx  + Ey  + F   = 0

For example, the circle template equation with center at (h, k)  e.g.

(x - h)²    + (y - k ) ²   =  r² 

Can be extracted from the general analytic equation by taking:

A = C = 1     B = 0,   D = -2h

E = - 2k,  F =  h²    +  k ²   -    r² 

The parabola specific equation:   x ²  =    4py 

Is obtained by using:

A = 1,  E = - 4p, and B = C = D = F = 0

Note that the terms:   Ax² ,      Bxy ,     Cy²

are the second degree or quadratic terms with  Bxy the "cross product" term

Most problems to do with this term involve eliminating the cross product using a rotation of Cartesian axes such that:

(x)
(y)  =

(cos Θ.   -    sin  Θ) (x')
(sin Θ.    +  cos Θ)  .(y')

Performing the matrix operation for rotation:

x = x' cos Θ   -    y' sin  Θ

y = x' sin Θ.   +  y'  cos Θ

What if  Θ  =  45 degrees?

Then  cos (45)  = sin (45)    =  1 / Ö 2


Then:  x = x' / Ö 2   -   y' / Ö 2  =    (x'   -   y') / Ö 2

y  = x' / Ö 2   -+  y' / Ö 2  =   ( x'   +   y' ) / Ö 2  


Now, if  we apply the rotation of axes equation to the general analytic equation for 2nd degree curves what do we get? It can be shown we have:

A'x' ²  +    B' x' y' +   C' y' ²   + D' x'  + E'y'  + F'    = 0


With the following relations to the original coefficients:

A'  =   A cos² Θ  +   B cos Θ sin  Θ  + C sin ² Θ

B'    = B (cos² Θ  - sin ² Θ) + 2 (C - A) sin  Θ cos Θ  

C =   A sin ² Θ  -   B  sin  Θ cos Θ +   C cos² Θ   

D'   =  D cos Θ + E sin  Θ  

 E' =   - D  sin  Θ   + E'  cos Θ

 F'    = F     

The technique for getting rid of the cross product terms is pretty straightforward, given an angle or rotation  Θ  can always be found such that the new cross product term is eliminated.  How to do this? Simply set  B' = 0 in the 2nd equation in the set above and solve for the angle Θ. Of course, it helps to have at hand a couple of trig identities:
 
cos² Θ  -    sin ² Θ    =    sin 2Θ 


And:

2  sin  Θ cos Θ      = sin 2Θ

So that, for example:

B' =  B cos 2Θ  +   (C - A) sin 2Θ     


So B' will vanish if we choose:

cot 2Θ  =   (A  - C)/ B


Example Problem: Eliminate the cross product term in the equation:

x²  +   xy  +  y²    =  3

And thereby identify the 2nd degree curve and graph  it.

Solution:

The given equation has: A = B = C = 1

Choose Θ  according to: cot 2Θ  =   (A  - C)/ B  =   (1 - 1)/ 1 = 0/1 = 0

Then:  2Θ  =  90 deg     so: Θ  =  45 deg    

Then:  x  =   (x'   -   y')  / Ö 2       

y = (x'   +   y' ) / Ö 2  
 
And:

3(x') ²  +  ( y') ²    =  6

Divide through by 6:

(x') ² / 2   +  ( y') ²  / 6  =  1

which is an ellipse with foci on the y' - axis. This is sketched below:

                           
                                   Fig. 2 - Ellipse for reference

Problems:


1)  Use the general analytic equation to write specific equations for the ellipse and hyperbola. Sketch the resulting curves to confirm your answers.


2) The discriminant:  B²   -  4AC     =   B'²   -  4A'C'     

has been found valid for any rotation of axes (i.e. for any angle Θ )  Thus, the quantity is invariant under a rotation of axes.

a) Show that for the parabola the discriminant:  B²   -  4AC    =  0

b) Show that for the ellipse:   B²   -  4AC    <  0

c) Show that for the hyperbola:   B²   -  4AC    >  0

3) Determine the equation to which:

x²  +   xy  +  y²    =  1

reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve.

4) Use the discriminant to classify each of the following curves as a circle, parabola, ellipse, or parabola:

a) x²  +   y²    +   xy  + x  -   y    =  3

b) x²  -    y²    =  1

c)   x²  +  3 y²    -  3 xy   -  6 y    =  7

d) x²  +   y²    +  3 x  -  2 y    =  3

Exactly What Is 'Escape Velocity'?

One of the frequent questions asked in general physics, mechanics and astrodynamics as well as general astronomy is to do with escape velocity, i.e. 'What is escape velocity?'  The answers given by various profs, lecturers, physics teachers can range from the purely tautological, i.e. "It is the velocity needed to escape from the Earth's gravity, or 11.2 kilometers per second", to the convoluted - involving the potential for a small, very slow moving object to gradually thrust its way to the critical speed. (With no thought to the fact that something like a "toy helicopter" would disintegrate long before achieving such speed, even it experienced prolonged acceleration over time, e.g. such as 1 m/s/s).

Man of the convoluted answers come in response to apparently innocent (but untutored questions) such as:

1) Couldn't an object of any mass move "free" of Earth's gravity ( at least free of falling back to Earth) whether it is moving away from Earth at 1 mph or 20,000 mph?  

2) Does escape velocity refer to a velocity that must be reached in order to "coast" out of gravity's effective reach?

3)Or is the escape velocity specific orbiting velocity?

In the case of questions (1) and (2) the person has no clue that the force of gravity, say arising from the Earth, doesn't just suddenly cease at a specific boundary. To appreciate this one can make use of the Newtonian force equation for (gravitational attraction) and set it equal to the weight (product of mass m by acceleration of gravity) at Earth's (mass M_E) surface, whence.

mg = G M_E m/ r²

 Then, solving for g, the gravitational acceleration:

g = GM_E/r²

The important aspect of this equation is the inverse square relation of g to the distance, r. One can thereby see that even at an enormous distance (say r = 10 million km) there will still be a minuscule g.  More important, this concept leads the way to an even more fundamental one that bears directly on the escape velocity. We call this the gravitational potential V.

We can pursue this manner of definition, and at the same time disallow any definitions that make use of objects which may start out at say, 1 m/s and experience continual thrust (acceleration) to finally reach the escape velocity. While this is certainly possible it is not regarded as a practical modus operandi for attaining v_esc economically and hence doesn't meet the criteria for acceptable  definition. (For example, starting out with such a low velocity and only allowing a = 1 m/s²     would take  11,200 secs or just over three hours to attain escape velocity - which is absurd.)
 
 Indeed, if anyone proposed this in an advanced celestial mechanics or astrodynamics class he'd likely be laughed out of the lecture room. 

This elicits the question: 'What does?'

Perhaps the best definition I have found which meets the criteria is the one from the Oxford Dictionary of Physics - which I reproduce below:

"The minimum speed needed by a space vehicle, rocket etc. to escape from the gravitational field of the moon, Earth or other celestial body. The gravitational force between a rocket of mass m and a celestial body of mass M is:

F = GMm/ r ²

Therefore, the gravitational potential energy of the rocket with respect to its possible position very far from the celestial body (on which it is resting) can be shown to be:

V = - GMm/ r

assuming by convention the potential energy is zero at an infinite distance from the celestial body.

If the rocket is to escape the gravitational field it must have a kinetic energy (K) that exceeds this potential energy, i.e.

mv_esc²/ r   must be greater than MmG/r

or v_esc >   [2MG/ r]^1/2


This is the value of the escape velocity"

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In the case of  v_esc with respect to Earth, we can work it out. The Earth's radius is 6396 km or call it 6.4 x 10⁶ m.

G, the Newtonian gravitational constant is G = 6.7 x 10^-11 Nm²/ kg²

The mass of the Earth is 6 x 10²⁴ kg

Then:

[2MG/ r]^1/2  =  [2 (6 x 10²⁴ kg) (6.7 x 10^-11 Nm²/ kg²)/ 6.4 x 10⁶ m ]^1/2


This comes out to v_esc = 11,200 m/s or 11.2 km/s


 We can frame this in terms of conservation of energy also, such that the kinetic (K)  plus potential energy (V)  for the gravitating system must equal zero.

Or:

0 = K + V  = mv² / r +  (-GMm/r) = mv² / r -  GMm / r

With a glance at the attached image one can see a plot with V, the gravitational  potential energy (GPE) vs. r.

As shown here, the zero of GPE is taken at infinity, i.e. V = 0 at r = oo. The dotted line applies to an object - say satellite- in a bound orbit, so that we see in any such case E < 0 which means the object is in a gravitational potential "well". All objects. i.e. planets or moons, bound to a more massive object in some orbit are said to be in a gravitational potential "well" and their value of V is negative.

By contrast, if the dotted line were situated above the zero line we would have an unbound or 'hyperbolic' orbit. This is only an orbit in the loosest sense, in that it fulfills one of the Keplerian conic sections most students learn in freshman calculus, i.e.
.

Also, the escape velocity is not the specific orbiting velocity since the energy needed to attain the former is greater than the latter. In general, for a body-object much less massive than say the Earth (m << M) we have the mean orbital speed v(0)

=  v / [2]^1/2

I.e. lower than escape velocity by a factor of root 2.

Most interested readers should be able to figure out from the Oxford definition and also the GPE curve provided  why the escape velocity cannot be less  than what it is, i.e. for the Earth! (Hint: Apply the conservation of energy)

Suggested insight problems:

1) Use the principle of conservation of energy to find the escape velocity for the planet Mars. Take Mars' mass M = 6.4 x 10²³ kg  and its radius R = 3390 km.

2) Mars' moon Deimos takes 30.3 hours to make one orbit. Using Kepler's 3rd law of planetary motion work out its mean distance (a) and thereby estimate the value of the gravitational potential V, of Deimos relative to Mars.

3) The condition for a gas molecule of mass m to escape from a planet of mass M and radius r is given as:

 6 [3kT / m]^1/2  >  [2MG/ r]^1/2

Where k is the Boltzmann constant, and T is the surface temperature in degrees Kelvin. Use the above condition to ascertain the extent to which primitive oxygen on Mars could have escaped. Take T = 300 K and use the Martian planetary properties in (1).

Note: calculations show that if the mean molecular speed is as much as 1/3 the velocity of escape then the planet will lose up to half the gas from its atmosphere (of that composition) in a few weeks. If the molecular speed is even one -fifth of  v_esc   the gas will disperse into space in a few hundred million years. The ability of fractions of gas to escape is directly a result of the Maxwell-Boltzmann distribution of gas velocities. Can you explain?