PARAMETRIC EQUATIONS AND POLAR …rfrith.uaa.alaska.edu/M201/Chapter10/Chap10_Sec6.pdfPARAMETRIC...
Transcript of PARAMETRIC EQUATIONS AND POLAR …rfrith.uaa.alaska.edu/M201/Chapter10/Chap10_Sec6.pdfPARAMETRIC...
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PARAMETRIC EQUATIONS AND POLAR COORDINATES
10
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PARAMETRIC EQUATIONS & POLAR COORDINATES
In Section 10.5, we defined the parabola in terms of a focus and directrix.
However, we defined the ellipse and hyperbola in terms of two foci.
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10.6Conic Sections
in Polar Coordinates
In this section, we will:
Define the parabola, ellipse, and hyperbola
all in terms of a focus and directrix.
PARAMETRIC EQUATIONS & POLAR COORDINATES
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CONIC SECTIONS IN POLAR COORDINATES
If we place the focus at the origin, then a conic section has a simple polar equation.
This provides a convenient description of the motion of planets, satellites, and comets.
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CONIC SECTIONS
Let F be a fixed point (called the focus) and lbe a fixed line (called the directrix) in a plane.
Let e be a fixed positive number (called the eccentricity.)
Theorem 1
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CONIC SECTIONS
The set of all points P in the plane such that
(that is, the ratio of the distance from Fto the distance from l is the constant e) is a conic section.
PFe
Pl=
Theorem 1
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CONIC SECTIONS
The conic is:
a) an ellipse if e < 1.
b) a parabola if e = 1.
c) a hyperbola if e > 1.
Theorem 1
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CONIC SECTIONS
Notice that if the eccentricity is e = 1, then
|PF| = |Pl|
Hence, the given condition simply becomes the definition of a parabola as given in Section 10.5
Proof
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CONIC SECTIONS
Let us place the focus F at the origin and the directrix parallel to the y-axis and d units to the right.
Proof
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CONIC SECTIONS
Thus, the directrix has equation x = dand is perpendicular to the polar axis.
Proof
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CONIC SECTIONS
If the point P has polar coordinates (r, θ), we see that:
|PF| = r|Pl| = d – r cos θ
Proof
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CONIC SECTIONS
Thus, the condition |PF| / |Pl| = e, or |PF| = e|Pl|, becomes:
r = e(d – r cos θ)
Proof—Equation 2
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CONIC SECTIONS
Squaring both sides of the polar equation and convert to rectangular coordinates, we get
x2 + y2 = e2(d – x)2 = e2(d2 – 2dx + x2)
or (1 – e2)x2 + 2de2x + y2 = e2d2
Proof
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CONIC SECTIONS
After completing the square, we have:
22 2 2 2
2 2 2 21 1 (1 )e d y e dx
e e e
+ + = − − −
Proof—Equation 3
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CONIC SECTIONS
If e < 1, we recognize Equation 3 as the equation of an ellipse.
Proof
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CONIC SECTIONS
In fact, it is of the form
where
2 2
2 2
( ) 1x h ya b−
+ =
2 2 2 2 22 2
2 2 2 21 (1 ) 1e d e d e dh a b
e e e= − = =
− − −
Proof—Equations 4
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CONIC SECTIONS
In Section 10.5, we found that the foci of an ellipse are at a distance c from the center, where
4 22 2 2
2 2(1 )e dc a b
e= − =
−
Proof—Equation 5
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CONIC SECTIONS
This shows that
Also, it confirms that the focus as defined in Theorem 1 means the same as the focus defined in Section 10.5
2
21e dc h
e= = −
−
Proof
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CONIC SECTIONS
It also follows from Equations 4 and 5 that the eccentricity is given by:
If e > 1, then 1 – e2 < 0 and we see that Equation 3 represents a hyperbola.
cea
=
Proof
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CONIC SECTIONS
Just as we did before, we could rewrite Equation 3 in the form
Thus, we see that:
2 2
2 2
( ) 1x h ya b−
− =
2 2 2wherece c a ba
= = +
Proof
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CONIC SECTIONS
By solving Equation 2 for r, we see that the polar equation of the conic shown here can be written as:
1 cosedre θ
=+
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CONIC SECTIONS
If the directrix is chosen to be to the left of the focus as x = –d, or parallel to the polar axis as y = ±d, then the polar equation of the conic is given by the following theorem.
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CONIC SECTIONS
A polar equation of the form
represents a conic section with eccentricity e.
The conic is: An ellipse if e < 1. A parabola if e = 1. A hyperbola if e > 1.
or1 cos 1 sin
ed edr re eθ θ
= =± ±
Theorem 6
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CONIC SECTIONS.
Theorem 6
The theorem is illustrated here.
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CONIC SECTIONS
Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line y = –6.
Example 1
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CONIC SECTIONS
Using Theorem 6 with e = 1 and d = 6, and using part (d) of the earlier figure, we see that the equation of the parabola is:
Example 1
61 sin
rθ
=−
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CONIC SECTIONS
A conic is given by the polar equation
Find the eccentricity. Identify the conic. Locate the directrix. Sketch the conic.
Example 2
103 2cos
rθ
=−
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CONIC SECTIONS
Dividing numerator and denominator by 3, we write the equation as:
From Theorem 6, we see that this represents an ellipse with e = 2/3.
103
231 cos
rθ
=−
Example 2
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CONIC SECTIONS
Since ed = 10/3, we have:
So, the directrix has Cartesian equation x = –5.
10 103 3
23
5de
= = =
Example 2
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CONIC SECTIONS
When θ = 0, r = 10.
When θ = π, r = 2.
Thus, the vertices have polar coordinates (10, 0) and (2, π).
Example 2
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CONIC SECTIONS
The ellipse is sketched here. Example 2
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CONIC SECTIONS
Sketch the conic
Writing the equation in the form
we see that the eccentricity is e = 2.
Thus, the equation represents a hyperbola.
Example 312
2 4sinr
θ=
+
61 2sin
rθ
=+
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CONIC SECTIONS
Since ed = 6, d = 3 and the directrix has equation y = 3.
The vertices occur when θ = π/2 and 3π/2; so, they are (2, π/2) and (–6, 3π/2) = (6, π/2).
Example 3
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CONIC SECTIONS
It is also useful to plot the x-intercepts.
These occur when θ = 0, π.
In both cases, r = 6.
Example 3
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CONIC SECTIONS
For additional accuracy, we could draw the asymptotes.
Note that r → ±∞ when 1 + 2 sin θ → 0+ or 0-
and 1 + 2 sin θ = 0 when sin θ = –½.
Thus, the asymptotes are parallel to the rays θ = 7π/6 and θ = 11π/6.
Example 3
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CONIC SECTIONS
The hyperbola is sketched here. Example 3
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CONIC SECTIONS
When rotating conic sections, we find it much more convenient to use polar equations than Cartesian equations.
We use the fact (Exercise 77 in Section 10.3) that the graph of r = f(θ – α) is the graph of r = f(θ) rotated counterclockwise about the origin through an angle α.
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CONIC SECTIONS
If the ellipse of Example 2 is rotated through an angle π/4 about the origin, find a polar equation and graph the resulting ellipse.
Example 4
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CONIC SECTIONS
We get the equation of the rotated ellipse by replacing θ with θ – π/4 in the equation given in Example 2.
So, the new equation is:10
3 2cos( / 4)r
θ π=
− −
Example 4
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CONIC SECTIONS
We use the equation to graph the rotated ellipse here.
Notice that the ellipse has been rotated about its left focus.
Example 4
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EFFECT OF ECCENTRICITY
Here, we use a computer to sketch a number of conics to demonstrate the effect of varying the eccentricity e.
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EFFECT OF ECCENTRICITY
Notice that:
When e is close to 0, the ellipse is nearly circular. As e → 1-, it becomes more elongated.
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EFFECT OF ECCENTRICITY
When e = 1, of course, the conic is a parabola.
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KEPLER’S LAWS
In 1609, the German mathematician and astronomer Johannes Kepler, on the basis of huge amounts of astronomical data, published the following three laws of planetary motion.
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KEPLER’S FIRST LAW
A planet revolves around the sun in an elliptical orbit with the sun at one focus.
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KEPLER’S SECOND LAW
The line joining the sun to a planet sweeps out equal areas in equal times.
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KEPLER’S THIRD LAW
The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
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KEPLER’S LAWS
Kepler formulated his laws in terms of the motion of planets around the sun.
However, they apply equally well to the motion of moons, comets, satellites, and other bodies that orbit subject to a single gravitational force.
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KEPLER’S LAWS
In Section 13.4, we will show how to deduce Kepler’s Laws from Newton’s Laws.
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KEPLER’S LAWS
Here, we use Kepler’s First Law, together with the polar equation of an ellipse, to calculate quantities of interest in astronomy.
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KEPLER’S LAWS
For purposes of astronomical calculations, it’s useful to express the equation of an ellipse in terms of its eccentricity e and its semimajor axis a.
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KEPLER’S LAWS
We can write the distance from the focus to the directrix in terms of a if we use
Equation 4:
So, ed = a(1 – e2).
2 2 2 2 22 2
2 2 2
2
(1 )(1 )
(1 )
e d a ea de e
a ede
−= ⇒ =
−
−⇒ =
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KEPLER’S LAWS
If the directrix is x = a, then the polar equation is:
2(1 )1 cos 1 cos
ed a ere eθ θ
−= =
+ +
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KEPLER’S LAWS
The polar equation of an ellipse with focus at the origin, semimajor axis a, eccentricity e, and directrix x = d can be written in the form
Equation 7
2(1 )1 cosa er
e θ−
=+
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PERIHELION AND APHELION
The positions of a planet that are closest to and farthest from the sun are called its perihelion and aphelion, respectively, and correspond to the vertices of the ellipse.
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PERIHELION AND APHELION DISTANCES
The distances from the sun to the perihelion and aphelion are called the perihelion distance and aphelion distance, respectively.
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PERIHELION AND APHELION
In this figure, the sun is at the focus F.
So, at perihelion, we have θ = 0 and, from Equation 7,
2(1 )1 cos 0
(1 )(1 )1
(1 )
a ere
a e ee
a e
−=
+− +
=+
= −
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PERIHELION AND APHELION
Similarly, at aphelion, θ = π and r = a(1 + e)
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PERIHELION AND APHELION
The perihelion distance from a planet to the sun is
a(1 – e)
and the aphelion distance is
a(1 + e)
Equation 8
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PERIHELION AND APHELION
a. Find an approximate polar equation for the elliptical orbit of the earth around the sun (at one focus) given that: The eccentricity is about 0.017 The length of the major axis is about 2.99 x 108 km.
b. Find the distance from the earth to the sun at perihelion and at aphelion.
Example 5
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PERIHELION AND APHELION
The length of the major axis is: 2a = 2.99 x 108
So, a = 1.495 x 108.
Example 5 a
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PERIHELION AND APHELION
We are given that e = 0.017
So, from Equation 7, an equation of the earth’s orbit around the sun is:
Approximately,
Example 5 a
81.49 101 0.017cos
rθ
×=
+
2 8 2(1 ) (1.495 10 )[1 (0.017) ]1 cos 1 0.017cosa er
e θ θ− × −
= =+ +
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PERIHELION AND APHELION
From Equation 8, the perihelion distance from the earth to the sun is:
Example 5 b
8
8
(1 ) (1.495 10 )(1 0.017)1.47 10 km
a e− ≈ × −
≈ ×
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PERIHELION AND APHELION
Similarly, the aphelion distance is:
8
8
(1 ) (1.495 10 )(1 0.017)1.52 10 km
a e+ ≈ × +
≈ ×
Example 5 b