Download Conic Sections in Polar Coordinates

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Conic Sections in Polar Coordinates
MATH 211, Calculus II
J. Robert Buchanan
Department of Mathematics
Spring 2016
Introduction
We have developed the familiar formulas for the parabola,
ellipse, and hyperbola from their definitions in rectangular
coordinates.
a(x − b)2 + c = y
(x − x0 )2 (y − y0 )2
+
= 1
a2
b2
(x − x0 )2 (y − y0 )2
−
= 1
a2
b2
(parabola)
(ellipse)
(hyperbola)
Today we will see that polar coordinates unify and simplify the
definitions of the conic sections.
Definition of the Conic Sections (1 of 2)
Consider a fixed point P (called the focus) in the plane and a
fixed line (called the directrix) not containing P. Let a curve be
defined as the set of all points in the plane whose distance from
the focus is a constant multiple e (called the eccentricity) of
their distance to the directrix.
d
e*d
directrix
P
Definition of the Conic Sections (2 of 2)
Theorem
The set of all points whose distance to the focus is the product
of the eccentricity e > 0 and the distance to the directrix is
1. an ellipse if 0 < e < 1,
2. a parabola if e = 1,
3. a hyperbola if e > 1.
Definition of the Conic Sections (2 of 2)
Theorem
The set of all points whose distance to the focus is the product
of the eccentricity e > 0 and the distance to the directrix is
1. an ellipse if 0 < e < 1,
2. a parabola if e = 1,
3. a hyperbola if e > 1.
Proof.
Assume the focus is at (0, 0) and the directrix is x = d > 0.
q
x 2 + y 2 = e(d − x)
Polar Coordinate Formulation
p
Making use of the identities: r = x 2 + y 2 and x = r cos θ we
can develop a single equation for all three conic sections.
q
x 2 + y 2 = e(d − x)
r
r
= e(d − r cos θ)
=
ed
1 + e cos θ

 0 < e < 1 : ellipse
e=1:
parabola

e>1:
hyperbola
Different Orientations of the Directrix
Theorem
The conic section with eccentricity e > 0, focus at (0, 0) and the
indicated directrix has the polar equation
ed
1. r =
, if the directrix is the line x = d > 0,
1 + e cos θ
ed
2. r =
, if the directrix is the line x = d < 0,
−1 + e cos θ
ed
3. r =
, if the directrix is the line y = d > 0,
1 + e sin θ
ed
, if the directrix is the line y = d < 0.
4. r =
−1 + e sin θ
Examples
Find polar equations for the conic sections with focus (0, 0) and
directrix d = x = −2 and eccentricities
1. e = 1/2,
2. e = 1,
3. e = 2.
y
4
e=1
e=2
2
e=12
-6
-4
-2
2
-2
-4
4
6
x
Solution
1. e = 1/2:
2. e = 1:
3. e = 2:
Solution
1. e = 1/2:
1
ed
2
2 (−2)
r=
=
=
1
−1 + e cos θ
2
−
cos θ
−1 + 2 cos θ
2. e = 1:
3. e = 2:
Solution
1. e = 1/2:
1
ed
2
2 (−2)
r=
=
=
1
−1 + e cos θ
2
−
cos θ
−1 + 2 cos θ
2. e = 1:
r=
3. e = 2:
(1)(−2)
2
=
−1 + (1) cos θ
1 − cos θ
Solution
1. e = 1/2:
1
ed
2
2 (−2)
r=
=
=
1
−1 + e cos θ
2
−
cos θ
−1 + 2 cos θ
2. e = 1:
r=
3. e = 2:
r=
(1)(−2)
2
=
−1 + (1) cos θ
1 − cos θ
(2)(−2)
4
=
−1 + (2) cos θ
1 − 2 cos θ
Examples
Find polar equations for the conic sections with focus (0, 0) and
eccentricity e = 1/2 and directrices
1. d = x = 4,
2. d = y = 1,
3. d = y = −2.
y
2
1
-4
-3
-2
-1
1
-1
-2
2
3
4
x
Solution
1. d = x = 4:
2. d = y = 1:
3. d = y = −2:
Solution
1. d = x = 4:
r=
2. d = y = 1:
3. d = y = −2:
1
ed
4
2 (4)
=
=
1
1 + e cos θ
2 + cos θ
1 + 2 cos θ
Solution
1. d = x = 4:
r=
1
ed
4
2 (4)
=
=
1
1 + e cos θ
2 + cos θ
1 + 2 cos θ
2. d = y = 1:
r=
3. d = y = −2:
1
ed
1
2 (1)
=
=
1
1 + e sin θ
2
+
sin θ
1 + 2 sin θ
Solution
1. d = x = 4:
1
ed
4
2 (4)
=
=
1
1 + e cos θ
2 + cos θ
1 + 2 cos θ
r=
2. d = y = 1:
r=
1
ed
1
2 (1)
=
=
1
1 + e sin θ
2
+
sin θ
1 + 2 sin θ
3. d = y = −2:
r=
1
ed
2
2 (−2)
=
=
1
−1 + e sin θ
2 − sin θ
−1 + 2 sin θ
Parametric Equations for Conic Sections (1 of 2)
Example
Find parametric equations for the conic section with equation
x2 y2
+
= 1.
4
9
Parametric Equations for Conic Sections (1 of 2)
Example
Find parametric equations for the conic section with equation
x2 y2
+
= 1.
4
9
Use the fundamental trigonometric identity: cos2 t + sin2 t = 1.
x
= 2 cos t
y
= 3 sin t
Parametric Equations for Conic Sections (1 of 2)
Example
Find parametric equations for the conic section with equation
x2 y2
+
= 1.
4
9
Use the fundamental trigonometric identity: cos2 t + sin2 t = 1.
x
= 2 cos t
y
= 3 sin t
We can see that
x2 y2
(2 cos t)2 (3 sin t)2
+
=
+
= 1.
4
9
4
9
Parametric Equations for Conic Sections (2 of 2)
Example
Find parametric equations for the conic section with equation
(x − 2)2 (y + 2)2
−
= 1.
9
25
Parametric Equations for Conic Sections (2 of 2)
Example
Find parametric equations for the conic section with equation
(x − 2)2 (y + 2)2
−
= 1.
9
25
Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1.
x
= 2 + 3 cosh t
y
= −2 + 5 sinh t
Parametric Equations for Conic Sections (2 of 2)
Example
Find parametric equations for the conic section with equation
(x − 2)2 (y + 2)2
−
= 1.
9
25
Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1.
x
= 2 + 3 cosh t
y
= −2 + 5 sinh t
We can see that
(2 + 3 cosh t − 2)2 (−2 + 5 sinh t + 2)2
(x − 2)2 (y + 2)2
−
=
−
= 1.
9
25
9
25
Kepler’s 2nd Law of Planetary Motion
Based on astronomical observations Johannes Kepler
hypothesized that the planets move in elliptical orbits with the
sun at one focus. His 2nd law of planetary motion states that
the orbits sweep out equal areas in equal times. This implies
that the planets speed up when they are close to the sun and
slow down when they are further away.
0.5
0
y
-0.5
-1
-1.5
-2
-1
-0.5
0
x
0.5
1
Calculation of Area and Arc Length
2
.
2 + sin θ
2
Z 1 π
2
A[0,π] =
dθ ≈ 0.9455994
2 0
2 + sin θ
2
Z
1 5.224895
2
A[3π/2,5.224895] =
dθ ≈ 0.9455995
2 3π/2
2 + sin θ
s
2
Z π
dr
2
L[0,π] =
r +
dθ ≈ 2.53
dθ
0
s
2
Z 5.224895
dr
L[3π/2,5.224895] =
r2 +
dθ ≈ 1.02
dθ
3π/2
Suppose r =
Homework
I
Read Section 9.7
I
Exercises 1–27 odd.
Related documents