Multivariable Calculus
MA22S1
Dr Stephen Britton
September 2013
Introduction
The aim of this course to introduce you to the basics of multivariable calculus. The early part is dedicated to discussing curves and conic sections in
two dimensions, and then moves to three dimensions, covering lines, planes
and quadric surfaces. After this, we will move on to discussing vector-valued
functions and the curves they represent. You will learn how to use differentiation to find tangent lines, normal vectors, slopes and more. The final
part of the course deals with multiple integrals, specifically double and triple
integrals to find areas and volumes.
These notes are based on the textbook Calculus, Late Transcendentals, 10th
edition, by Howard Anton, Irl Bivens and Stephen Davis, and published by
Wiley.
1
1
Curves and Conic Sections in Two Dimensions
To begin with in this course, we will only consider two directions. Until
section 3, we can assume that everything is described in the xy-plane.
1.1
Parametric Curves
If we consider a generic curve C in the xy-plane, is there a way to parameterise the motion using a single variable? We define the parametric
equations of the motion of a particle as
x = f (t) ,
y = g(t) ,
(1)
where t is the parameter which describes where we are on the curve. The
curve itself is called the graph of the parametric equations, or alternatively,
the trajectory of the particle. An example is given in Figure 1. Generally
y
3
2
Hx,yL
1
-2
2
-1
-2
4
6
8
10
x
C
-3
Figure 1: Trajectory C.
speaking, t represents time in this course, but it could be any parameter that
makes sense in a given problem. Often, t will be restricted to an interval
[a, b], but if no interval is given, assume that the range is (−∞, ∞). This
will be written
x = f (t) , y = g(t) , (a ≤ t ≤ b) .
(2)
Problems will often ask you to sketch the graph of parametric equations.
To do this, eliminate t. The resulting equation involving x and y should be
something you can recognize.
2
Example: Sketch the graph of the equations
x = 2 cos t ,
(0 ≤ t ≤ 2π) .
y = 2 sin t ,
Solution: The easiest way to understand what this looks like is to notice
that we can eliminate t by the doing the following
x2 + y 2 = (2 cos t)2 + (2 sin t)2 = 4(cos2 t + sin2 t) = 4 .
(3)
This is clearly the equation of a circle of radius 2. To see what portion of
the circle is graphed, lets look at a few values:
t
t=0:
π
t= :
2
t=π:
3π
t=
:
2
t = 2π :
(x, y)
(2, 0)
(0, 2)
(−2, 0)
(0, −2)
(2, 0)
In other words, we have an entire circle, as is sketched in Figure 2.
Example: Sketch the graph of the equations
x = t − 1,
y = 2t + 3 ,
(0 ≤ t ≤ 2π) .
Solution: To eliminate t, solve both equations for t, i.e.
t = x + 1,
1
t = (y − 3) .
2
We then equate these and simplify to get one equation.
1
x + 1 = (y − 3)
2
⇒2x + 2 = y − 3
⇒y = 2x + 5 .
3
2
Hx,yL
1
y
t
-2
-1
x
1
2
-1
-2
Figure 2: Circle of radius 2.
y
15
10
5
-4
-2
2
4
x
Figure 3: Plot of y = 2x + 5.
We now can easily see that this is a line with slope 2 which intersects the
y-axis at y = 5. It is plotted in Figure 3.
Next, we define the orientation of a graph to be the direction of increasing
parameter. In the example of the line above, the orientation is up the slope
as that is the direction of increasing t. A curve with orientation is called a
parametric curve. For the example of the circle, we had an anti-clockwise
orientation. If we replaced t with −t, then x(−t) = 2 cos(−t) = 2 cos t,
y(−t) = 2 sin(−t) = −2 sin t, which has clockwise orientation.
4
1.1.1
Expressing equations parametrically
We can always describe the equation y = f (x) parametrically. The simplest
way to do this is to take x = t and then y = f (t).
Example: y = x2 + 1.
Solution: x = t ⇒ y = t2 + 1.
Of course, for something like a circle you can of course do the same. However,
there are often better parameterisations to chose. In the case of a circle of
radius a, the parameterisation x = a cos t, y = a sin t is often convenient.
1.1.2
Tangent lines for parametric curves
Recall that the tangent line to a curve y = f (x) at a point P is the line
that passes through P that has slope m = dfdt(x) . For a parametric curve, we
can find the slope using the parameter t via the chain rule:
dy
dy/dt
=
.
dx
dx/dt
(4)
There are three special types of tangent lines:
• If dy/dt = 0 and dx/dt 6= 0, the slope is 0 and we have a horizontal
tangent line.
• If dx/dt = 0 and dy/dt 6= 0, the slope is infinite and we have a vertical
tangent line.
• If dx/dt = 0 and dy/dt = 0, the slope is indeterminate and we have a
singular point. These require special treatment.
Let’s look at an example.
Example: Find the tangent line of x = t, y = −t2 for t = 2.
Solution: Of course, we can see that this is equivalent to y = −x2 , but
let’s do this using the new method. First, the point we are interested in is
P0 = (2, −4). We know that a line has an equation y = y0 + m(x − x0 ), and
we can find the slope of the tangent lines since
dy
dy/dt
−2t
=
=
= −2t .
dx
dx/dt
1
5
(5)
Therefore, at P0
m|P0 = −2(2) = −4 .
(6)
The equation of the tangent line is then
y = −4 − 4(x − 2) = 4 − 4x .
(7)
This is sketched in Figure 4.
y
-4
-2
2
4
x
H2,-4L
-5
-10
-15
Figure 4: Tangent line to the function y = −x2 .
1.1.3
Arc Length of a Parametric Curve
For a parametric curve
x = x(t) ,
y = y(t) ,
(a ≤ t ≤ b) ,
(8)
the arc length is given by
Z
b
L=
a
s
dx
dt
2
+
dy
dt
2
dt .
Example: Find the circumference of the circle
x = a cos t ,
y = a sin t , (0 ≤ t ≤ 2π) .
6
(9)
Solution: We find
Z 2π p
Z
2
2
L=
(−a sin t) + (a cos t) dt =
0
2π
at dt = 2πa .
0
If the curve doubles over, this will not work: in other words no point on the
curve should be graphed by two values of t. In our example of the circle, if
we integrated from 0 to 3π we would find 3πa, which is obviously not the
circumference, because the values between 2π and 3π repeat those between
0 and π.
1.2
Polar Coordinates
Up until now we have been dealing with the coordinates x and y, which are
known as Cartesian coordinates or rectangular coordinates. We can
equally describe a plane using any pair of coordinates so that each position
on the plane is uniquely described by the pair. In particular a useful set of
coordinates are polar coordinates.
1.2.1
Defining Polar Coordinates
To define polar coordinates, we begin by fixing a point O, called the origin
or pole. We then define a half-line (or ray) which begins at 0 and continues
to infinity in a given direction, called the polar axis. This is shown in
PHr,ΘL
r
Θ
OriginHOL
Polar axis
Figure 5: Polar coordinates
Figure 5. The distance from the origin to the point P is called the radial
coordinate, r, and the angle that the line |OP | makes with the polar axis
is called the angular coordinate, θ, more commonly known as the polar
7
angle. It is important to notice that the polar angle returns to its original
position when the angle is 2π. Therefore, any angle greater than or equal
to 2π is equivalent to an angle in the range (0 ≤ θ ≤ 2π). In fact, more
generally, the angles θ − 2π n, θ and θ + 2π n are equivalent if n is an integer.
1.2.2
Relationship to Cartesian Coordinates
It is quite easy to make a link between polar coordinates and Cartesian
coordinates. The trick is to make the polar axis coincide with the x-axis.
Then, we can see from Figure 6 that the following relationship holds
y
PHr,ΘL
y=r sin Θ
r
Θ
O
x=r cos Θ
x
Figure 6: Polar coordinates in terms of Cartesian coordinates
x = r cos θ ,
y = r sin θ ,
(10)
y
.
x
(11)
which can also be written in the form
r 2 = x2 + y 2 ,
tan θ =
√
Example: Change the Cartesian coordinates (4, 4 3) into polar coordinates.
Solution:
√
r2 = 42 + (4 3)2 = 64 ⇒ r = 8 ,
8
and
√
4 3 √
π
tan θ =
= 3⇒θ= .
4
3
The polar coordinates are therefore (r, θ) = (8, π/3).
Example: Change the polar coordinates (3, 3π/4) into Cartesian coordinates.
Solution:
x = 3 cos
3π
3
= −√ ,
4
2
and
y = 3 sin
3
3π
=√ ,
4
2
√
√
The Cartesian coordinates are therefore (x, y) = (−3/ 2, 3/ 2).
If you are asked to graph an equation in polar coordinates, the easiest thing
1.0
А2
0.5
А4
4
Π
-1.0
-0.5
0.5
0
3
1.0
3
2
-0.5
1
2
1
3А2
-1
-1.0
r=0
1
2
3
4
-1
Figure 7: Unit circle and the half-line θ = π4 .
to do is to simply plot some points and see what happens. Later on, you may
begin to recognise some of the graphs and be able to plot them from memory.
For now, let’s look at some simple examples. If we fix r, say to r = 1, we get
a circle, see Figure 7. Note also that once we hit θ = 2π, the graph repeats,
and so we do not need to continue. Also in the same figure we see the plot
of θ = π/4. We can either imagine that we can take negative values of r,
9
or as we would normally expect we could impose r ≥ 0 and plot a half-line,
as is done here. Finally, let’s look at an example where both coordinates
vary, say r = sin θ. Then we get the plot in Figure 8, where we see that
the plot repeats values after θ = π, because the Cartesian coordinates are
x = sin t cos θ, y = sin θ sin θ = sin2 θ ≥ 0.
А2
1.0
3А4
-1.0
А4
0.5
-0.5
0.5
Θ=0 or Θ=Π
1.0
Figure 8: Plot of r = sin θ.
1.2.3
Symmetry Tests
Symmetry tests can be performed in polar coordinates by appropriate choice
of shifts.
• A curve in polar coordinates is symmetric about the x-axis if replacing
θ with −θ leaves the equations unchanged.
• A curve in polar coordinates is symmetric about the y-axis if replacing
θ with π − θ leaves the equations unchanged.
• A curve in polar coordinates is symmetric about the origin if replacing
r with −r or equivalently θ with θ + π leaves the equations unchanged.
Let’s look at an example.
Example: Test the symmetry properties of r = sin θ.
Solution: If we replace θ with −θ, we get sin(−θ) = − sin θ, so it is not symmetric about the x-axis. If we replace θ with π − θ, we get sin(π − θ) = sin θ,
so it is symmetric about the y-axis. Finally, if we replace θ with θ + π we
10
1.0
1.0
Hr, ΘL
Hr, Π-ΘL
Hr, ΘL
0.5
-2
0.5
-1
1
-2
2
-1
1
-0.5
2
-0.5
Hr, -ΘL
-1.0
-1.0
1.0
Hr, ΘL
0.5
-2
-1
Hr, Θ+ΠL
or H-r, ΘL
1
2
-0.5
-1.0
Figure 9: Symmetry test
get sin(θ + π) = − sin θ, or if we replace r with −r we get the same, so it is
not symmetric about the origin.
1.2.4
Families of curves
Circles: The following describe circles of radius a,
r = a,
r = 2a cos θ ,
r = 2a sin θ .
(12)
The first of these is a circle centered on the origin, the second centred on
(a, 0), and the last centred on (0, a). Figure 10 shows these.
Rose curves: The following describe rose curves for a > 0 and n integer,
r = a sin nθ ,
r = a cos nθ .
(13)
If n is odd, the pattern is drawn when the values (0 ≤ θ < π) are taken, and
for n even, the pattern is drawn when the values (0 ≤ θ < 2π) are taken.
See what the look like: Section 10.2 of the text book has images.
11
1.5
1.5
1.0
1.0
0.5
0.5
0.5
-1.5
-1.0
-0.5
0.5
1.0
1.0
1.5
2.0
2.5
1.5
-0.5
-0.5
-1.0
-1.0
r=2a Cos Θ
r=a
-1.5
-1.5
2.5
2.0
1.5
1.0
0.5
-1.5
-1.0
-0.5
r=2a Sin Θ
0.5
1.0
1.5
-0.5
Figure 10: Families of circles
Limaçons: The following describe limaçons for a > 0 and b > 0,
r = a ± b sin nθ ,
r = a ± b cos nθ .
(14)
If a = b, these becomes cardioids. Again, see textbook for images.
Spirals: Although there are many types of spiral, the only one we are interested in is the “Archimedean spiral”, which is the one you most likely
imagine when someone says the word spiral. It has the equation
√
r = a θ.
(15)
12
1.2.5
Tangent Lines in Polar Coordinates
If we take a general curve of the form r = f (θ) and write it in terms of its x
and y coordinates, we would have
x = f (θ) cos θ ,
y = f (θ) sin θ .
(16)
This is now a parametric curve in parameter θ, and so we can find the slope
of the tangent line provided f (θ) is differentiable. We have
dx
dr
= −f (θ) sin θ + f 0 (θ) cos θ = −r sin θ +
cos θ ,
dθ
dθ
dr
dy
= f (θ) cos θ + f 0 (θ) sin θ = r cos θ +
sin θ ,
dθ
dθ
(17)
and the slope is
dy
=
dx
dy
dθ
dx
dθ
=
dr
r cos θ + dθ
sin θ
.
dr
−r sin θ + dθ cos θ
(18)
Example: Find the slope of the tangent line to the circle r = 2 cos θ at
θ = π/3.
Solution: We get
dr
dθ
= −2 sin θ, and so
2 cos2 θ − 2 sin2 θ
2 cos 2θ
dy
=
=
= − cot 2θ .
dx
−2 sin θ cos θ − 2 sin θ cos θ
−2 sin 2θ
Therefore, the slope at θ = π/3 is
m = − cot
1.2.6
2π
1
=√ .
3
3
Tangent Lines to Polar Curves at the Origin
At the origin, we have r = 0 with angle θ0 , so assuming that
origin, the equation becomes
dr
0 + dθ
sin θ0
dy
=
= tan θ0 .
dr
dx
0 + dθ cos θ0
dr
dθ
6= 0 at the
(19)
The line with this slope is the line θ = θ0 , and so this line is tangent to the
curve at the origin. To recall what this line looks like, return to section 1.2.2.
13
Example: Find the tangent lines at the origin of the four-petal rose
r = sin 2θ .
Solution: We must find all values of θ less than 2π for which r = 0. This
gives us θ0 = 0, θ0 = π/2, θ0 = π, θ0 = 3π/2. These are the tangent lines
and are shown in Figure 11. Note that the lines θ0 = 0 and θ0 = π coincide,
as do θ0 = π/2 and θ0 = 3π/2.
1.0
Π 3Π
Θ= ,
2 2
0.5
Θ=0, Π
-1.0
-0.5
0.5
-0.5
-1.0
Figure 11: r = sin 2θ
14
1.0
1.2.7
Arc Length in Polar Coordinates
To find the arc length in terms of polar coordinates, we recall formula (9)
and substitute the results of (17), giving
2 2
dx
dy
+
dθ
dθ
2 2
dr
dr
cos θ + r cos θ +
sin θ
= −r sin θ +
dθ
dθ
dr
(20)
= r2 (sin2 θ + cos2 θ) + 2r cos θ sin θ
dθ
2
dr
dr
− 2r cos θ sin θ +
(sin2 θ + cos2 θ)
dθ
dθ
2
dr
2
,
=r +
dθ
which results in
Z
β
s
r2
L=
+
α
dr
dθ
2
dθ
(21)
Let’s look at an example.
Example: Find the arc length of the logarithmic spiral r = aebθ between
θ = 0 and θ = π.
Solution:
Z πp
(aebθ )2 + (abebθ )2 dθ
L=
Z0 π √
π a √
a√
=
a 1 + b2 ebθ dθ =
1 + b2 ebθ 0 =
1 + b2 (ebπ − 1) .
b
b
0
1.2.8
(22)
Area in Polar Coordinates
In order to calculate area in polar coordinates, we need to find a way to
parameterise the area. The first thing to remember is that any polar curve
can be written in the form r = f (θ), and therefore we can parameterise in
terms of θ. Then, if we choose two rays θ = α and θ = β such that the
angles satisfy α < β ≤ α + 2π (in other words, the angles are different and
are less than 2π apart, so that the area is only covered once), provided f (θ)
15
is continuous in this region, R, we can find the area. This concept is shown
in Figure 12. We can obtain the area under the curve r = f (θ) by diving
6
5
r=fHΘL
4
3
R
2
Θ=Α
Θ=Β
-4
1
-2
0
2
4
Figure 12: Area in polar coordinates
this region into wedges using a set of rays,
θ = θ1 , θ = θ2 , · · · θ = θn−1 ,
which divide the region into small areas A1 , A2 , · · · An with angles ∆θ1 , ∆θ2 , · · · ∆θn .
This is shown in Figure 13. The total area is then the sum of these small
6
5
r=fHΘL
4
A2
3
DΘ2
2
A1
DΘ1
An
Θ=Β
-4
-2
DΘn
Θ=Α
1
0
2
4
Figure 13: Area in polar coordinates divided into wedges
areas,
A = A1 + A2 + · · · + An =
n
X
Ak .
(23)
k=1
Each of these is the area of a sector, which has area given by 21 r2 θ. Using
r = f (θ), and θ = θk for any ray in the wedge of area Ak , we arrive at the
16
discrete version of the area formula
A=
n
X
Ak ∼
k=1
n
X
1
k=1
2
[f (θk )]2 ∆θk .
(24)
We can take the limit where these angles ∆θk vanish to give an integral form
A = lim
∆k →0
n
X
1
k=1
2
2
Z
β
[f (θk )] ∆θk =
α
1
[f (θ)]2 dθ .
2
(25)
Thus, the area of a region R under a continuous curve r = f (θ) and between
θ = α and θ = β is
Z β
Z β
1
1 2
2
A=
[f (θ)] dθ =
r dθ .
(26)
α 2
α 2
The most difficult parts of this is generally finding the limits of integration.
Sometimes you will be told explicitly.
Important note: Always draw the region to help you identify the limits.
Let’s look at a simple example.
Example: Find the area of the circle r = 2a sin θ.
Solution: This looks like the third types of circle Figure 10. Since we want
the area of the whole circle, the limits are 0 ≤ θ ≤ π since the entire circle
is above the x-axis. Note that for a circle centered on the origin r = a, the
limits would be 0 ≤ θ ≤ 2π. The area is
Z π
Z π
1
2
2
(2a sin θ) dθ = 2a
sin2 θ dθ
A=
0
0 2
Z π
1
1 π
2
2
= 2a
(1 − cos 2θ) dθ = 2a
θ 0 − 0 = πa2 .
2
0 2
We can often make use of the symmetry of the region to simplify a problem.
For example, the upper and lower semi-circles of a full circle are mirror images
and obviously have the same area. Hence
Z 2π
Z π
1 2
1 2
a dθ = 2
a = πa2 ,
(27)
2
2
0
0
17
is clearly true. Let’s look at a more complicated example to help understand
this.
Example: Find the area of the cardioid r = 4 + 4 cos θ outside the circle
r = 6.
Solution: The region is the coloured region in Figure 14. The rays θ = θ1
5
-5
Θ=Θ2
5
-5
Θ=Θ1
Figure 14: Area of the cardiod r = 4 + 4 cos θ outside the circle r = 6
and θ = θ2 represent the limits of integration. We can find these limits by
finding the intersections of the circle with the cardiod,
4 + 4 cos θ = 6 ⇒ cos θ =
1
π
⇒θ=± .
2
3
We can now see that there are two equally valid approaches: integrate from
− π3 to π3 , or use symmetry to take twice the area from 0 to π3 . Obviously,
we must subtract the area of the circle in this region from the area of the
cardiod in this region. Let’s look at both methods.
18
Integration over whole region:
Z π/3
Z π/3
1 2
1
2
(4 + 4 cos θ) dθ −
(6) dθ
A=
−π/3 2
−π/3 2
Z π/3
1
=
(16 + 32 cos θ + 16 cos2 θ − 36) dθ
2
−π/3
Z π/3
(16 cos θ + 8 cos2 θ − 10) dθ
=
−π/3
Z
π/3
(16 cos θ + 4(1 + cos 2θ) − 10) dθ
=
−π/3
π/3
= (16 sin θ + 4(θ + 2 sin 2θ) − 10θ)−π/3
√ !
√ !
√
√
π
π 1 3
π
3
π 1 3
3
= 16
+4
+
− 10
+4 − −
+ 10
− − 16
2
3 2 2
3
2
3 2 2
3
√
= 18 3 − 4π .
Notice that the first and second terms in the penultimate line are in fact the
same. This is due to the symmetry about the x-axis.
Using symmetry:
Z π/3
A=2
(16 cos θ + 8 cos2 θ − 10) dθ
0
√ !
√
√
3
π 1 3
π
= 2 16
+4
+
− 10
= 18 3 − 4π .
2
3 2 2
3
The procedure for equating the equations of the curves can fail to find some
intersections. This is why it is always important to draw diagrams to ensure
that you know how many intersection points to expect.
1.3
Conic Sections
A conic section (or conic) is any curve that can be obtained as the set
of points of intersection between a plane and a double-napped circular cone.
These are the ellipse, parabola and hyperbola (note that a circle is a special
case of an ellipse). The degenerate conic sections arise when the plane
19
passes through the vertex of the cone, and are the point, line and intersecting
lines. The non-degenerate conic sections are shown in Figures 15 and the
degenerate conic sections are shown in Figure 16.
-1
-2
2
1
0
1
0
-1
-2
2
2
2
1
1
0
0
-1
-1
-2
-2
-2
-2
-1
-1
0
0
1
1
2
-2
0
-1
1
2
2
2
2
1
1
0
0
-1
2
-1
1
-2
0
-2
-1
-2
0
-1
-2
1
-1
2
Figure 15: Conic sections
20
0
1
-2
2
-2
0
-1
2
1
-2
2
2
1
1
0
0
-1
-1
-2
0
-1
2
1
-2
-2
-2
-1
-1
0
0
1
1
2
2
2
1
0
-1
2
-2
-2
1
-1
0
0
-1
1
2
-2
Figure 16: Degenerate conic sections
21
1.3.1
Types of Conic Section
A parabola is the set of points in the plane equidistant from a fixed line
called a directrix and a fixed point called the focus. A parabola is symmetric about the axis perpendicular to the directrix which runs through the
focus, called the axis of symmetry. The parabola intersects this axis at its
vertex. See Figure 17.
6
5
Axis
4
Focus3
2
Directrix
1
-2
-1
1
2
Figure 17: A parabola
An ellipse is the set of all points in the plane that such that the sum of
the distance from a point on the ellipse to two fixed points called foci is a
constant. This constant must be greater than the distance between the foci.
The midpoint of the line line joining the foci is called the centre. Note that
a circle is a special case of an ellipse when the distance between the foci is
0, i.e. the foci coincide with the centre. The line segment across the ellipse
through the foci is called the major axis, and the line segment across the
ellipse perpendicular to this through the centre is called the minor axis.
Finally, the endpoints of the major axis are called vertices. See Figure 18.
-2
1.0
1.0
0.5
0.5
-1
Focus
1
Centre
-0.5
-2
2
-1
Focus
1
-0.5
-1.0
-1.0
Figure 18: An ellipse
22
Major axis
Minor axis
2
A hyperbola is the set of all points in the plane that such that the difference of the distance from a point on the hyperbola to two fixed points called
foci is a positive constant. This constant must be less than the distance
between the foci. This difference must be positive, and as a result there are
two branches which curve around the different foci. The midpoint of the
line segment between the foci is called the centre, and the line through the
foci is called the focal axis. The line perpendicular to the focal axis and
through the centre is called the conjugate axis. The intersections of the
hyperbola with the focal axis are called vertices. A hyperbola also has a
pair of associated lines called asymptotes, which are line such that as the
hyperbola move towards ±∞ away from the centre, the distance between the
asymptotes and the hyperbola approaches zero. See Figure 19.
4
4
2
2
-5
-5
5
5
-2
-2
-4
-4
Figure 19: A hyperbola. The points marked are the centre (middle) and foci
(left and right). The horizontal red line is the focal axis and the vertical red
line is the conjugate axis.
1.3.2
Equations of parabolas
We denote the distance between the vertex and focus of a parameter as p.
Then, the distance from the vertex to the directrix is also p. It is sometimes
convenient to denote the focus as F (x, y). The standard positions of a
parabola are those when the vertex coincides with the origin. Figure 20
sketches the standard parabolas. You should therefore recognise any equation
of the form
y 2 = ±4px , or x2 = ±4py ,
(28)
as the standard equations of a parabola. To understand these equations,
we return to the definition of a parabola. The distance from the directrix to
23
y
y
2
2
1
1
FHp,0L
0
0.5
1.0
1.5
2.0
FH-p,0L
x
-2.0
-1.5
-1.0
-1
-1
2
2
y =-4px
y =4px
-2
-2
y
y
4
-2
-1
1
3
-1
2
-2
1
-2
x
-0.5
FH0,pL
-1
x
FH0,-pL
-3
1
-1
2
2
x
-4
x2 =-4py
2
x =4py
-5
Figure 20: Standard equations of a parabola
a point must be the same as the distance to the focus. Look at Figure 21.
The distance to the directrix
to Q is always simply y + p. The distance from
p
the focus, F , to Q is (y − p)2 + x2 . Therefore
(y − p)2 + x2 = (y + p)2
⇒y 2 − 2py + p2 + x2 = y 2 + 2py + p2
⇒x2 = 4py ,
(29)
as you would expect from the diagram. To sketch a parabola, determine
which of the four types in equation (28) it is, and then use Figure 20 to
determine which way it curves. Determine p from the equation and plot a
couple of convenient points to ensure you are close. Useful points are the
intersections of the parabola with a line parallel to the directrix running
through the focus, see Figure 22. Please note that this is basically the same
method given in the textbook, but I think the explanation is less confusing.
Also, notice that we are always assuming that the vertex of the parabola is
at the origin. If this wasn’t the case, we would have to find it.
Example: Find the focus of the parabola y 2 = −16x and sketch the
24
y
4
3
Q
2
FH0,pL
1
-2
-1
1
-1
2
x
y=-p
-2
Figure 21: Equations a parabola
y
4
3
2
H-2p,pL
-2
H0,pL
H2p,pL
1
-1
1
-1
2
x
y=-p
-2
Figure 22: Sketching a parabola
parabola, focus and directrix.
Solution: We see that we have a parabola of type y 2 = −4px and so p = 4.
Since we are looking at the second type in Figure 20, the focus is at (−4, 0),
and the parabola should be to the left of the y-axis. It is shown in Figure
23.
More generally, we can have a parabola which has its vertex at any point.
They have the following equations:
Vertex (h, k), parallel to x-axis
(y − k)2 = 4p(x − h) , opens to the right.
(y − k)2 = −4p(x − h) , opens to the left.
25
(30)
y
15
H-4,8L
10
5
H-4,0L
-8
-6
-4
x=4
-2
2
4
x
-5
H-4,-8L
-10
-15
Figure 23: Sketching the parabola y 2 = −16x
Vertex (h, k), parallel to y-axis
(x − h)2 = 4p(y − k) , opens above.
(x − h)2 = −4p(y − k) , opens below.
(31)
You should be able to recognise these equations as parabolas and reduce
appropriate quadratic equations to these forms. It is often necessary to
complete the squares to find the correct form of the equation.
Example: Determine if y 2 − 6y − 2x + 1 = 0 represents a parabola and if
so find p and the vertex position.
Solution: We complete the square to get
y 2 − 6y − 2x + 1 = 0
⇒(y − 3)2 − 9 − 2x + 1 = 0
⇒(y − 3)2 − 2x − 8 = 0
⇒(y − 3)2 = 2(x + 4) .
Hence, p =
1.3.3
1
2
and the vertex is at (−4, 3).
Equations of ellipses
We denote the length of the major axis of an ellipse by 2a and the length of
the minor axis by 2b, with the distance between the foci denoted by 2c. In
this context, it is sometimes convenient to denote the foci as F1 (x1 , y1 ) and
F2 (x2 , y2 ). a is referred to as the semi-major axis and b as the semi-minor
26
axis. This is just terminology: obviously a number is not the same as a line!
These numbers uniquely define the ellipse. Their ratio give the shape, and
the magnitude gives the size of the ellipse. This is shown in the first diagram
in Figure 24. If we consider the red triangles in the second diagram, we see
2
c
2
c
Q
1
1
b
-3
-2
-1
1
2
b
-3
3
b
-1
a
-2
-1
P
1
c
c
2
3
-1
a
-2
-2
Figure 24: The semi-major axes and semi-minor axes for ellipses
√
that the distance from the two foci to Q is simply 2 b2 + c2 . Since the sum of
the distances from the foci to any point is the same, this should be the same
as the sum of the distances from the foci to P , which is (a + c) + (a − c) = 2a.
We therefore find the relations
√
√
(32)
a = b2 + c2 , or c = a2 − b2 .
As for a parabola, the easiest types of ellipse to imagine are those centred on
the origin. These are the standard positions of the ellipse, which give rise
to the standard equations. These standard ellipses are shown in Figure
25. You should recognise any equation of the forms
x2 y 2
+ 2 = 1,
a2
b
or
x2 y 2
+ 2 = 1,
b2
a
(33)
as a standard ellipse. To see where these come from, consider Figure 26. The
27
y
3
a
2
F1 H0,cL
1
b
y
2
1
-2
b
-1
1
2
x
b
-1
a F1 H-c,0L
-3
-2
F2 H0,-cL
F2 Hc,0L a
-1
1
2
3
x
-2
a
-1
b
-2
x2 y2
+ =1
a2 b2
-3
x2 y2
+ =1
b2 a2
Figure 25: Standard ellipses
sum of the distances from the foci to Q should be 2a. We therefore have
p
p
(x + c)2 + y 2 + (x − c)2 + y 2 = 2a
p
p
⇒ (x + c)2 + y 2 = 2a − (x − c)2 + y 2
p
⇒(x + c)2 + y 2 = (x − c)2 + y 2 − 4a (x − c)2 + y 2 + 4a2 (squaring)
p
⇒4cx = −4a (x − c)2 + y 2 + 4a2
cx p
= (x − c)2 + y 2
⇒a −
a
cx 2
⇒ a−
= (x − c)2 + y 2 (squaring)
(34)
a
2
c
⇒a2 − 2cx + 2 x2 = x2 − 2cx + c2 + y 2
a
1
⇒a2 − c2 = 2 x2 a2 − c2 + y 2
a
x2
y2
⇒1 = 2 + 2
a
a − c2
x2 y 2
⇒1 = 2 + 2 ,
a
b
28
1.0
Q
0.5
-2
-1
1
H-c,0L
2
Hc,0L
-0.5
-1.0
Figure 26: Equations of ellipses
due to the relations (32). This is clearly the correct equation for the ellipse
in Figure 26. To sketch an ellipse, you will want to determine whether the
major axis is the x or y axis, i.e. which type of parabola in Figure 25 we
are dealing with. Determine a and b, and plot the points where the ellipse
intersects the x- and y-axes.
More generally, we want to consider ellipses with centres that are not the
origin. In this case, the equations look like (assuming b < a)
Ellipse with centre (h, k), and major axis parallel to x-axis
(35)
(x − h)2 (y − k)2
+
=
1
,
a2
b2
Ellipse with centre (h, k), and major axis parallel to y-axis
(x − h)2 (y − k)2
+
= 1,
b2
a2
(36)
and to sketch the ellipse we follow more or less the same procedure with the
new centre and using the points where the ellipse meets its major and minor
axes for reference.
Example: Determine if 9x2 + 4y 2 − 18x + 24y + 9 = 0 represents an ellipse.
Sketch the resulting curve.
29
Solution: Completing the squares, we get
9x2 + 4y 2 − 18x + 24y + 9 = 0
⇒(3x − 3)2 − 9 + (2y + 6)2 − 36 + 9 = 0
⇒9(x − 1)2 + 4(y + 3)2 = 36
(x − 1)2 (y + 3)2
⇒
+
= 1,
4
9
(37)
which is an ellipse centred on (1, −3),
√ with the√major axis parallel to the
y-axis, and a = 3, b = 2. Also, c = 32 − 22 = 5. It is sketched in Figure
27.
y
-1
1
-1
2
3
x
F1
-2
-3
Centre
-4
-5
F2
-6
Figure 27: The ellipse
30
(x−1)2 +
4
+
(y+3)2
9
=1
1.3.4
Equations of hyperbolas
When defining hyperbolas, we denote the distance between the vertices by
2a, the distance between the foci by 2c, and in analogy to the ellipse, we
define a constant b via the formulae
√
√
(38)
b = c2 − a2 , or c = a2 + b2 .
These quantities are drawn in Figure 28. We see that b is defined in such a
4
c
2
4
c
-5
b
-5
5
a
-2
a
2
a
V1
c
V2
5
-2
-4
-4
Figure 28: Defining a hyperbola
way that the asymptotes intersect the circle with diameter 2c. We sometimes
find it convenient to denote the foci as F1 (x1 , y1 ) and F2 (x2 , y2 ), and the
vertices as V1 (x1 , y1 ) and V2 (x2 , y2 ). If we look at one of these vertices, say
V2 , we see that the difference of the distances from V2 to the foci is
(c − a) + 2a − (c − a) = 2a .
(39)
Since the difference of the distances from the foci to every point is the same,
it is always 2a. If the centre of the hyperbola is at the origin and the foci are
on either the x- or y-axis, the hyperbola is in one of the standard positions,
with the resulting standard equations. The parabola are shown in Figure
29 and the equations that you should recognise as standard hyperbola at the
origin are
x2 y 2
− 2 = 1,
a2
b
b
with asymptotes y = ± x ,
a
31
y 2 x2
or
− 2 = 1,
a2
b
a
or y = ± x .
b
(40)
y
b
y=- x
a
b
y= x
a
y
5
F2 H0,cL
b
F1 H-c,0L
-5
a
y= x
b
5
-a
a
F2 Hc,0L
5
a
y=- x
b
-b
-5
x
-b
a
b
5
x
-a
-5
x2 y2
- =1
a2 b2
-5
F1 H0,-cL
y2 x2
- =1
a2 b2
Figure 29: Standard hyperbolas
The asymptotes are given the limit x, y → ∞ of
y2
x2
=
− 1,
b2
a2
or
y2
x2
=
1
+
,
a2
b2
(41)
or
y2
a2
=
,
x2
b2
(42)
which gives the ratios
y2
b2
=
,
x2
a2
or
a
b
y = ± x , or y = ± x .
(43)
a
b
To draw a hyperbola from a standard equation, locate the foci and thus the
focal axis. Next, locate the vertices and the asymptotes. If necessary, plot
some extra points and then draw the hyperbola passing through the vertices,
any extra points and approaching the asymptotes. More generally, we want
to consider hyperbolas with centres that are not the origin. In this case, the
equations look like
Hyperbola with centre (h, k), and focal axis parallel to x-axis
(44)
(x − h)2 (y − k)2
−
=
1
,
a2
b2
32
Hyperbola with centre (h, k), and focal axis parallel to y-axis
(45)
(y − k)2 (x − h)2
−
=
1
,
a2
b2
Please note that when we write the hyperbolas this way, the asymptotes for
the first case are the lines with slope ± ab passing through (h, k) and for the
second first case the asymptotes are the lines with slope ± ab passing through
(h, k).
Example: Determine if x2 − 4y 2 + 2x + 8y − 7 = 0 represents a hyperbola,
and if show, sketch the resulting curve.
Solution: We have
x2 − 4y 2 + 2x + 8y − 7 = 0
⇒(x + 1)2 − 1 − 4(y − 1)2 + 4 − 7 = 0
⇒(x + 1)2 − 4(y − 1)2 = 4
(x + 1)2 (y − 1)2
−
= 1.
⇒
4
1
This is clearly a hyperbola centred on (−1, 1). Its focal axis is parallel to
the x-axis. The vertices are length a way from the centre, and since a = 2
by comparing with equation (44), the √
vertices are at (−3, 1) and (1, 1). To
find the
foci,
recall
that
they
are
c
=
a2 + b2 away from
√
√ the centre. Here
√
2
2
c = 2 + 1 = 5, and so the foci are at (−1 ± 5, 1). Finally, the
asymptotes can be found by finding the lines that have slope ± 12 that pass
through (−1, 1). Using
3
2
F 1 H-1-
H-1,1L
5 ,1L
F 2 H-1+
5 ,1L
1
-6
-4
-2
2
4
-1
Figure 30: The hyperbola
(x+1)2
4
y = mx + b ,
33
−
(y−1)2
1
=1
we find that for m =
asymptotes are
1
,
2
b =
3
2
and for m = − 12 , b =
1
.
2
Therefore the
x 3
x 1
+ , y=− + .
2 2
2 2
This parabola is shown in Figure 30.
y=
1.3.5
Reflection properties for conic sections
Reflection property of parabolas: Consider a tangent line at a point P
on the parabola. The angles between the tangent line and the line through P
parallel to the focal axis and between the tangent line and the line connecting
P to the focus are equal.
Reflection property of ellipses: Consider a tangent line at a point P on
the ellipse. The angles between the tangent line and the lines joining P to
the foci are equal.
Reflection property of hyperbolas: Consider a tangent line at a point P
on the hyperabola. The angles between the tangent line and the lines joining
P to the foci are equal.
These reflection properties are drawn in Figure 31.
1.3.6
Applications of conic sections
Some applications of conic sections include:
Parabola: Parabolic reflector - used in headlights on cars, etc. to project a
focussed beam of light.
Ellipse: Planetary orbits - Kepler’s problem. More on this later.
Hyperbola: Scattering of atomic particles - Rutherford’s experiment.
1.4
1.4.1
Rotated conic sections
Quadratic equations
All conic sections are special cases of the general quadratic equation
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 .
(46)
If A, B, C 6= 0 it is quadratic in x and y, and if they vanish, it reduces to a
linear equation in x and y. Obviously B = 0 gives rise to the conic sections
34
y
4
1.0
3
Α
P
Α
0.5
Α
P
-2
2
Α
1
F
-2
-1
1
2
-1
1
F1
x
-0.5
2
F2
-1
-1.0
-2
P
1.5
1.0
F1
-4
0.5
-2
Α
Α
2
F2
4
-0.5
-1.0
-1.5
Figure 31: Reflection properties
we have already met when appropriate constants are chosen. However, if
B 6= 0, then we have a cross-product term, which is new. The effect of
this term is to rotate the axes of the conic section so that they no longer
align with the x and y axes. To understand why this is true, consider the
ellipse described by the foci F1 (1, 2), F2 (−1, −2) and a = 3, which is drawn
in Figure 32. Based on the information we have, the point in this figure
should satisfy
p
p
(x − 1)2 + (y − 2)2 + (x + 1)2 + (y + 2)2 = 2a = 6 ,
(47)
which we square to give
p
p
2x2 + 2 (x − 1)2 + (y − 2)2 (x + 1)2 + (y + 2)2 + 2y 2 + 10 = 36 .
(48)
If we now write
p
2
p
2 (x − 1)2 + (y − 2)2 (x + 1)2 + (y + 2)2 = (26 − 2y 2 − 2x2 )2 , (49)
when we simply we get
8x2 − 4xy + 5y 2 = 36 ,
35
(50)
y
PHx,yL
2
H1,2L
1
-3
-2
-1
1
2
3
x
-1
H-1,-2L
-2
Figure 32: Rotated ellipse
which has the xy term we expected.
A useful tool to help identify conic sections is the discriminant,
δ = B 2 − 4AC .
δ < 0 represents an ellipse, with A = C and B = 0 being the special case of
a circle. δ = 0 represents a parabola. Finally, δ > 0 represents a hyperbola.
1.5
Rotation of the x and y axes
How do we relate these rotated conic sections to those we are familiar with?
To do this, we must understand how to perform a rotation of the plane. If
we begin with a coordinate system in polar coordinates
x = r cos(θ + α) ,
y = r sin(θ + α) ,
(51)
for θ constant such that the polar axis at θ + α = 0 coincides with the x-axis,
we can easily imagine that we could equally construct a coordinate system
which has it’s polar axis at α = 0, which would be
x0 = r cos α ,
y 0 = r sin α .
36
(52)
y
1.2
y'
1.0
P
0.8
0.6
r
x'
0.4
0.2
0.5
1.0
x
-0.2
Figure 33: Rotated coordinate axes
These two systems are shown in Figure 33. We can use trigonometric identities to rewrite the unprimed coordinates as
x = r cos θ cos α − r sin θ sin α ,
y = r sin θ cos α + r cos θ sin α ,
(53)
which means that we can relate the coordinate systems by
x = x0 cos θ − y 0 sin θ ,
y = x0 sin θ + y 0 cos θ .
(54)
These are known as the rotation equations. We can also invert these to
get
x0 = x cos θ + y sin θ ,
y 0 = −x sin θ + y cos θ .
(55)
We can eliminate a cross-product term by choosing a convenient coordinate
system x0 y 0 . The trick is to find the correct angle to rotate by.
Theorem: For the equation
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 ,
37
(56)
such that B 6= 0, the x0 y 0 -coordinate system obtained by rotating the xy-axes
through an angle θ satisfying
cot 2θ =
A−C
,
B
(57)
then the equation in x0 y 0 -coordinates becomes
A0 x2 + C 0 y 2 + D0 x + E 0 y + F 0 = 0 .
(58)
Proof: If we substitute equation (55) into equation (56), we get
A(x0 cos θ − y 0 sin θ)2 + B(x0 cos θ − y 0 sin θ)(x0 sin θ + y 0 cos θ)
+ C(x0 sin θ + y 0 cos θ)2 + D(x0 cos θ − y 0 sin θ)
+ E(x0 sin θ + y 0 cos θ) + F = 0 ,
(59)
which means
A0
B0
C0
D0
E0
F0
= A cos2 θ + B cos θ sin θ + C sin2 θ
= B(cos2 θ − sin2 θ) + 2(C − A) sin θ cos θ
= A sin2 θ − B sin θ cos θ + C cos2 θ
= D cos θ + E sin θ
= −D sin θ + E cos θ
=F.
(60)
A−C
cos 2θ
= cot 2θ =
,
sin 2θ
B
(61)
B 0 = 2B cos 2θ + 2(C − A) sin θ cos θ = 0
(62)
If we have
then
as required.
Example: Rotate to the coordinates that coincide with the axes of the conic
section x2 + 4xy − 2y 2 − 6 = 0. What type of conic section is it?
Solution: We need to rotate by θ, where
1 − (−2)
3
=
4
4
⇒ θ = 0.463648 rad = 26.5651o
⇒ sin θ = 0.447214 , cos θ = 0.894427 .
cot 2θ =
38
Using these, we can substitute into equation (60) to get
A0 = 2 ,
B0 = 0 ,
C 0 = −3 ,
F 0 = −6 ,
where of course, D0 = E 0 = 0 since there are no linear terms initially. The
equation in the new variables is therefore
2x02 − 3y 02 = 6
x02 y 02
⇒
−
= 1,
3
2
which is a hyperbola.
1.6
Conic Sections in Polar Coordinates
To find how to represent conic sections in polar coordinates, first we must
state a theorem.
Focus-Directrix Property of Conics: Suppose that a point P moves in
the plane, with the motion determined by a fixed point (the focus) and a
fixed line (the directrix) such that the focus does not lie on the directrix.
Then, if the point moves in such a way that its distance to the focus divided
by its distance to the directrix is a constant e, called the eccentricity, then
the curve the point traces is a conic section. The eccentricity determines the
conic section as follows:
(a) a parabola if e = 1 ,
(b) an ellipse if 0 < e < 1 ,
(c) a hyperbola if e > 1 .
(63)
In the case of a parabola, this is fairly easy to see. For a parabola y = 4px
curving to the right, we take the usual focus and directrix x = −p. However,
for an ellipse or a hyperbola we must define the directrix. We will take the
directrix to be the line x = a2 /c if we consider the standard shapes in Figure
34. Below we will call the point P , the focus F and the directrix D. In this
notation, we require
PF
= e.
(64)
DP
39
y
y
2
D
FHx,yL
D
0.5
-1
1
2
FHc,0L
3
x
-2
-1
1
FHc,0L
-0.5
-1
e=1
x=-p
PHx,yL
1.0
1
2
x
-1.0
0<e<1
-2
x=a2 c
-1.5
y
PHx,yL
1.5
D
1.0
FHc,0L
0.5
-4
-2
2
4
x
-0.5
-1.0
e>1
x=a2 c
-1.5
Figure 34: Conic sections defined and eccentricities
For a parabola, we already know that the distance P F to the focus is equal
to the distance to the directrix P D, so e = 1. For the ellipse, Recalling the
fifth line of equation 34, we have
2
p
c
a
(x − c)2 + y 2 =
−x ,
(65)
a c
where the left-hand side is P F , and the bracketed term on the right is P D.
Therefore, for an ellipse,
c
e= ,
(66)
a
which given 0 < c < a gives 0 < e < 1, unless a = 0 or c = 0. What would
the conic section be in these situations? Finally, for the hyperbola, we know
that the distance to the further focus minus the distance to the nearer focus
is 2a. Therefore, in analogy with equation (34), we get
p
p
(x + c)2 + y 2 − (x − c)2 + y 2 = 2a ,
(67)
so performing the same procedures, we arrive at
p
c
a2
2
2
(x − c) + y =
x−
,
a
c
40
(68)
where the left-hand side is again P F and the right-hand side is P D since
the directrix is the opposite side to the point compared with the case of the
ellipse. Therefore, we also have
e=
c
,
a
(69)
for a hyperbola.
Also note that for an ellipse, eccentricity can be regarded as a measure of the
flatness of the ellipse: as e approaches 0, the ellipse becomes circular; as e
approaches 1 the ellipse flattens out. Earth’s orbit has e = 0.017. See Figure
35 for plot of ellipses of different eccentricites.
1.0
0.5
-1.0
F
-0.5
0.5
1.0
1.5
2.0
e=0
e=0.2
e=0.4
e=0.6
e=0.8
-0.5
-1.0
Figure 35: Ellipses with same focus and same semi-major axis, but different
eccentricities
1.6.1
Polar equations of conics
In order to derive the polar equations of conics, it is convenient to chose
the pole to coincide with the focus, and the polar axis to be either parallel
or perpendicular to the directrix. In addition, there are two choices for the
direction the polar axis points. As a result there are four possibilities that
one might consider. We will choose the directrix to be perpendicular to
the polar axis, which will point towards the directrix. Figure 36 shows the
construction. The distance between the pole and directrix is d, see Figure
36. d is constant is set by our choice of pole. We know that
P F = eP D .
41
(70)
PHx,yL
D
r
Θ
F
r cos Θ
Pole
d
Figure 36: Polar coordinates defined for conics
For the situation in the diagram, we have P F = r and P D = d − r cos θ, and
therefore
ed
.
(71)
r=
1 + e cos θ
This can be used to represent a parabola, ellipse or hyperbola. In summary,
for the different cases, we have
(a) Directrix right of pole
ed
r=
1 + e cos θ
(b) Directrix left of pole
ed
r=
1 − e cos θ
(72)
(c) Directrix above pole
ed
r=
1 + e sin θ
(d) Directrix below pole
ed
r=
1 − e sin θ
These are sketched in Figure 37. To sketch conic sections from their polar
coordinates, we need to identify the relevant parameters. For a parabola, d
tell us the distance from the directrix to the focus, and so we can immediately
sketch it.
Example: Sketch r =
2
.
1−cos θ
Solution: This looks like (b) of (72) with d = 2 and e = 1.
42
HaL
HbL
D
D
F
F
HcL
HdL
D
F
F
D
Figure 37: Choices of polar coordinates for conics
For an ellipse, if we consider the first diagram in Figure 38, we see that
r0 = a − c ,
and so
1
a = (r1 + r0 ) ,
2
r1 = a + c ,
(73)
1
c = (r1 − r0 ) .
2
(74)
We also find
r0 r1 = a2 − c2 = b2 ⇒ b =
√
r0 r1 .
(75)
For an hyperbola, if we consider the second diagram in Figure 38, we see
that
r0 = c − a , r1 = c + a ,
(76)
and so
1
c = (r1 + r0 ) .
2
1
a = (r1 − r0 ) ,
2
We also find
r0 r1 = c2 − a2 = b2 ⇒ b =
Finally let’s do a quick example.
Example: Sketch r =
2
.
1+2 sin θ
43
√
r0 r1 .
(77)
(78)
5
a
b
2
r1
r0
c
-5
5
1
a
b
-3
-2
-1
1
c
2
3
-5
-1
a
a
r1
-2
r0
Figure 38: Ellipses and hyperbolas in polar coordinates
Solution: This looks like (c) of (72). We have d = 1 and e = 2. This
is a hyperbola with directrix length 1 above the pole. Because e > 1, the
denominator causes complicated behavior. At θ = 7π/6, sin θ = −1/2,
causing the denominator to be 0, and a switch from r being positive to r
being negative. At θ = 3π/2, sin θ = −1 and is the turning point of the
upper branch of the curve. At θ = 11π/6, r returns to being positive. In
3
11А6
5
y=2
y=43
y=23
2
3А2
1
А2
2
3
4
5
1
6
-5
-3
А6
-2
-1
1
2
3
-1
Figure 39: Rough sketch of a hyperbola in polar coordinates
short, we get r0 by setting θ = π/2 and r1 from θ = 3π
as these represent
2
the vertices of the hyperbola, i.e. the turning points of the branches. Note
that the upper branch appears for r < 0 because in the plot of the function
44
it is below the axis. The function and the hyperbola are drawn in Figure 39.
Then,
2
2
2
2
r0 =
= , r1 = = 2.
(79)
=
1 + 2 sin 3π/2 | − 1|
1 + 2 sin π/2
3
Therefore
2
1
a = (r1 − r0 ) = ,
2
3
1.6.2
√
√
2 3
b = r0 r1 =
,
3
1
4
c = (r1 + r0 ) = .
2
3
(80)
Planetary motion
While you might see more of this in a mechanics course, we will here state
Kepler’s Laws of Planetary Motion:
1. Law of Orbits: The motion of each planet traces an ellipse with the
Sun at one of the foci.
2. Law of Areas: The line joining the Sun to the centre of the planet
sketches out equal areas in equal times.
3. Law of Periods: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
Kepler’s second law is shown graphically in Figure 1.6.2. These laws are valid
1.0
0.5
-2
-1
Earth
1
-0.5
2
Sun
-1.0
Figure 40: The Law of Areas. The two red areas represent two areas traced
out during equal time periods. Of course, this is not to scale.
for any celestial body under the influence of a single gravitational force. As
45
a result, these are only ever approximations as all the planets also exert
gravitational forces on each other, but it is a very good approximation. The
closest point between the bodies during the orbit is called the perigee (i.e.
r0 ), and the farthest point is called the apogee (i.e. r1 ), which are also
known as the perihelion and aphelion when the Sun is orbited, from the
Greek word for the Sun, Helios. Kepler’s third law may be denoted
3
T = a2 ,
(81)
provided a is measured in astronomical units (1AU=1.5 × 108 km), and T is
1.0
Apogee
-2
D
0.5
Perigee
-1
1
2
-0.5
-1.0
-1.5
ae
a
ae
Figure 41: The length parameters for an orbit
measured in Earth years so that T = 1 is the period of the Earth’s orbit and
a = 1 is the distance to the sun. Normally, orbits are specified by eccentricity
and the semi-major axis, so using c = ea and
d=
a
a(1 − e2 )
a
− c = − ea =
,
e
e
e
(82)
we get
(a) r =
(1 − e2 )
1 + e cos θ
(b) r =
(1 − e2 )
1 − e cos θ
(83)
2
(c) r =
(1 − e )
1 + e sin θ
2
(d) r =
(1 − e )
,
1 − e sin θ
and
r0 = a(1 − e) ,
r1 = a(1 + e) .
(84)
The various length parameters are shown in Figure 1.6.2. If there is time,
we will consider Kepler’s laws in greater detail later.
46