14.1
PARAMETRIC EQUATIONS
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Programming a Robot Using a Parameter
Example 1
If t is time in minutes, describe the path followed by a robot given by
x = 2t, y = t for 0 ≤ t ≤ 5.
Solution
At time t = 0, the robot’s position is given by
x = 2 · 0 = 0, y = 0, so it starts at the point (0, 0).
One minute later, at t = 1, its position is given by
x = 2 · 1 = 2, y = 1, so it has moved to the point (2, 1).
At time t = 2, its position is given by
x = 2 · 2 = 4, y = 2, so it has moved to the point (4, 2).
y
5
Robot stops here
The path followed by the robot is given by
t=5
4
starts
here
t=4
3
t=3
2
1
t=0
(0, 0) → (2, 1) → (4, 2) → (6, 3) → (8, 4) → (10, 5).
t=2
t=1
2
4
6
8
10
x
At time t = 5, the robot stops at the point
(10, 5) because we have restricted the
values of t to the interval 0 ≤ t ≤ 5. In the
figure we see the path followed by the
robot; it is a straight line.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Programming a Robot Using a Parameter
Example 2 (a)
A robot begins at the point (1, 0) and follows the path given by the
equations x = cos t, y = sin t where t is in minutes, 0 ≤ t ≤ 6.
(a) Describe the path followed by the robot.
Solution (a)
At time t = 0, the robot’s position is given by
x = cos 0 = 1, y = sin 0 = 0, so it starts at the point (1, 0).
One minute later, at t = 1, its position is given by (cos1,sin1) ≈ (.54, .84)
Replacing t by 2, 3, 4, 5, and 6, the path followed by the robot is
(1,0) → (0.54,0.84) → (−0.42,0.91) → (−0.99,0.14) → (−0.65,−0.76)
y
→ (0.28,−0.96) → (0.96,−0.28)
t=1
t=2
t=3
t=0
x
Robot stops here
t=4
t=5
t=6
In the figure we see the path followed
by the robot; it is circular with a radius
of one meter. At the end of 6 minutes
the robot has not quite returned to its
starting point at (1, 0).
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Eliminating the Parameter t
Example 2 (b)
A robot begins at the point (1, 0) and follows the path given by the
equations x = cos t, y = sin t where t is in minutes, 0 ≤ t ≤ 6.
(b) What happens when you try to eliminate the parameter t ?
Solution (b)
One way to eliminate t from this pair of equations is to use the
Pythagorean identity, cos2t + sin2t = 1. Since x = cos t and y = sin t, we can
substitute x and y into this equation: x2 + y2 = 1. This is the equation of a
circle, centered at the origin, with radius 1. Solving for y2, we have
y2 = 1 – x2 and, solving for y, we get:
y
y   1  x 2 or y   1  x 2
The first solution represents the upper
half of the circle, while the second
solution represents the bottom half
of the circle.
x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Different Motions Along the Same Path
Example 4
Describe the motion of the robot that follows the path given by:
(a) x = cos(−t), y = sin(−t) for 0 ≤ t ≤ 6 (b) x = cos t, y = sin t for 0 ≤ t ≤ 10
Solution
(a) The robot travels around
(b) The robot travels around the
the circle in the clockwise
circle more than once but less
direction, opposite to that in
than twice, coming to a stop
Examples 2 and 3.
southwest of the landing site.
y
t=8
t=5
Robot
stops here
t=4
t=6
t=0
t=3
t=2
t=1
t=2
y
t=1
t=7
t=9
t=3
x
t=0
x
t=6
t=10
Robot
stops here
t=4
t=5
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
The Archimedean Spiral
In polar coordinates, the Archimedean spiral is the graph of
the equation introduced on page 352: r = θ.
Since the relationship between polar coordinates and
Cartesian coordinates is x = r cos θ and y = r sin θ, we can
write the Archimedean spiral r = θ as x = θ cosθ and y = θ sinθ.
Replacing θ (an angle) by t (a time), we obtain the parametric
y
equations for the
t = 5π/2
spiral in the figure:
5
x = t cos t
t = π/2
t = 2π t = 4π
t=π
t = 3π
y = t sin t.
x
5
t=0
5
10
5
t = 3π/2
10
t = 7π/2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Graphs to Parameterize a Curve
Example 5
The first two graphs show functions, f(t) and g(t). Describe the motion of
the particle whose coordinates at time t are given by x = f(t), y = g(t).
x
y
f(t)
1
0
y
g(t)
1
1
2
3
4
t
1
0
1
2
3
4
t
0
1
Solution Curve
x
Solution
As t increases from 0 to 1, the x-coordinate increases from 0 to 1, while
the y-coordinate stays fixed at 0. The particle moves along the x-axis from
(0, 0) to (1, 0). As t increases from 1 to 2, the x-coordinate stays fixed at x
= 1, while the y-coordinate increases from 0 to 1. Thus, the particle moves
along the vertical line from (1, 0) to (1, 1). Between times t = 2 and t = 3, it
moves horizontally backward to (0, 1), and between times t = 3 and t = 4 it
moves down the y-axis to (0, 0). Thus, it traces out the square.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
14.2
IMPLICITLY DEFINED
CURVES AND CIRCLES
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Conic Sections
The curves known as conic sections include two
we are already familiar with, circles and parabolas.
They also include ellipses and hyperbolas. Conic
sections are so-called because, as was
demonstrated by the Greeks, they can be
constructed by slicing, or sectioning, a cone. Conic
sections arise naturally in physics, since the path of
a body orbiting the sun is a conic section. We will
study them in terms of parametric and implicit
equations. As we have already studied parabolas,
we now focus on circles, ellipses and hyperbolas.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Circles
Example 2
Graph the parametric equations
x = 5 + 2 cos t and y = 3 + 2 sin t.
Solution
We see that x varies between 3 and 7 with a midline of 5 while y
varies between 1 and 5 with a midline of 3. The figure gives a graph
of this function. It appears to be a circle of radius 2 centered at the
y
point (5, 3).
5
4
center is (5, 3)
3
radius = 2
2
1
1
2
3
4
5
6
7
x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Parametric and Nonparametric
Equations of a Circle
For r > 0, the parametric equations of a circle of
radius r centered at the point (h, k) are:
x = h + r cos t y = k + r sin t
0 ≤ t ≤ 2π.
An implicit equation for the circle of radius r
centered at the point (h, k) is:
(x − h)2 + (y − k)2 = r2.
This is called the standard form of the equation
of a circle.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Equations for Circles
Example 3 +
Write the equation of this circle in standard form.
x = 5 + 2 cos t and y = 3 + 2 sin t.
Solution
We have h = 5, k = 3, and r = 2. This gives
(x − 5)2 + (y − 3)2 = 22 = 4.
y
5
4
center is (5, 3)
3
radius = 2
2
1
1
2
3
4
5
6
7
x
If we expand the equation for the circle
in Example 3, we get a quadratic
equation in two variables:
x2 − 10x + 25 + y2 − 6y + 9 = 4, or
x2 + y2 − 10x − 6y + 30 = 0.
Note that the coefficients of x2 and y2
(in this case, 1) are equal. Such
equations often describe circles; we put
them into standard form by completing
the square.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Equations for Circles
Example 4
Describe in words the curve defined by the equation
x2 + 10x + 20 = 4y − y2.
Solution
Rearranging terms gives
x2 + 10x + y2 − 4y = − 20
We complete the square for the terms involving x and (separately)
for the terms involving y:
(x2 + 10x + 25) + (y2 − 4y + 4) − 25 − 4 = −20
(x + 5)2 + (y – 2)2 − 29 = −20
(x + 5)2 + (y – 2)2 = 9 = 32
This equation is a circle of radius r = 3 with center (h, k) = (−5, 2).
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
14.3
ELLIPSES
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Parametric Equations for an Ellipse
The parametric equations of an ellipse centered
at (h, k) are:
x = h + a cos t y = k + b sin t 0 ≤ t ≤ 2π.
We usually take a > 0 and b > 0.
In the special case where a = b, these equations give a
circle of radius r = a. Thus, a circle is a special kind of
ellipse. The parametric equations of an ellipse are
transformations of the parametric equations of the
unit circle, x = cos t, y = sin t. These transformations
have the effect of shifting and stretching the unit circle.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
The Graph of an Ellipse
Example 1
Graph the ellipse given by
x = 7 + 5 cos t, y = 4 + 2 sin t,
Solution
The center of the ellipse is (h, k) = (7, 4).
The value a = 5 determines the horizontal
“radius” of the ellipse, while b = 2
determines the vertical “radius”. The
value of x varies from a maximum of 12
to a minimum of 2 about the vertical
midline x = 7. Similarly, the value of y
varies from a maximum of 6 to a
minimum of 2 about the horizontal
midline y = 4. The ellipse is symmetric
about the midlines x = 7 and y = 4.
0 ≤ t ≤ 2π.
Graph of the ellipse
and midlines
x=7
y
6
(7,4)
y=4
4
2
2
4
6
8
10
12
14
x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
The Major and Minor Axes of an Ellipse
Example 2
Graph the ellipse given by
x = 7 + 2 cos t, y = 4 + 5 sin t,
0 ≤ t ≤ 2π.
How is this ellipse similar to the one in Example 1? How is it different?
Solution
This ellipse has the same center (h, k) =
(7, 4) as the ellipse in Example 1.
However, as we see in the figure, the
vertical axis of this ellipse is the longer
one and the horizontal axis is the
shorter one. The longer axis of an
ellipse is called the major axis and the
shorter axis is called the minor axis. In
Example 1, the horizontal axis was the
major axis, but in this example the
vertical axis is the major axis.
Graph of the ellipse
and its midlines
y
10
x=7
8
6
(7,4)
4
y=4
2
2
4
6
8
10
x
12
2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Eliminating the Parameter t in the
Equations for an Ellipse
The implicit equation for an ellipse centered at
(h, k) and with horizontal axis 2a and vertical
2
2
(
x

h
)
(
y

k
)
axis 2b is:

1
a2
b2
As with circles, we can eliminate t from the parametric
equations for an ellipse by first rewriting the equations
for x and y as follows:
x = h + a cos t → x – h = a cos t → (x – h)/a = cos t
y = k + b sin t → y – k = b sin t → (y – k)/b = sin t
Then the result follows by applying the Pythagorean
Functions Modeling Change:
Identity.
A Preparation for Calculus,
4th Edition, 2011, Connally
14.4
HYPERBOLAS
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Introduction to Hyperbolas
We now use another form of the Pythagorean identity to
parameterize a curve. With tan t = sin t/cos t and sec t =
1/cos t, on page 370 we showed that
sec2 t − tan2 t = 1.
We use this identity to investigate the curve described by
the parametric equations
x = sec t and y = tan t.
By eliminating the parameter t, we get
x2 − y2 = 1.
This implicit equation looks similar to the equation for a
unit circle x2 + y2 = 1. However, the curve it describes is
very different. Solving for y2, we find that
y2 = x2 − 1.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Graphing the Unit Hyperbola
The hyperbola with the equation
y2 = x2 − 1 or x2 − y2 = 1
is called the Unit Hyperbola by analogy to the unit circle. The graphs
of other hyperbolas will be viewed as transformations of the unit
hyperbola.
y
5
A graph of x2 − y2 = 1
3
The X-shaped graph of the
asymptotes y2 = x2 (y = ± x)
x
5
3
3
5
has been dashed in.
3
Note that , in order for y2 to be
positive (and y real), |x| ≥ 1.
5
It has two branches, one to the right of the y-axis and one to the
left. As x grows large in magnitude, either toward +∞ or − ∞, the
graph approaches the asymptotes, y2 = x2.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
A General Formula for Hyperbolas
The parametric equations for a hyperbola
centered at (h, k) and opening to the left
and right are
x = h + a sec t y = k + b tan t 0 ≤ t ≤ 2π.
We usually take a > 0 and b > 0.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Graphing Hyperbolas
Example 2
Graph the equation
( y  7) 2 ( x  4) 2

1
25
9
Graph of the hyperbola
opening up and down
and its asymptotes
y
25
Solution
Since h = 4 and k = 7, we have shifted
the unit hyperbola 4 units to the right
and 7 units up and it will open up and
down. Since we have a2 = 9 (a = 3) and
b2 = 25 (b = 5), we have stretched the
unit hyperbola horizontally by a factor
of 3 and vertically by a factor of 5. The
equations of the asymptotes are
( x  4)
( y  7)

9
25
2
2
Note the red box with labels,
demonstrating the roles of a and b.
15
5
6
2a=6
(4,7) 2b=10
4
x
14
10
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Eliminating the Parameter t in the
Equations for a Hyperbola
The implicit equation for a hyperbola,
( x  h) 2 ( y  k ) 2

1
2
2
a
b
describes a hyperbola that opens left and right. Its
asymptotes are diagonal lines through the corners of a
rectangle of width 2a and height 2b centered at the
point (h, k). The graph with equation
( y  k ) 2 ( x  h) 2

1
2
2
b
a
has a similar shape, except that it opens up and down.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
14.5
GEOMETRIC PROPERTIES
OF CONIC SECTIONS
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Geometric Definitions
of Circles and Ellipses
A circle of radius r is the set of points in the plane a
distance r from a given point.
An ellipse is the set of points in the plane for which the
sum of the distances to the foci is constant.
• Each of the two points is called a focus or focal point
of the ellipse.
• The two focal points lie on the major axis, and are
equally spaced about the minor axis.The constant
sum of distances is the length of the major axis.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Locating the Focal Points of an Ellipse
Example 2
By finding a formula for c in terms of a and b, determine the focal points
(±c, 0) for the ellipse where a and b are positive and a > b:
x2/a2 + y2/b2 = 1
Solution
From the geometric definition of an ellipse,
y
the sum of the distances to the two focal
b
points is the same at every point on the
curve. Thus, if the point is (a, 0), the sum of
a
a
the distances to (c, 0) and (-c, 0) = (a – c) + (a
x + c) = 2a. Similarly, if the point is (0, b), the
-a -c
c a
sum of the distances to (c, 0) and (-c, 0) is the
same:
c 2  b 2  c 2  b 2  2a,
-b
Graph of x2/a2 + y2/b2 = 1
with focal points
making the distance from (0, b) to (c, 0) “a.”
Using the Pythagorean Theorem a2 = b2 + c2
or
c  a 2  b2 .
Functions Modeling Change:
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4th Edition, 2011, Connally
The Focal Points of an Ellipse
Location of the Focal Points of an Ellipse
The focal points of the ellipse x2/a2 + y2/b2 = 1
with a > b > 0 are (±c, 0) where
c  a b .
2
2
The focal points lie on the major axis.
If b > a, the focal points are (0, ±c) with
c  b a .
2
2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Reflective Properties of Ellipses
Reflective property of an ellipse
A ray that originates at one focus of an ellipse is
reflected off the ellipse to the other focus.
Point of
reflection
●
Focus 1: Point
of origination
Focus 2:
End point
of reflection
This reflective property,
is used in applications
involving light, sound, and
shock waves
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Reflective Property of an Ellipse
Example 3
An elliptical whispering gallery has a major axis of 100 m
and minor axis of 40 m. Where should two speakers
position themselves so as to most easily hear each other?
Solution
Imagine the room centered at the origin with major axis
on the x-axis. Then the positive x-intercept is a = 50 and
the positive y-intercept is b = 20. The focal points are
located at
c   a 2  b2   502  202   2100  45.8.
The speakers should position themselves approximately
4.2 m in from the ends of the major axis.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Geometric Definition of Hyperbolas
A hyperbola is the set of points in the plane for which
the difference of the distances to two points is
constant. Each of the two points is called a focus or
focal point of the hyperbola.
Location of the Focal Points of a Hyperbola
The focal points of the hyperbola
given by equation x2/a2 − y2/b2 = 1
F1
are (±c, 0) where
c  a b .
2
y
●
F2
A
x
2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Locating the Focal Points of a Hyperbola
Example 4
(a) Find the focal points and the asymptotes for the hyperbola given by
x2/9 − y2/16 = 1.
(b) Graph the hyperbola.
(c) For each of the two x-intercepts, find the difference of the distances to
the two focal points.
Solution
Since a = 3 and b = 4, c   a 2  b2   9  16  5. So the focal points
are located at (5,0) and (-5,0) and the asymptote equations are y = ±4/3 x
Plot of the hyperbola with asymptotes
y
4
5
3
3
4
5
x
The intercept (3, 0) is distance 2
from one focus and distance 8
from the other focus. The
difference of the distance to the
two focal points is 6. Similarly, the
difference of the distances to the
intercept (−3, 0) is also 8 − 2 = 6.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Reflective Properties of Hyperbolas
Reflective property of hyperbolas:
A beam aimed at the far focus behind the hyperbola is
reflected off the curve to hit the focus in front of the
hyperbola. Alternately, a beam originating from the focus
in front of the hyperbola is reflected off the curve so that it
appears to have originated from the focus behind the
y
hyperbola.
Point of
This reflective property,
is used in navigational
systems and is associated
with sonic booms.
reflection
●
x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Reflective Property of a Hyperbola
Example 5
A hyperbolic mirror has equation x2/16 − y2/9 = 1. Find the
equation of the line along which a beam of light could be sent out
from point A = (15, 50) to reflect off the hyperbola and arrive at the
focus closest to point A.
Solution
The focal points for this hyperbola are (±c, 0) with c  16  9  5.
The focal point closest to A is (5, 0) and the focal point farthest away
is (−5, 0). Since we want the beam of light to reflect off the
hyperbola and arrive at (5, 0), we aim the beam at the other focal
point, (−5, 0). The line from point A to (−5, 0) has slope m where:
m = (50 – 0)/(15 – (-5)) = 50/20 = 2.5 and its equation is
y = 2.5(x − (−5)) = 2.5x + 12.5.
A beam of light sent along the line y = 2.5x + 12.5 toward the point
(−5, 0) is reflected off the hyperbola to the point (5, 0).
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Geometric Definition of Parabolas
A parabola is the set of points in the plane
for which the distance from a point is
equal to the distance from a fixed line. The
point is called the focus and the line is
called the directrix .
Functions Modeling Change:
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Location of the Focus and
Directrix of a Parabola
The focus of the parabola y = ax2 is the
point (0, c) where c = 1/(4a).
The directrix is the line y = −c.
Even though they
may not look it, the
two red line
segments are
the same length
y
Focus
x
Directrix
Functions Modeling Change:
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4th Edition, 2011, Connally
Locating the Focus and Directrix of a Parabola
Example 7
(a) Find the focus and the directrix of the parabola y = 3x2.
(b) Show that the point (1, 3) on the parabola is equidistant from the focus
and the directrix.
Solution
We have a = 3 so the focus is at the point (0, 1/12) and the directrix is the
line y = −1/12.
We have the distance from (1, 3) to the focus (0, 1/12):
2
2
2
1

(
3

1
/
12
)

1

(
35
/
12
)
 37 / 12.
d1 =
The distance from (1, 3) to the directrix y = −1/12 is the difference in the
y-values, since the directrix is a horizontal line:
d2 = 3 + 1/12 = 37/12.
Thus the distance from the point (1, 3) to the focus equals the distance
from the point (1, 3) to the directrix. This property holds for all points
(x, y) on the parabola.
Functions Modeling Change:
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4th Edition, 2011, Connally
Reflective Properties of Parabolas
Reflective property of parabolas
All beams parallel to the axis of symmetry
reflect off the parabola to the focus.
Giant parabolic mirrors are
used in telescopes to focus
light and radio waves from
outer space. Car headlights
have parabolic reflectors
with the lightbulb at the
focus.
Focus
Point of
reflection
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Parabolic Mirrors in Car Headlights
Example 8
The parabolic reflector behind a car head light is 16 cm
across and 3 cm deep. Where should the bulb be
positioned?
Solution
The shape of the parabolic reflector is made by rotating a
parabola y = ax2 about the y-axis. Since the rim of the
reflector lies on the parabola, the point (8, 3) satisfies the
equation y = ax2 so 3 = a(82) and a = 3/64. The focus is
located at: c = 1/(4a) = 1/(4(3/64)) = 16/3.
The bulb should be placed 16/3 cm from the vertex of the
parabola along the axis of symmetry.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
14.6
HYPERBOLIC FUNCTIONS
Functions Modeling Change:
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Hyperbolic Sine and
Hyperbolic Cosine
Hyperbolic Sine and Hyperbolic Cosine
x
x
x
x
e e
e e
cosh x 
sinh x 
2
2
From these definitions, we can see that
• cosh 0 = 1, sinh 0 = 0
• cosh(−x) = cosh x, sinh(−x) = −sinh x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Plots of the Hyperbolic Sine and Cosine
The figures show graphs of cosh x and sinh x together with
dashed-line graphs of ½ ex and ½ e−x or -½ e−x. The graphs
suggest that cosh x is even and sinh x is odd and that cosh x
has a y-intercept of 1,whereas sinh x has a y-intercept of 0:
y
y
y = cosh x
4
4
y = -½ e−x
2
y = ½ ex
2
y = sinh x
y = ½ ex
3
3
x
3
3
x
y = -½ e−x
2
2
4
4
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Long-Term Behavior of the Hyperbolic Cosine
Example 4
Describe and explain the behavior of cosh x as x →∞ and then
as x → − ∞.
Solution
From the figure, we see that as x →∞, the graph of cosh x
resembles the graph of ½ ex. Similarly, as x → − ∞, the graph of
cosh x resembles the graph of ½ e-x
Since e−x → 0 as x →∞ and ex → 0 as x → − ∞, we can predict
these results algebraically:
x
x
e e
1
 ex
2
2
e x  ex
1 x
As x  , cosh x 
 e
2
2
y
As x  , cosh x 
4
y = cosh x
y = -½ e−x
3
2
y = ½ ex
3
x
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Identities Involving cosh x and sinh x
cosh2 x − sinh2 x = 1
This can be shown using the fact that ex · e−x = 1:
2
2
e e  e e 
  

(cosh x)  (sinh x)  
 2   2 
e 2 x  2e x  e  x  e 2 x e 2 x  2e x  e  x  e 2 x


4
4
e 2 x  2  e 2 x  e 2 x  2  e 2 x

4
1
x
2
x
x
x
2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Defining the
Hyperbolic Tangent Function
Hyperbolic Tangent
tanh x = sinh x/cosh x.
y
1
3
3
x
1
Graph of the hyperbolic tangent
and its horizontal asymptotes
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Parameterizing the Hyperbola
Using Hyperbolic Functions
Consider the curve parameterized by the equations
x = cosh t, y = sinh t,
−∞ < t < ∞.
We can eliminate the parameter t by using the identity just derived:
x2 − y2 = cosh2 x − sinh2 x = 1
This is the implicit equation for the unit hyperbola.
Notice that since cosh t > 0, we have x > 0, so the parameterization gives
only the right branch of the hyperbola.
x = cosh t, y = sinh t
y
As t →∞, both cosh t and sinh t approach
t=2
½ et, so for large values of t we see that
y=x
cosh t ≈ sinh t. Thus, as t increases the
t=0
t=1
curve draws close to the diagonal line y =
x
x. Similarly, as t → − ∞, we know that x →
t=-1
½ e−t and y → − ½ e−t, which means that y
y=-x
→−x. Thus, as t → − ∞, the curve draws
t=-2
close to the asymptote y = −x.
4
2
1
2
3
4
2
4
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally