299
Section 10.7 – Parametric Equations
Objective 1:
Defining and Graphing Parametric Equations.
Recall when we defined the x(rcos(θ), rsin(θ))
and y-coordinates on a circle of
radius r as a function of θ. The
value of x and y depended on
r
the value of θ. Notice that the
θ
graph of the equation of the
2
2
2
circle, x + y = r , is not a function
since you can find a vertical line
that passes through more than
one point of the curve, But, by
defining x = rcos(θ) and y = rsin(θ)
on the interval [0, 2π), each
coordinate is a function of θ.
This means we can view each coordinate as a separate curve.
x = rcos(θ) is the curve of the cosine function with amplitude r and
y = r sin(θ) is the curve of the sine function with amplitude r. When we
graph the points (x, y) on the graph to obtain the circle of radius r, the
variable θ no longer is part of the equation for the plane curve. We call the
variable θ in this case a parameter and the equations x = rcos(θ) and
y = rsin(θ) parametric equations.
Definition
Let x = f(t) and y = g(t) be two functions defined on the interval I. The set of
all points (x, y) = (f(t), g(t)) for all t in I is called the plane curve. The
equations x = f(t) and y = g(t) for all t in I are called the parametric
equations and t is called the parameter.
Suppose the parametric equations are defined on the interval a ≤ t ≤ b. As
we plot the points (x, y) as t goes from a to b, the successive points trace
out the graph in a certain direction. The direction the graph is traced out as
t goes from a to b is called the orientation. Thus, in the case of the circle
of radius r discussed above, as t goes from 0 to 2π, the graph gets traced
out in a counterclockwise direction so the orientation would be
counterclockwise. To indicate the orientation on a graph, we draw arrows
on the plane the curve showing the direction that curve is traced out.
300
Graph the curve whose parametric equations are given and show its
orientation:
Ex. 1
x = 0.5t,
y = 0.25t2 – 4 ≤ t ≤ 2
Solution:
Let's begin by making a table of values and deriving a set of points
(x, y) to plot:
t
x = 0.5t
y = 0.25t2
(x, y)
–4
–2
4
(– 2, 4)
–3
– 1.5
2.25
(– 1.5, 2.25)
–2
–1
1
(– 1, 1)
–1
– 0.5
0.25
(– 0.5, 0.25)
0
0
0
(0, 0)
1
0.5
0.25
(0.5, 0.25)
2
1
1
(1, 1)
Now, plot the points and sketch the graph:
To verify the graph, we can get rid
of the parameter. In the equation
x = 0.5t, we can solve for t to get
t = 2x. Now, substitute 2x in for t
in the equation y = 0.25t2 and
simplify:
y = 0.25(2x)2
y = 0.25(4x2)
y = x2
Since – 4 ≤ t ≤ 2, then
– 4 ≤ 2x ≤ 2 or – 2 ≤ x ≤ 1. Thus,
y = x2 for – 2 ≤ x ≤ 1
Notice that it matches our graph.
Objective 2: Finding a rectangular equation for a curve defined
parametrically.
In the last example, we took care to pay attention to the domain of the
function after getting rid of the parameter. Many times, the graph of the
curve after getting rid of the parameter will contain more points than the
parameterized curve so we will need to pay careful attention to which piece
of the curve actually belongs on the graph.
301
Find the rectangular equation of the curve and sketch the graph:
Ex. 2
x = 3cos(t), y = 4sin(t); t ≥ 0
Solution:
First, solve each of the parametric equations for the trigonometric
function:
x = 3cos(t)
y = 4sin(t)
x
y
cos(t) =
sin(t) =
3
2
2
Recall, that cos (t) + sin (t) = 1,
so
or
€
2
2
( ) +( )
x
3
x2€ y 2
+
9
16
y
4
=1
4
=1
€
Thus, the graph is an ellipse
€ with center
€
at (0, 0), major axis
running along the y-axis,
€vertices (0, 4) and (0, – 4) and
b-intercepts (3, 0) and (– 3, 0),
Since t starts at zero, the graph
starts at (3, 0) and traces out in a counterclockwise fashion.
Suppose t had been restricted to a much smaller interval. Although, the
process would be exactly the same as we did above, we may need to
eliminate part of the graph depending on the restriction on t. If 0 ≤ t ≤ π
was the restriction, we would only have the upper half of the graph. If
π
3π
≤t≤
was the restriction, we would only have the left side.
2
2
π
2
0≤t≤π
€
€
€
€
≤t≤
3π
2
302
Ex. 3
x = 3sin(t), y = 2cos(t);
–π≤t≤0
Solution:
First, solve each of the parametric equations for the trigonometric
function:
x = 3sin(t)
y = 2cos(t)
x
y
sin(t) =
cos(t) =
3
2
2
Recall, that cos (t) + sin (t) = 1,
2
so
x2€
y2
+
9
4
or
2
( ) +( )
y
2
x
3
=1
2
=1
€
Thus, the graph is an ellipse
€ with center
€
at (0, 0), major axis
running along the x-axis,
€vertices (3, 0) and (– 3, 0) and
b-intercepts (0, 2) and (0, – 2),
Since t starts at – π, the graph
starts at (0, – 2) and traces out in a clockwise fashion. It ends at
(0, 2).
€
Ex. 4
x = sec(t), y = tan(t);
Solution:
Recall, that tan2(t) + 1 = sec2(t),
so
(y)2 + 1 = (x)2
or
2
2
€
x –y =1
Thus, the graph is a hyperbola
with center at (0, 0), transverse
axis running along the x-axis,
vertices (1, 0) and (– 1, 0) and
oblique asymptotes y = ± x
π
Since t starts at – , the graph
–
π
6
≤t≤
€
6
starts at (sec(–
=(
2 3
3
3
π
6
), tan(–
π
6
))
,–
) ≈ (1.15, – 0.58)
€ 3
and traces out in a clockwise fashion. It ends at
π€
π €
(sec( ), tan( )) = (2, 3 ) ≈ (2, 1.73).
3
€
3
€
€
€
€
π
3
303
Objective 3
Time as a parameter.
A common use of parametric equations is when t represents the time. We
can describe the path of an object using time as the parameter. The
advantage with using parametric equations in this situation is that not only
can we determine the path of the object, but also the time that the object
reaches a certain point on the path. In calculus, we can derive the
parametric equations that determine the path of a projectile through the air.
Parametric Equations of a Projectile Motion
Let an object be propelled upward at an
angle θ with the horizontal from a height h
vo
above the horizontal. If the initial speed is
θ
vo, then, neglecting for air resistance,
h
the resulting motion is called Projectile
Motion and the parametric equations
for this motion are:
1
x = (vocos(θ))t
and y = – gt2 + (vosin(θ))t + h
2
where t is the time in seconds and g is the acceleration due to gravity (on
Earth, g ≈ 32 ft/s2 or 9.8 m/s2).
€
Solve the following:
Ex. 5 Standing on top of a hill that was six feet above the horizontal,
George Johnson hits a golf ball at 125 ft/sec at an angle of 35˚ from
the horizontal.
a) Find the parametric equations that describe the position of the ball
as a function of time.
b) How long was the ball in the air?
c) Find the horizontal distance the ball travelled before hitting the
ground.
d) When is the ball at its maximum height? What is that maximum
height?
Solution:
a)
The initial velocity vo = 125 ft/sec. Since the velocity is given in
ft/sec, we will use g = 32 ft/sec2. The angle θ = 35˚ and h is 6 ft.
x = (vocos(θ))t = 125cos(35˚)t ≈ 102.3940t
y=–
1 2
gt
2
+ (vosin(θ))t + h = –
≈ – 16t2 + 71.6971t + 6
€
€
1
•32t2
2
+ (125sin(35˚))t + 6
304
b)
Thus, our parametric equations are:
x ≈ 102.3940t
and y ≈ – 16t2 + 71.6971t + 6
When the ball hits the ground, the height the ball above the
ground is 0. To find how long the ball was in the air, we need to
first find when the ball hit the ground. Setting y = 0 and solving
yields:
y ≈ – 16t2 + 71.6971t + 6 = 0 (use the quadratic formula)
−71.6971± (71.6971)2 −4(−16)(6)
−b± b2 −4ac
t=
≈
2(−16)
2a
−71.6971±74.3268
=
≈ 4.563 seconds or – 0.082
−32
c)
d)
Since the negative answers does not sense, the ball was in the
€
€ seconds.
air for about 4.563
Since the ball was in the air for approximately 4.563 seconds,
€then the horizontal distance travelled is:
x ≈ 102.3940(4.563) ≈ 467.2489 ft.
Notice that the equation for y is a quadratic equation with the
coefficient of the squared term being negative. Thus, the
maximum value (k) will occur at the vertex (h, k).
−71.6971
−b
h=
≈
≈ 2.2405 seconds
2a
€
seconds
2(−16)
Plugging t = 2.2405 into y, we get the maximum height:
y ≈ – 16(2.2405)2 + 71.6971(2.2405) + 6 ≈ 86.3198 ft
Thus, the ball reached a maximum height of about 86.3198 ft at
t€≈ 2.2405 seconds.
If you graph the function using
equations from part a, we need
to remember the x-values are
the horizontal distance travelled
and not the time.
Ex. 6 At 3 pm, Juan gets in his car and drives towards Atlanta at an
average speed of 55 mph. Two hour later, LaTonya leaves from the
same location and travels along the same route to Atlanta at 77 mph.
How long will it take LaTonya to catch Juan? Use a simulation of two
motions to verify your answer.
Solution:
Let t be the time that Juan is on the road. Since LaTonya leaves two
hours later, her time on the road is t – 2. Since distance = rate × time,
305
then the distance that Juan travels is x1 = 55t and the distance that
LaTonya travels is x2 = 77(t – 2). At the point when LaTonya catches
Juan, each will have travelled the same distance, so set x1 = x2 and
and solve for the time t:
x1 = x2
55t = 77(t – 2)
(distribute)
55t = 77t – 154
(subtract 77t from both sides)
– 22t = – 154
(divide both sides by – 22)
t = 7 hours.
LaTonya's time on the road is t – 2 = 5.
Thus, it takes LaTonya 5 hours to catch Juan.
To verify the answer, we will use a simulation of two motions. To do
this, we will graph the distance each of them travelled at different
times until their positions become aligned. To make it easier to see
the alignment on the graph, we will let y1 = 2 and y2 = 4. Thus, on our
graph, the line that is generated at y = 2 will represent the distance
Juan has travelled and the line generated at y = 4 will represent the
distance that LaTonya travelled.
t = 2, t – 2 = 0
t = 3, t – 2 = 1
t = 4, t – 2 = 2
t = 5, t – 2 = 3
t = 6, t – 2 = 4
t = 7, t – 2 = 5
306
Notice the distance are aligned at t = 7 hours in the simulation. This
verifies our answer.
Objective 4:
Finding Parametric Equations.
Given an equation in rectangular form defined as a function y = f(x), the
most basic parametric equations to use is the let x = t and y = f(t) for t in the
domain of f. Sometimes, the conditions of the application dictate that such
a set of parametric equations will not work. In that case, our choice of
parametric equations must allow for the variable x to still be able to match
all the values in the domain of f. For instance, if the domain of f is [– 3, 2],
we cannot choose x = t2 since this would only allow for values of x ≥ 0.
Find two different sets of parametric equations for the following:
Ex. 7
y = – 2x2 + 3
Solution:
Let x = t. Then y = – 2t2 + 3, so the parametric equations are:
x = t, y = – 2t2 + 3,
–∞<t<∞
Another pair that would work would be to let x = t3. Then y = – 2t6 + 3,
so the parametric equations are:
x = t3, y = – 2t6 + 3,
–∞<t<∞
Find parametric equations for the following object in motion:
Ex. 8
x2
4
+ y2 = 1
where the parameter t is on seconds and
a) the motion around the ellipse is clockwise, begins at the point (– 2, 0),
and requires 2 seconds to make one revolution.
b) the motion around the ellipse is counterclockwise, begins at the point
(0,€1), and requires 4 seconds to make one revolution.
Solution:
a) The equation is an ellipse
with center at (0, 0), major
along the x-axis, vertices of
(– 2, 0) and (2, 0) and
b-intercepts (0, 1) and (0, – 1).
Since the ellipse begins at
(– 2, 0), then at t = 0, x = – 2
and y = 0.
307
Thus, x = – 2cos(ωt) and y = sin(ωt) or y = – sin(ωt). Since the
orientation is clockwise, then as the curve leaves the point (– 2, 0),
the y-values will be positive, so y = sin(ωt). Hence, our equations are:
x = – 2cos(ωt) and y = sin(ωt)
Since, as the graph starts to get traced out, the x-values are negative
and the y-values are positive, so ω has to be greater than 0. Also, it
takes 2 seconds to make one revolution, the period is 2 =
2π
ω
which
2π
ω
which
implies ω = π. Therefore, our parametric equations are:
x = – 2cos(πt) and y = sin(πt)
0≤t≤2
b) The equation is an ellipse
€
with center at (0, 0), major
along the x-axis, vertices of
(– 2, 0) and (2, 0) and
b-intercepts (0, 1) and (0, – 1).
Since the ellipse begins at
(0, 1), then at t = 0, x = 0
and y = 1.
Thus, x = 2sin(ωt) or – 2sin(ωt)
and y = cos(ωt). Since the
orientation is counterclockwise,
then as the curve leaves the
point (0, 1), the x-values will be
negative, so x = – 2sin(ωt). Hence, our equations are:
x = – 2sin(ωt) and y = cos(ωt)
Since, as the graph starts to get traced out, the x-values are negative
and the y-values are positive, so ω has to be greater than 0. Also, it
takes 4 seconds to make one revolution, the period is 4 =
implies ω =
π
2
. Therefore, our parametric equations are:
x = – 2sin(
π
2
t) and y = cos(
π
2
t)
0≤t≤4
€
There are some curves that cannot be represented easily in terms of x and
y, but are€very easy to represent using parametric equations. One example
is a curve called a cycloid. Its path is traced out by a point on the rim of a
€ a as the circle rolls
€ along a fixed line without slipping. To
circle of radius
derive the parametric equations, let P = (x, y) be a point on the rim of a
circle of radius a such that P starts at the origin. We will allow the circle to
roll along the positive x-axis:
308
Y
O
P
C
t
B
X
A
2a
Let C be the center the circle and t be the angle of how much the circle has
rotated. Since the circle has rolled the distance from O to A, d(O, A), then
the amount of the circle that has travelled on the ground is the arc length
from A to P. Since there is no slippage, then
arc length AP = d(O, A)
The arc length AP = rθ = at since the radius r = a and t is the angle defined
above. This implies that d(O, A) = at.
Also, at = d(O, A) = d(O, X) + d(X, A) which implies d(O, X) = at – d(X, A)
In examining the diagram, d(X, A) = d(P, B) which in the length of the
opposite side of t in triangle PCB. Thus PB = asin(t) which implies
d(O, X) = at – d(X, A) = at – asin(t) = a(t – sin(t))
Since d(O, X) = x, then x = a(t – sin(t))
Now, d(A, C) = d(A, B) + d(B, C) or d(A, B) = d(A, C) – d(B, C),
but d(A, C) = a and d(B, C) is the adjacent side to t in triangle PCB which
means d(B, C) = acos(t). Substituting these relationships in, we get:
d(A, B) = a – acos(t) = a(1 – cos(t))
Since y = d(A, B), then y = a(1 – cos(t))
π
Although we have proved this for t being less than or equal to , these
equations can be extended to any real number t.
2
Parametric Equations for a Cycloid
x = a(t – sin(t)), y = a(1 – cos(t)), – ∞ < t < ∞. The period
will be 2πa.
€
Suppose an object is moving without friction from a starting point P to an
ending point Q that is lower than P, but not directly below P. If gravity is the
only force acting on it, then the quickest way it can get to Q is along the
path of an inverted cycloid. This is known as the curve of quickest decent.
Another example is suppose that several objects are placed on an inverted
at different locations of cycloid at the same time and began sliding at the
same moment. If none of the objects were placed on the lowest point of the
cycloid, they all will reach the lowest point at the same time.