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Curves Defined by Parametric Equations - Classwork
Until now, we have been representing graphs by single equations involving variables x and y. We will now study
problems with which 3 variables are used to represent curves. These equations are called parametric equations.
Suppose a golfer strikes a golf ball that is propelled into the air at an angle of 45o. If the initial velocity of the ball
$x 2
+ x.
is 64 feet per second, the object follows the parabolic path given by y =
128
However, although you have the path of the object, you do not know when the object is at a given time. In order to
do this, we introduce a third variable t, called a parameter. Both variables x and y are written as a function of t,
and you obtain the parametric equations:
x = 32 t 2
and
y = $16 t 2 + 32 t 2
From this set of equations, we can determine that at the time t = 0, the ball is at the point (0, 0). Similarly at the
time t = 1, the ball is at the point 32 2 , 32 2 $ 16 .
(
)
Definition of a Plane Curve
If f and g are continuous functions of t on an interval H, then the equations
x = f ( t ) and y = g( t )
are called parametric equations, with t being the parameter. The set of
points ( x, y ) obtained as t varies over the interval ( t1, t 2 ) is called the graph
of the parametric equations. The parametric equations and the graph taken
together is called a plane curve.
When sketching a curve by hand represented by parametric equations, you
use increasing values of t. Thus the curve will be traced out in a specific
direction. This is called the orientation of the curve. You use arrows to
show the orientation.
Example 1) Sketch the curve described by the parametric equations:
3t
x = t 2 $ 1 and t = , $ 2 < t < 3
2
t
0<
0@
=
@
<
K
x
y
MasterMathMentor.com
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Stu Schwartz
Example 2) Sketch the curve described by the parametric equations:
x = 4 t 2 $ 1 and t = 3t, $ 1 < t < 1.5
t
0@
0HB
=
HB
@
@HB
x
y
Note that both examples trace out the exact same graph. But the speed is different. Example 2’s graph is traced
out more rapidly. Thus in applications, different parametric equations can be used to represent various speed at
which objects travel along paths. Finding a rectangular equation that represents the graph of a set of parametric
equations is called eliminating the parameter. Here is a simple example of eliminating the parameter.
Example 3) Eliminate the parameter in x = t 2 + 3
y = 2t
and
• Solve for t in the second equations
• Substitute in the second equations and simplify
Example 4) Sketch the curve represented by the equations x =
• Use the x equation, solve for t
1
t
and y =
, t > $4
t+4
t+4
• Substitute t into the y equation and eliminate the complex fractions.
• Realize that t n-4, meaning that lim x ( t ) = 0
t +,
Example 5) Sketch the curve represented by
x = 5 sin 8
and
y = 3 cos8
0 < 8 < 2!
• Solve for cos8 and sin 8 in both equations.
• Use the fact that %*9 < 8 + ;$%< 8 = @ to form an equation using only x and y.
• This is a graph of an ellipse centered at (0, 0) with vertices at (5,0 ) and (-5, 0) and minor axis
endpoints at (0, 3), (0, -3). Note that the ellipse is traced counterclockwise as 8 goes from 0 to 2".
• What would occur if the equations were x = 5 sin 8
MasterMathMentor.com
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and
y = 5 cos8 I 999999999999
Stu Schwartz
Using the technique in the example above you can conclude that the graph of the parametric equations
x = h + a sin 8
( x $ h)
a
2
2
+
and
( y $ k)
y = k + b cos8
0 < 8 < 2! is the ellipse (traced clockwise) given by
2
= 1 which has center ( h8 k ) and endpoints ( a, 0), ($a, 0), (0, b) and (0, $b) .
b2
Eliminating the parameter is an aid to curve sketching. If the parametric equations represent the path of a moving
object, the graph alone is not sufficient to describe the object’s motion. You still need the parametric equations to
tell you the position, direction, and speed at a given time.
Example 6) Find a set of parametric equations to represent the graph of y = x $ x 2 using each of the following
dy
parameters: a) t = x
b. the slope m = at the point ( x, y )
dx
a) This one is easy. Just let t = x in the equation:
dy
b) Since m = , differentiate the equation and solve for x
dx
Now substitute for x in the original equation.
Note that when graphed in parametric mode, the curve has a right to left orientation determined by
the direction of increasing values of slope m. For part a), the curve has the opposite orientation.
Example 7) At any time t with = < t < @= , the coordinates of P are given by the parametric equations:
x = t $ 2 sin t and
y = 2 $ 2 cos t
Sketch this using your calculator.
t=3
t=4
t=2
t=10
t=8
t=1
Z P$*9,% ;$--&%#$9:*92 ,$ *9,&2&- 5.6"&% $3 , .-&
%+$)9H C, , = @8 P +.% ;$$-:*9.,&% $3 (0HLN8H O<)\ .,
,+*% *9%,.9,8 P *% +&.:*92 .67$%, :"& 9$-,+H
Z 4+& 3"66 ;"-5& *% 9$, ,+& 2-.#+ $3 . 3"9;,*$9\ %$7&
1 5.6"&% +.5& 7$-& ,+.9 $9& ( 5.6"&H
Z 4+& #*;,"-& %+$)% ,+& 10 .9: (0 .1&% /", 9$ ,0.1*%H
Z 4+& /"66&,% $9 ,+& 2-.#+ .##&.- ., &V".6 ,*7&
*9,&-5.6% /", 9$, ., &V".6 :*%,.9;&% 3-$7 &.;+ $,+&/&;."%& P %#&&:% "# .9: %6$)% :$)9 .% *, 7$5&%H
U& )*66 %$$9 %&& +$) ,$ ;.6;"6.,& ,+& %#&&: $3 .
#.-.7&,-*; ;"-5& ., . #$*9,H
t=9
t=5
t=7
t=0
t=6
Example 8) Parametric curves may have loops, cusps, vertical tangents and other peculiar features.
and
0<t<5
Graph a) x = 2 cos t + 2 cos( 4 t )
y = sin t + sin( 4 t )
0 < t < 2!
b) x = sin(5 t ) and y = sin(6 t )
t=5
a.
MasterMathMentor.com
t=0
b.
This is called a Lissa_ou curve.
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Stu Schwartz
Curves Defined by Parametric Equations - Homework
1. Consider the parametric equations x = t
and
y = 2t $ 1
a) Complete the table
t
x
y
0
1
2
3
4
b) Plot the points (x, y) in the table and sketch a graph of the parametric equations. Indicate the
orientation of the graph.
c) Find the rectangular equation by eliminating the parameter.
0$! ! 3
2. Consider the parametric equations x = 4 cos8 and y = 6 sin 2 8 on 2 , 5
1 2 24
a) Complete the table
t
"
-2
"
-4
0
"
4
"
2
x
y
b) Plot the points (x, y) in the table and sketch a graph of the parametric equations. Indicate the
orientation of the graph.
c) Find the rectangular equation by eliminating the parameter.
3. In the following exercises, eliminate the parameter and confirm graphically that the rectangular equations
yield the same graph as the parametrics. Be sure you take domain and range of the parametric into account.
a. x = 4 t $ 1 and
c. x = 3 t and
MasterMathMentor.com
y = 2t + 3
y = 3$ t2
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b. x = t + 3 and
y = t2
d. x = t 2 $ t and
y = t2 + t
Stu Schwartz
e. x = t $ 2 and
g. x = sec 2 8
y=
and
t
t $2
f. x = t $ 3
and
y=t+3
y = tan 2 8
h. x = cos8
and
y = 4 sin 8
4. Use your calculators to graph the curve represented by the parametric equations. Indicate the orientation of the
curve. Identify any points at which the curve is not smooth. Do not take these problems lightly. `our
task is to come up with an appropriate window to view them. Let your t run from 0 to 2!, 4!, 8!,
etc.
a. Cycloid: The curve traced by a point on the
circumference of a circle as it rolls on a straight line.
x = 2(8 $ sin 8 ) and y = 2(1 $ cos8 )
b. Prolate Cycloid: Same as a) except the point
goes below the line (railroad track)
x = 28 $ 4 sin 8 and y = 2 $ 4 cos8
Scale
Scale
k
k
,
,
k
k
l
l
c. Hypocycloid: x = 3 cos3 8 and y = 3 sin 3 8
,
,
d. Curtate cycloid: x = 28 $ sin 8 and y = 2 $ cos8
Scale
Scale
k
k
,
,
k
k
l
l
e. Witch of Agnesi: x = 2 cot 8 and y = 2 sin 2 8
f. Folium: x =
Scale
k
k
MasterMathMentor.com
,
,
l
l
- 152 -
l
l
3t
3t 2
=
and
y
1+ t3
1+ t3
Scale
k
k
l
l
,
,
,
,
Stu Schwartz
l
l
Using Calculus With Parametric Equations - Classwork
If a smooth curve B is given by the equations
x = f ( t ) and y = g( t ) , then the slope of B at ( x8 y) is
dy
dy
= dt , dx = 0
dt
dx dx
dt
Note that the slope formula is in terms of t, not x. so you need
to find the value of t corresponding to the point (x, y)
Example 1) Find
dy
for the curve given by x = cos t and y = $ sin t
dx
Find the slope of this curve at the point (1, 0) and (0, -1)
dy
is a function of t , you can use the rule above repeatedly to find
dt
higher@order derivatives. For instance:
Because
d 0 dy 3
2
d y d 0 dy 3 dt 21 dx 54
=
=
dx
dx 2 dx 21 dx 54
dt
Find the second derivative
d2y
of the parametric equation above.
dx 2
Example 2) for the curve given by x = t and y =
MasterMathMentor.com
d 0d2y 3
2
5
d 3 y d 0 d 2 y 3 dt 1 dx 2 4
= 2
5 = dx
dx 3 dx 1 dx 2 4
dt
1 2
(t $ 2t ) , find the slope and concavity at (2, 4)
2
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Stu Schwartz
T1.7#6& K) 4+& #.-.7&,-*; ;"-5& *% 2*5&9?
x = t $ 2 sin t and y = 2 $ 2 cos t
a) find the slope at t = 1.
b) Where is the curve horizontal?
c) where is the curve vertical?
We know we can find the arc length of a curve B given by y = h ( x ) over the interval [ x 0 , x1 ] by
x2
s=
#
2
x2
2
1 + [ h :( x )] dx =
% dy (
1 + ' * dx
& dx )
#
x1
x1
If B#is represented by the parametric equations x = f ( t )
x2
s=
#
x1
% dy (
1 + ' * dx , you can write
& dx )
#
x1
x2
Arc Length
2
x2
2
1 + [ h :( x )] dx =
#
s =
x1
2
% dy (
1 + ' * dx =
& dx )
2
b
=
(dx dt ) + (dy
#
2
(dx dt )
a
b
=
and y = g( t ), a < t < b
#
a
2
2
x2
% dy dt (
1+ '
* dx
& dx dt )
#
x1
2
dt ) % dx (
' *dt =
& dt )
2
% dx ( % dy (
' * + ' * dt =
& dt ) & dt )
b
b
#
a
2
(dx dt )
2
2
+ ( dy dt ) % dx (
' *dt
% dx (
& dt )
' *
& dt )
# [ f :(t )] + [g:(t )]
2
dt
a
Note that this formula only works when the curve does not intersect itself
on the interval a < t < b and the curve must be smooth.
Exmple 4) A circle of radius r has rectangular equation x 2 + y 2 = r 2 and parametric equations
x = r cos8 y = r sin 8 . Show that its arc length (circumference) is 2" using
a) Rectangular arc length formula
MasterMathMentor.com
b) Parametric arc length formula
- 154 -
Stu Schwartz
Example 5) A circle of radius 1 rolls around the circumference of a larger circle of radius 5. The epicycloid
traced by a point on the circumference of the smaller circle is given by
x = 5 sin t $ sin 5 t
and
y = 5 cos t $ cos 5 t
.) G-.#+ ,+& #.-.7&,-*; &V".,*$9% 3$t = 0 to 2!
[$6, 6], [$6, 6]
]*&) *, ^$$7 %V".-&H
b) Note the curve has cusp points when t = 0 and t = ! 2 . Between these points, dx dt and dy dt are
not simultaneously zero (cusps). So the portion of the curve generated from t = 0 and t = ! 2 is
smooth. So, to find the total distance traveled by the point we can find the arc length in the 1st
quadrant and multiply the result by 4. While it may be possible just to integrate from 0 to 2" (and it is in
this case, it can be dangerous. It is easier to integrate on sections you are sure it is differentiable.
Area of a surface of revolution
If a smooth curve B is given by x = f ( t ) and y = g( t ), a < t < b , and B does not intersect itself, then the area
of the surface of revolution about the coordinate axes is given by:
t= b
S = 2!
# g(t )
t= a
t= b
S = 2!
# f (t )
t= a
2
2
2
2
% dx ( % dy (
' * + ' * dt - Revolution about x - axis : g( t ) 9 0
& dt ) & dt )
% dx ( % dy (
' * + ' * dt - Revolution about y - axis : f ( t ) 9 0
& dt ) & dt )
Example 6) Find the surface area when x = 4 sin t and y = 4 cos t
MasterMathMentor.com
- 155 -
0<t<
!
is rotated about the x axis.
6
Stu Schwartz
Using Calculus With Parametric Equations - Homework
2
In the following exercises, find dy dx and d y 2 and evaluate each at the indicated value of the parameter.
dx
1. x = 3t
and
3. x = 2 t $ 2
5. x = t
y = 4t + 1
and
and
y = t2 $ 4t
y = t $1
2. x = t
t =2
t =1
t =2
and
y = 4t $1
4. x = 2 cos t
and
y = 2 sin t
6. x = sin 3 t
and
y = cos3 t
t=4
t=
t=
5!
4
3!
4
In the following 2 exercises, find the equation of the tangent line at the indicated points on the curve.
7. x =
2
and y = 2 sin 2 8 at (2,1)
tan 8
MasterMathMentor.com
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Stu Schwartz
(
8. x = $4 cos8 and y = 3 + 2 sin 8 at $2,3 + 3
)
In the following exercises, find all points of horizontal and vertical tangency to the curve.
9. x = 2 $ t and
y = t2
10. x = t + 4 and
y = 2t 2 + 6t + 1
11. x = 8 cos2 8 and
13. x = t 2 + t $ 2
and
MasterMathMentor.com
y = $4 sin 8
y = t 3 $ 3t
12. x = 28 and
14. x = 2 cos8 and
- 157 -
y = 2(1 $ cos8 )
y = $2 sin 28
Stu Schwartz
Find the arc length of the given curve on the indicated interval. Calculators permitted on 15 and 16.
15. x = t 2 $ t and
y = 4t3 + 2
17. x = 2e$ t sin t and
18.
x=
t5
1
+ 3
10 6 t
MasterMathMentor.com
and
y = 2e$ t cos t
y=t
16. x = t
,]
[$11
[1, 3] .
and
y=3 t
[1, 2]
[0, !]
Explain how this complicated problem can be made easier.
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Stu Schwartz
19. The path of a soccer ball is modeled by the equations x = (100 cos 30 o ) t and y = (100 sin 30 o ) t $ 16 t 2
where x and y are measured in feet. Graph the path of the projectile and use the integration capabilities
of the calculator to approximate the arc length of the path and the difference between the arc length
and the range of the soccer ball.
3t
3t 2
=
and
y
, t 9 0 , sketch it on your calculator.This graph
1+ t3
1+ t3
is called a folium. It looks like a leaf (foliage).
20. Given the parametric equations x =
a) Determine what values of t give a horizontal tangent.
b) Approximate the arc length of the closed loop. Set up the integral and use your calculator appropriately
21. Find the area of the surface generated by revolving the curve about the given axes. Use calculator on a).
a)
x = t .9: y = N $ At 8
*) x $ .1*%
MasterMathMentor.com
[=8 <]
b.
**) y $ .1*%
x = L ;$%8 .9: y = L %*9 8 8
*) x $ .1*%
- 159 -
[=8 ! < ]
**) y $ .1*%
Stu Schwartz
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