Download Parameterization of Curves in Two Dimensions and more Schemes and Mind Maps Mathematics in PDF only on Docsity!
Parametric Equations
Part 1: Vector-Valued Functions
Now that we have introduced and developed the concept of a vector, we are ready to use vectors to deÖne functions. To begin with, a vector-valued function is a function whose inputs are a parameter t and whose outputs are vectors r (t). In 2 dimensions, a vector-valued function is of the form
r (t) = hf (t) ; g (t)i
Moreover, the set of position vectors of the form r (t) = hf (t) ; g (t)i for t in [a; b] forms a curve whose orientation is in the direction in which the parameter is increasing. The position r (a) is called the initial point of the curve, and the position r (b) is the curveís terminal point.
The function r (t) = hf (t) ; g (t)i ; t in [a; b] ; is called a parameterization of the curve, and in order to sketch the curve, we often let
x = f (t) ; y = g (t) ; t in [a; b]
and then eliminate the parameter t in hopes of obtaining a familiar equation in x and y:
EXAMPLE 1 Sketch the curve parameterized by
r (t) = 1 t^2 ; 2 t^2 t^4 ; t in [0; 1]
Solution: To begin with, x = 1 t^2 ; which means that t^2 = 1 x: Since y = 2t^2 t^4 ; that means that
y = 2 (1 x) (1 x)^2 (1)
which simpliÖes to y = 1 x^2. Moreover, the initial and terminal points are
initial: (t = 0) x = 1 02 = 1; y = 2 0 04 = 0 terminal: (t = 1) x = 1 12 = 0; y = 2 1 14 = 1
That is, the graph of r (t) = 1 t^2 ; 2 t^2 t^4 for t in [0; 1] is the section of the graph of y = 1 x^2 between the initial point (1; 0) and the terminal point (0; 1) :
Let C be the curve parameterized by r (t) = hf (t) ; g (t)i, t in [a; b] : If f (t) and g (t) are continuous on [a; b] ; if r (a) = r (b) ; and if r (t) has a well-deÖned orientation, then the curve C is a closed curve.
EXAMPLE 2 Sketch the curve parameterized by u () = hcos () ; sin ()i for in [0; 2 ] :
Solution: Since x = cos () and y = sin () ; we begin with
cos^2 () + sin^2 () = 1
Thus, the identity implies that
x^2 + y^2 = 1
which is the unit circle. Moreover, the initial and terminal points are initial: ( = 0) x = cos (0) = 1; y = sin (0) = 0 terminal: ( = 2) x = cos (2) = 1; y = sin (2) = 0
In addition, we may also on occasion use identities such as 2 sin (t) cos (t) = sin (2t) and ete t^ = 1.
EXAMPLE 3 Sketch the curve parameterized by r (t) = h3 cos (t) ; 2 sin (t)i for t in [0; 2 ] :
Solution: Since x = 3 cos (t) and y = 2 sin (t) ; we begin with
cos^2 (t) + sin^2 (t) = 1
Substituting cos (t) = x= 3 and sin (t) = y= 2 thus yields (^) x 3
(^) y 2
If t = 0 or t = 2; then x = 3 and y = 0: Thus, the initial and terminal points are the same and correspondingly the graph of r (t) = h3 cos (t) ; 2 sin (t)i for t in [0; 2 ] is the entire ellipse with counterclockwise orientation.
x^2 9
y^2 4
Unfortunately, a parameterization may not deÖne an orientation for a curve, in that the parameterization may trace the curve from the initial point to a point on the curve and then may retrace the curve back to where it started, as we will see in the next example.
EXAMPLE 4 Find the Cartesian equation and sketch the curve r (t) = hcos (2t) ; cos (t)i for t in [0; 2 ].
Solution: Since x = cos (2t) and y = cos (t) ; we use a double angle identity:
cos (2t) = 2 cos^2 (t) 1 x = 2 y^2 1
Moreover, r (0) = hcos (0) ; cos (0)i = h 1 ; 1 i ; r (=2) = hcos () ; cos (=2)i = h 1 ; 0 i ; and r () = hcos (2) ; cos ()i = h 1 ; 1 i : Also, r (3=2) =
hcos (3) ; cos (3=2)i = h 1 ; 0 i and r (2) = hcos (4) ; cos (2)i = h 1 ; 1 i : Thus, the curve starts at (1; 1) ; progresses to ( 1 ; 1) ; and then retraces the curve in returning to (1; 1) : Consequently, the ori- entation is not well-deÖned.
EXAMPLE 5 Find the Cartesian equation of the curve
r () = 2 sin () cos () ; 2 sin^2 () ; in [0; ]
Solution: Since 2 sin () cos () = sin (2) and 2 sin^2 () = 1 cos (2) ; we can rewrite r as
r () = hsin (2) ; 1 cos (2)i
which implies that x = sin (2) and y = 1 cos (2) : Since
sin^2 (2) + cos^2 (2) = 1
and since cos (2) = 1 y; we must have
x^2 + (y 1)^2 = 1
Thus, r () = 2 sin () cos () ; 2 sin^2 () is a circle with radius 1
then its standard parameterization is given by
r () = p + R u ()
where p = (p; q) is the position vector of the center. Straightforward vector arithmetic subsequently yields
r () = hp + R cos () ; q + R sin ()i (2)
where is in [0; 2 ].
EXAMPLE 6 Determine the standard parameterization of a circle of radius 1 centered at (0; 1) ; and then verify that it is a circle by eliminating the parmeter.
Solution: Substituting p = 0; q = 1; and R = 1 into (2) yields
r () = hcos () ; 1 + sin ()i
To verify this, notice that x = cos () and y = 1 + sin () implies that x^2 + (y 1)^2 = cos^2 () + sin^2 () = 1 Notice that the same curve is parameterized in example 5.
An object is said to be in uniform circular motion if it is moving in a circle with a parameterization (2) in which
=! (t t 0 )
where the constant! is the angular speed in radians per second and t 0 is the time at which vanishes (i.e., is equal to 0). That is, the parameterization of the position of an object in uniform circular motion is given by
r (t) = hp + R cos (! (t t 0 ) ; q + R sin (! (t t 0 )))i
Alternatively, uniform circular motion shows us that a curve (such as a circle) has inÖnitely many parameterizations. Examples 5 and 6 are also illustrations of this fact.
EXAMPLE 7 Two objects are moving on a circle with radius 2 centered at (3; 4) with angular speeds of 2 radians per second and 3 radians per second, respectively. Parameterize the motion of each object and compare the two given that both are at = 0 initially.
Solution: To begin with, = 0 initially means that t 0 = 0: Thus, if we denote the two motions by r 1 (t) and r 2 (t) ; then
r 1 (t) = h3 + 2 cos (2t) ; 4 + 2 sin (2t)i r 2 (t) = h3 + 2 cos (3t) ; 4 + 2 sin (3t)i
Moreover, the second object will traverse the circle 3 times for every 2 traversals of the circle by the Örst object, as is shown in the Ögure below.
The fundamental period of a uniform circular motion r (t) is the smallest positive number T for which r (t + T ) = r (t) (3)
That is, T is the time it takes an object in uniform circular motion to make one complete cycle around the circle. Since = 2 radians is the angle corresponding to one complete cycle, it follows that
! =
T
Thus, if we are given the period of a uniform circular motion, we can use (4) to calculate its angular speed.
Many oriented curves are not deÖned by a simple equation in x and y: For example, the curve
r (t) = hcos (3t) ; sin (t)i ; t in [ 20 ; 20]
is much too complicated to be represented by a simple equation in x and y; as is shown below:
In fact, one of the most important oriented curves in mathematicsó the cy- cloid ó is a curve in which the parameter cannot be eliminated.
EXAMPLE 9 Parameterize a cycloid, which is the path traced out by a point on the edge of a wheel of radius R as it rolls along a horizontal path at a constant speed.
Solution: To begin with, let us suppose that the point begins at the origin, and let us also suppose that x and y are the xy-coordinates
of the point on the circle at time t: If is the angle at time t formed by a radius to the point and a vertical line, then the point will have traveled a distance R at time t:
Moreover, the distance the center of the circle has moved will be the same as the distance the point on the rim has moved, so that the center of the circle will have moved a distance R horizontally at time t: The horizontal distance from the point to a vertical radius is R cos () ; so that
x + R cos () = R =) x = R R cos ()
Similarly, the vertical distance from the horizontal line to the center is the sum of the y-coordinate and R sin () : However, since the radius of the circle is R; we must have
y + R sin () = R =) y = R R sin ()
Finally, the center of the circle is moving at a constant speed, which implies that there is a constant! such that = !t: Thus, the para- meterization of a cycloid is given by
x = R!t R cos (!t) y = R R sin (!t)
where R and! are constants.
Exercises
Find the Cartesian equation in x and y of the graph of the given vector-valued function. Then sketch the graph showing orientation, if the orientation is well-
- Prove the following: If r (t) = hf (t) ; g (t)i ; t in [a; b] ; is a parameterization of a curve C; then so also is
rk (t) =
f
t k
; g
t k
; t in [ka; kb]
for every number k > 0 :
- Prove the following: If r (t) = hf (t) ; g (t)i ; t in [a; b] ; is a parameterization of a curve C; then
r (t) = hf ( t) ; g ( t)i ; t in [ b; a]
is a parameterization of C with the opposite orientation.
- The curve parameterized by r (t) = ht + 1 sin (t) ; 1 cos (t)i ; t in [0; 4 ] ; is an example of a Cycloid of Clairaut. Sketch it using a graphing calculator or a computer algebra system.
- The curve parameterized by r (t) = 2 cos (t) + 2 cos^2 (t) ; 2 sin (t) + sin (2t) ; t in [0; 2 ] ; is called a cardoid. Sketch it using a graphing calculator or a com- puter algebra system.
- The curve parameterized by r (t) = cos^3 (t) ; sin^3 (t) ; t in [0; 2 ] ; is called an astroid.
- (a) Sketch it using a graphing calculator or a computer algebra system. (b) Show that the equation of the astroid is
jxj^2 =^3 + jyj^2 =^3 = 1
- The set of curves parameterized by r (t) = ae t=^10 cos (t) ; be t=^10 sin (t) ; t in [0; 8 ] ; for a and b constants is a family of spirals which spiral toward the origin.
- (a) Sketch the curves corresponding to several values for a and b using a graphing calculator or a computer algebra system. (b) Show that the curve satisÖes the system of di§erential equations
dx dt
x y; x (0) = a dy dt
y + x; y^0 (0) = b
(c) We say that the system of equations in (b) has a stable equilibrium at (0; 0) : Use (a) to explain what this term means for this system of di§erential equations.
- The earth orbits the sun in a nearly circular orbit with a period of 365. days and a radius of R = 93; 000 ; 000 miles: What is the parameterization of the earthís orbit about the sun?
- Even as you read this, you are moving in a circle as the earth rotates. The radius R of your circle of motion depends on your latitude, ; where
R = 3963 cos ( )
Find out what your latitude is, use a period of 24 hours, and determine the parameterization for your approximate circle of motion.
- What is the parameterization of the circle in the animation shown below:
- What is the angular speed of the circle in problem 43? (Hint: it takes 20 seconds for the blue dot to move all the way around the circle).
- Write to Learn: A Reuleaux Triangle is a curve formed by circular arcs connecting vertices of an equilateral triangle with sides of length r; where each
boundary of are the same distance apart.
How "wide" is a Reuleaux triangle? Why is it of constant width?
