lecture 8 (parametric equations)

7/31/2019 Lecture 8 (Parametric Equations)

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Focus-Directrix Equations and Parametric Curves

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54Elementary Analysis 2

Focus-Directrix Equations and Parametric Curves 1/ 1


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What are parametric equations?

Consider a particle moving along the curve below



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Clearly, the curve is not a graph of any function.



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Worse, we may not be able to find a relation relating the x and the ycomponentof the particle at a given time t.



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But we can express the x and the ycomponent of the particle as functions of thevariable t.



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But we can express the x and the ycomponent of the particle as functions of thevariable t.

In this process, we say that we are parametrizingthe curve above and the

equations definingxand yas functions oftare called parametric equations.



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Examples

The curve defined byy= x2 can be parametrized as follows



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Examples

The curve defined byy= x2 can be parametrized as follows

x= t y= t2



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Examples

x= cos t y= sin t



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Examples

x= cos t y= sin t

x2 = cos2 t


l


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Examples

x= cos t y= sin t

x2 = cos2 t

y2

=sin2 t


E l


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Examples

x= cos t y= sin t

x2 = cos2 t

y2

=sin2 t

Thus,

x2+y2 = 1


E l


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Examples

x= cos t y= sin t

x2 = cos2 t

y2

=sin2 t

Thus,

x2+y2 = 1


E l


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Examples

x= cos t y= sin t

x2 = cos2 t

y2

=sin2 t

Thus,

x2+y2 = 1

Note: The particle determined by the above parametric equations move on the

circle in a clockwise motion.


Examples


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Examples

x= cos t y= sin t

x2 = cos2 t

y2

=sin2 t

Thus,

x2+y2 = 1

Note: The particle determined by the above parametric equations move on the

circle in a clockwise motion. Here, we say that the direction of increasing

parameteris clockwise.


Examples


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Examples

Another parametrizations of the previous circle...


Examples


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Examples


x=cos t

y= sin t


Examples


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Examples


x=cos t

y= sin t


Examples


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Examples


x=cos t

y= sin t

circle in a counter-clockwise

motion


Examples


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Examples


x=cos t

y= sin t


motion

x= 2cos ty= 2sint


Examples


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Examples


x=cos t

y= sin t


motion

x= 2cos ty= 2sint


Examples


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p


x=cos t

y= sin t


motion

x=

2cos t

y= 2sint

larger circle...


Examples


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p

x=

h+

rcos t

y= k+ rsint


Examples


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p

x=

h+

rcos t

y= k+ rsint


Examples


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p

x=

h+

rcos t

y= k+ rsint

translated circle...


Examples


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x=

h+

rcos t

y= k+ rsint


x= cos2ty= sin2t


Examples


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x=

h+

rcos t

y= k+ rsint


x= cos2ty= sin2t


Examples


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x=

h+

rcos t

y= k+ rsint


x= cos2ty= sin2t

faster circle...


Examples


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Consider

x= sec t

y= tan t

2 t

2


Examples


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Consider

x= sec t

y= tan t

2 t

2

Then,

x2 = sec2 t

y2 = tan2 t


Examples


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Consider

x= sec t

y= tan t

2 t

2

Then,

x2 = sec2 t

y2 = tan2 tThus,

x2y2 = 1


Examples


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Consider

x= sec t

y= tan t

2 t

2

Then,

x2 = sec2 t

y2 = tan2 tThus,

x2y2 = 1


Exercises


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Describe the curve given by the parametric equations x= 2cos tand y= 3sint.


Exercises


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Describe the curve given by the parametric equations x= 2cos tand y= 3sint.

What can be said about the parametric equations x= cos t,y=sintandx= cos2t,y=sin2t?


Slope of parametrized curves


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A parametrized curve C: x=f(t),y= g(t) is said to be smoothiffand g aredifferentiable and do not vanish simultaneously.




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IfCis a smooth curve we can determinedy

dx

and dx

dy

as follows




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IfCis a smooth curve we can determinedy

dx

and dx

dy

as follows

dy

dx=

dydtdxdt

dx

dy=

dxdtdydt


Example


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Find the equation of the tangent line to the right branch of the hyperbola

x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).

Solution: Note that when t= 4


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).

Solution: Note that when t= 4 we have the point (

2,1).


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


2,1). Thus,


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


2,1). Thus,

dy

dx= dy/dt

dx/dt= sec

2 t

sec ttan t= csc t


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


2,1). Thus,

dy

dx= dy/dt

dx/dt= sec

2 t

sec ttan t= csc t

Thus,

dy

dx

(

2,1)= csc t

t=/4

=

2


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


2,1). Thus,

dy

dx= dy/dt

dx/dt= sec

2 t

sec ttan t= csc t

Thus,

dy

dx

(

2,1)= csc t

t=/4

=

2

Hence, the equation is

y

1=

2x

2


Example


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x=

sec t,y=

tan t,

2 t

2at the point (

2,1).


2,1). Thus,

dy

dx= dy/dt

dx/dt= sec

2 t

sec ttan t= csc t

Thus,

dy

dx

(

2,1)= csc t

t=/4

=

2

Hence, the equation is

y

1=

2x

2y=

2x1


Example


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Findd2y

dx2ifx= t t2,y= t t3.


Example


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Findd2y


Solution:


Example


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Findd2y


Solution:

y = dydx

= 13t2

1

2t


Example


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Findd2y


Solution:

y = dydx

= 13t2

1

2t

Thus,

d2y

dx2= dy

dx


Example


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Findd2y


Solution:

y = dydx

= 13t2

1

2t

Thus,

d2y

dx2= dy

dx= dy

/dtdx/dt


Example


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Findd2y


Solution:

y = dydx

= 13t2

1

2t

Thus,

d2y

dx2= dy

dx= dy

/dtdx/dt

=(12t)(6t)(13t2)(2)

(12t)212t


Example


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Findd2y


Solution:

y = dydx

= 13t2

1

2t

Thus,

d2y

dx2= dy

dx= dy

/dtdx/dt

=(12t)(6t)(13t2)(2)

(12t)212t =

6t26t+2(12t)3


Length of a parametrized curve


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Let C: x=f(t),y= g(t), a t b. Then the length ofCis given by


Length of a parametrized curve


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Let C: x=f(t),y= g(t), a t b. Then the length ofCis given by

L=

b

a

dxdt

2+

dydt

2dt


Example


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Find the length of the

astroid

x= cos3 ty= sin3 t0 t 2


Example


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astroid



Example


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astroid


L = 8/4

0

(3cos2 tsin t)2+ (3sin2 tcos t)2 dt


Example


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astroid


L = 8/4

0


=12

/4

0

2cos tsin t dt


Example


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astroid


L = 8/4

0


=12

/4

0

2cos tsin t dt

= 12/4

0sin2t dt


Example


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astroid


L = 8/4

0


=12

/4

0

2cos tsin t dt

= 12/4

0sin2t dt

= 6 [cos2t]/40


Example


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astroid


L = 8/4

0


=12

/4

0

2cos tsin t dt

= 12/4

0sin2t dt

= 6 [cos2t]/40= 6


lecture 8 (parametric equations)

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