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Focus-Directrix Equations and Parametric Curves
Institute of Mathematics, University of the Philippines Diliman
Mathematics 54Elementary Analysis 2
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Consider a particle moving along the curve below
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What are parametric equations?
Clearly, the curve is not a graph of any function.
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What are parametric equations?
Clearly, the curve is not a graph of any function.
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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What are parametric equations?
Clearly, the curve is not a graph of any function.
Worse, we may not be able to find a relation relating the x and the ycomponentof the particle at a given time t.
But we can express the x and the ycomponent of the particle as functions of thevariable t.
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What are parametric equations?
Clearly, the curve is not a graph of any function.
Worse, we may not be able to find a relation relating the x and the ycomponentof the particle at a given time t.
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
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l
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Examples
x= cos t y= sin t
x2 = cos2 t
y2
=sin2 t
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E l
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Examples
x= cos t y= sin t
x2 = cos2 t
y2
=sin2 t
Thus,
x2+y2 = 1
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E l
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Examples
x= cos t y= sin t
x2 = cos2 t
y2
=sin2 t
Thus,
x2+y2 = 1
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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.
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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.
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Examples
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Examples
Another parametrizations of the previous circle...
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Examples
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Examples
Another parametrizations of the previous circle...
x=cos t
y= sin t
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Examples
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Examples
Another parametrizations of the previous circle...
x=cos t
y= sin t
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Examples
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Examples
Another parametrizations of the previous circle...
x=cos t
y= sin t
circle in a counter-clockwise
motion
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Examples
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Examples
Another parametrizations of the previous circle...
x=cos t
y= sin t
circle in a counter-clockwise
motion
x= 2cos ty= 2sint
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Examples
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Examples
Another parametrizations of the previous circle...
x=cos t
y= sin t
circle in a counter-clockwise
motion
x= 2cos ty= 2sint
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Examples
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p
Another parametrizations of the previous circle...
x=cos t
y= sin t
circle in a counter-clockwise
motion
x=
2cos t
y= 2sint
larger circle...
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Examples
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p
x=
h+
rcos t
y= k+ rsint
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Examples
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p
x=
h+
rcos t
y= k+ rsint
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Examples
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p
x=
h+
rcos t
y= k+ rsint
translated circle...
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Examples
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x=
h+
rcos t
y= k+ rsint
translated circle...
x= cos2ty= sin2t
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Examples
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x=
h+
rcos t
y= k+ rsint
translated circle...
x= cos2ty= sin2t
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Examples
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x=
h+
rcos t
y= k+ rsint
translated circle...
x= cos2ty= sin2t
faster circle...
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Examples
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Consider
x= sec t
y= tan t
2 t
2
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Examples
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Consider
x= sec t
y= tan t
2 t
2
Then,
x2 = sec2 t
y2 = tan2 t
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Examples
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Consider
x= sec t
y= tan t
2 t
2
Then,
x2 = sec2 t
y2 = tan2 tThus,
x2y2 = 1
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Examples
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Consider
x= sec t
y= tan t
2 t
2
Then,
x2 = sec2 t
y2 = tan2 tThus,
x2y2 = 1
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Exercises
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Describe the curve given by the parametric equations x= 2cos tand y= 3sint.
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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?
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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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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.
IfCis a smooth curve we can determinedy
dx
and dx
dy
as follows
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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.
IfCis a smooth curve we can determinedy
dx
and dx
dy
as follows
dy
dx=
dydtdxdt
dx
dy=
dxdtdydt
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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).
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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).
Solution: Note that when t= 4
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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).
Solution: Note that when t= 4 we have the point (
2,1).
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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).
Solution: Note that when t= 4 we have the point (
2,1). Thus,
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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).
Solution: Note that when t= 4 we have the point (
2,1). Thus,
dy
dx= dy/dt
dx/dt= sec
2 t
sec ttan t= csc t
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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).
Solution: Note that when t= 4 we have the point (
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
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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).
Solution: Note that when t= 4 we have the point (
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
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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).
Solution: Note that when t= 4 we have the point (
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
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
y = dydx
= 13t2
1
2t
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
y = dydx
= 13t2
1
2t
Thus,
d2y
dx2= dy
dx
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
y = dydx
= 13t2
1
2t
Thus,
d2y
dx2= dy
dx= dy
/dtdx/dt
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
y = dydx
= 13t2
1
2t
Thus,
d2y
dx2= dy
dx= dy
/dtdx/dt
=(12t)(6t)(13t2)(2)
(12t)212t
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Example
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Findd2y
dx2ifx= t t2,y= t t3.
Solution:
y = dydx
= 13t2
1
2t
Thus,
d2y
dx2= dy
dx= dy
/dtdx/dt
=(12t)(6t)(13t2)(2)
(12t)212t =
6t26t+2(12t)3
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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
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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
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
L = 8/4
0
(3cos2 tsin t)2+ (3sin2 tcos t)2 dt
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
L = 8/4
0
(3cos2 tsin t)2+ (3sin2 tcos t)2 dt
=12
/4
0
2cos tsin t dt
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
L = 8/4
0
(3cos2 tsin t)2+ (3sin2 tcos t)2 dt
=12
/4
0
2cos tsin t dt
= 12/4
0sin2t dt
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
L = 8/4
0
(3cos2 tsin t)2+ (3sin2 tcos t)2 dt
=12
/4
0
2cos tsin t dt
= 12/4
0sin2t dt
= 6 [cos2t]/40
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Example
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Find the length of the
astroid
x= cos3 ty= sin3 t0 t 2
L = 8/4
0
(3cos2 tsin t)2+ (3sin2 tcos t)2 dt
=12
/4
0
2cos tsin t dt
= 12/4
0sin2t dt
= 6 [cos2t]/40= 6
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