

ALGEBRA


Parametric equations

Parametric
equations definition

When
Cartesian coordinates of a curve or a surface are represented as
functions of the same variable (usually written t),
they are
called the parametric equations. 
Thus,
parametric equations in the xyplane 
x
= x(t)
and y
= y(t) 
denote
the x
and y
coordinate of the graph of a curve in the plane. 

::
The parametric equations of a line

If
in a coordinate plane a line is defined by the point P_{1}(x_{1},
y_{1}) and the
direction vector s
then, the position or (radius)
vector r
of any point P(x,
y) of the line 
r
= r_{1} + t · s,

oo
< t < + oo
and
where, r_{1}
= x_{1}i + y_{1} j
and s
= x_{s}i + y_{s} j, 
represents the vector equation of the line. 
Therefore,
any point of the line can be reached by the radius vector 
r
= xi + y j
= (x_{1} + x_{s}t) i
+ (y_{1} + y_{s}t) j 
since the scalar quantity t
(called the
parameter) can take any real value from 
oo
to +
oo. 
By
writing the scalar components of the above vector equation obtained is 
x
= x_{1} + x_{s} · t 
y
= y_{1} + y_{s} · t 
the parametric
equations of the line. 




To
convert the parametric equations into the Cartesian coordinates solve
given equations for t
and equate obtained expressions, thus




Therefore,
the parametric equations of a line passing through
two points P_{1}(x_{1},
y_{1}) and P_{2}(x_{2},
y_{2})

x
= x_{1} + (x_{2} 
x_{1}) t 
y
= y_{1} + (y_{2} 
y_{1}) t 

Parametric
curves have a direction of motion 
When
plotting
the points of a parametric curve by increasing t, the
graph of the function is traced
out in the direction of motion.


Example:
Write
the parametric equations of the line y
= (1/2)x
+ 3 and sketch its graph.

Solution: Since 

Let
take the xintercept as
the given point P_{1}, so 
for
y
= 0 => 0 = (1/2)x
+ 3, x
= 6 therefore,
P_{1}(6,
0). 
Substitute
the values, x_{1}
= 6, y_{1}
= 0, x_{s}
= 2,
and y_{s}
= 1
into the parametric equations of a line

x
= x_{1} + x_{s} · t,
x
= 6 + 2t 
y
= y_{1} + y_{s} · t,
y = t 

The direction of motion (denoted by red arrows) is given by increasing t.


Example:
Write
the parametric equations of the line through points, A(2,
0) and B(2,
2) and sketch the graph.

Solution: Plug the coordinates x_{1}
= 2, y_{1}
= 0, x_{2}
= 2, and y_{2}
= 2
into the parametric equations of a line

x
= x_{1} + (x_{2} 
x_{1}) t,
x
= 2
+ (2 + 2) t
= 2 +
4t,
x
= 2 +
4t, 
y
= y_{1} + (y_{2} 
y_{1}) t,
y = 0
+ (2 
0) t
= 2t,
y
= 2t. 
To
convert the parametric equations into the Cartesian coordinates solve

x
= 
2 +
4t for
t
and plug into y
= 2t 
therefore, 



The direction of motion (denoted by red arrows) is given by increasing t.



Use
of parametric equations, example

Intersection point of a line and a plane
in three dimensional space

The point of intersection is a common point of a line and a plane. Therefore, coordinates of intersection must
satisfy both equations, of the line l
and the plane P


where, (x_{0},
y_{0},
z_{0}) is a
given point of the line and s =
ai + bj + ck
is direction vector of the line, and
N
=
Ai + Bj + Ck
is the normal vector of the given plane. 
Let
transform equation of the line into the parametric form 

Then,
the parametric equation of a line, 
x =
x_{0} + at,
y = y_{0}
+ bt and
z
= z_{0} + ct 
represents coordinates of any point of
the line expressed as the function of a variable parameter t
which makes
possible to determine any point of the line according to a given condition.

Therefore, by plugging these variable coordinates of a point of the line into
the given plane determine what value
must have the parameter t,
this point to be the common point of the line and the plane.


Example:
Given is a line 

and a plane
4x 
13y + 23z 
45 = 0, find the

intersection
point of the line and the plane.


Solution:
Transition from the symmetric to the parametric form
of the line


by plugging these variable coordinates
into the given plane we will find the value of the parameter t
such that these
coordinates represent common point (the intersection) of the line and the plane, thus

x =
t +
4, y
= 4t 
3 and z
= 4t 
2 => 4x 
13y + 23z 
45 = 0 
which
gives, 4 · (t +
4)  13 · (4t 
3) + 23 · (4t 
2) 
45 = 0 => t = 1. 
Thus,
for t = 1
the point belongs to the line and the plane, so 
x =
t +
4 = 
1+ 4 = 3, y =
4t 
3 = 4 · 1 
3 = 1 and z =
4t 
2 = 4 · 1 
2 = 2. 
Therefore, the intersection point
A(3,
1,
2) is the point which
belong to both, the line and the plane, prove. 


::
The parametric equations of a
quadratic
polynomial,
parabola

The parametric equations of the parabola,
whose axis of symmetry is parallel
to the yaxis

The quadratic polynomial y
=
a_{2}x^{2}
+ a_{1}x + a_{0}
or y

y_{0}
= a_{2}(x
 x_{0})^{2},
V(x_{0},
y_{0}) 

are
the coordinates of translations of the source quadratic y
=
a_{2}x^{2},
can be transformed to the parametric form
by
substituting x
 x_{0}
= t.

Therefore,
x
=
t + x_{0} 
y
=
a_{2}t^{2} + y_{0 }_{
}are the
parametric equations of the
quadratic polynomial. 

Example:
Given
are the parametric equations, x
=
t + 1
and y
=
 t^{2} + 4,
draw the graph of the curve.

Solution: The equation x
=
t + 1
solve for t
and plug into y
=
 t^{2} + 4,
thus

t
=
x
 1
=> y
=
 t^{2} + 4,
y
=
 (x
 1)^{2} + 4 
i.e., y

4
=
 (x
 1)^{2}
or y
=
 x^{2} + 2x + 3 translated parabola with the vertex V(x_{0},
y_{0}),
or V(1,
4).




The parametric equations of the parabola, whose axis of symmetry is
parallel to the xaxis

The
quadratic expression y^{2}
=
2px,
where p
is the distance between focus and directrix, represents
the source or the
vertex form of the conic section called parabola
with the vertex at the origin whose
axis
of symmetry coincide with the xaxis.

If
we rewrite the above equation into x
= ay^{2} then,


represents
the translation of the parabola in the direction of the coordinate axes
by x_{0}
and y_{0}
i.e., the vertex V(x_{0},
y_{0}). 
Thus,
the parabola, whose axis of symmetry is parallel to the xaxis,
can be transformed to parametric form by
substituting y
 y_{0}
= t
into the above equation. 
Therefore, x
=
at^{2}
+ x_{0} 
y
= t
+ y_{0
}are the parametric equations
of the parabola y^{2}
=
2px. 

Example:
Given
is the parabola x
=
 y^{2} + 2y + 3,
write its parametric equations and draw the graph.

Solution: Rewrite given equation by calculating the coordinates of
translations x_{0}
and y_{0}
or using completing the
square method, we get x

4
=
 (y
 1)^{2},
where the vertex V(x_{0},
y_{0}),
so V(4,
1). 
By
substituting y
 1
= t
obtained are 
x
=

t^{2}
+ 4 
y
= t
+ 1
the parametric equations of the parabola. 



::
The parametric equations of a
circle

The parametric equations of a
circle
centered at the origin with radius r

The parametric equations
of a
circle
centered at the origin with radius r, 
x
= r cos t 
y
= r sin t 
where, 0
< t < 2p. 
To
convert the above equations into Cartesian
coordinates, square
and add both equations, so we get 
x^{2}
+ y^{2}
= r^{2} 
as sin^{2}
t
+ cos^{2}
t = 1. 



The parametric equations of a
translated circle with center (x_{0}, y_{0}) and radius r

The parametric equations
of a
circle with center at (x_{0},
y_{0})
and
radius r, 
x
= x_{0}
+ r cos t 
y
= y_{0}
+ r sin t 
where, 0
< t < 2p. 
If
we write
the above equations, 
x
 x_{0}
= r cos t 
y
 y_{0}
= r sin t 
then
square and add them, we get the
equation of the
translated circle in Cartesian coordinates, 
(x
 x_{0})^{2}
+
(y
 y_{0})^{2}
= r^{2}. 




::
The parametric equations of an
ellipse

The parametric equations of an ellipse
centered at the origin

Recall
the construction of a point of an ellipse using two concentric circles of radii equal to lengths of the

semiaxes a and b, with the center at the origin as shows the
figure,
then 
x
= a cos t 
y
= b sin t 
where, 0
< t < 2p. 
To
convert the above parametric equations into Cartesian
coordinates, divide the first
equation by
a
and the second by b, then square and add them, 

thus,
obtained is the standard equation of the ellipse. 



The parametric equations of a
translated
ellipse with center at (x_{0}, y_{0})

The parametric equations
of a translated ellipse with
center (x_{0},
y_{0})
and semiaxes a and b, 
x
= x_{0}
+ a cos t 
y
= y_{0}
+ b sin t
where, 0
< t < 2p. 
To
convert the above parametric equations into Cartesian
coordinates, we write them as 
x
 x_{0}
= a cos t 
y
 y_{0}
= b sin t 
and divide the first
equation by
a
and the second by b, then square and add
both equations, so we get 
















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