

Parametric Equations 
Parametric
equations definition 
The parametric equations of a line

The
parametric equations of a line passing through two points 
The direction of
motion of a parametric curve 
Evaluation
of parametric equations for given values of the parameter 
Sketching parametric
curve 
Eliminating the
parameter from parametric equations 
Parametric
and rectangular forms of equations
conversions 
Use
of parametric equations, example 





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.
So 

by
equating 


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, 





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 


and 



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 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. 








Precalculus
contents C 



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