Parametric Equations Parametric equations definition The parametric equations of a line
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 xy-plane
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 P1(x1, y1) and the direction vector s then, the position or (radius) vector r of any point P (x, y) of the line
r = r1 + t · s,    - oo < t < + oo   and where,  r1 = x1i + y1 j  and  s = xsi + ys j,
represents the vector equation of the line.
Therefore, any point of the line can be reached by the
r = xi + y j = (x1 + xst) i + (y1 + yst) 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 = x1 + xs · t y = y1 + ys · 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 P1(x1, y1) and P2(x2, y2)
 x = x1 + (x2 - x1) t y = y1 + (y2 - y1) 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 x-intercept as the given point P1so
for   y = 0   =>   0 = (-1/2)x + 3,   x = 6  therefore,  P1(6, 0).
Substitute the values, x1 = 6y1 = 0, xs = 2 and  ys = -1 into the parametric equations of a line
x = x1 + xs · t,      x = 6 + 2t
y = y1 + ys · 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 x1 = -2y1 = 0, x2 = 2,  and  y2 = 2 into the parametric equations of a line
x = x1 + (x2 - x1) t,      x = -2 + (2 + 2) t = -2 + 4t,       x = -2 + 4t,
y = y1 + (y2 - y1) 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, (x0, y0, z0) 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 = x0 + at,    y = y0 + bt   and   z = z0 + 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 + 4y = 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.   Pre-calculus contents C 