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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 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 radius vector
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 and equate obtained expressions, thus

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 P1, so
for   y = 0   =>   0 = (-1/2)x + 3,   x = 6  therefore,  P1(6, 0).

Substitute the values,  x1 = 6y1 = 0xs = 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,

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, (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 (the intersection) 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.
 :: The parametric equations of a quadratic polynomial, parabola
The parametric equations of the parabola, whose axis of symmetry is parallel to the y-axis
The quadratic polynomial  y = a2x2 + a1x + a0   or   y - y0a2(x - x0)2,   V(x0, y0)

are the coordinates of translations of the source quadratic  y = a2x2,  can be transformed to the parametric form by substituting  x - x0 = t.

   Therefore,        x = t + x0
                         y = a2t2 + y0        are the parametric equations of the quadratic polynomial.

Example:  Given are the parametric equations,  x = t + 1  and  y = - t2 + 4, draw the graph of the curve.

Solution:  The equation  x = t + 1 solve for t and plug into  y = - t2 + 4, thus

t = x - 1  =>   y = - t2 + 4,     y = - (x - 1)2 + 4

i.e., y - 4 = - (x - 1)2  or  y = - x2 + 2x + 3 translated parabola with the vertex V(x0, y0), or V(1, 4).

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

The quadratic expression  y2 = 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 x-axis.

If we rewrite the above equation into  x = ay2  then,

represents the translation of the parabola in the direction of the coordinate axes by x0 and y0 i.e., the vertex V(x0, y0).
Thus, the parabola, whose axis of symmetry is parallel to the x-axis, can be transformed to parametric form by substituting  y - y0 = t  into the above equation.
    Therefore,      x = at2 + x0
                         y = t + y0          are the parametric equations of the parabola  y2 = 2px.
Example:  Given is the parabola  x = - y2 + 2y + 3, write its parametric equations and draw the graph.
Solution:  Rewrite given equation by calculating the coordinates of translations x0 and y0 or using completing the square method, we get  x - 4 = - (y - 1)2,  where the vertex V(x0, y0), so V(4, 1).
By substituting  y - 1 = t  obtained are
                                              x = - t2 + 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
x2 + y2 = r2
as  sin2 t + cos2 t = 1.
The parametric equations of a translated circle with center (x0, y0) and radius r

The parametric equations of a circle with center at       (x0, y0) and radius r,

x = x0 + r cos t
y = y0 + r sin t
where,   0 < t < 2p.
If we write the above equations,
x -  x0  = r cos t
y -  y0  = r sin t
then square and add them, we get the equation of the translated circle in Cartesian coordinates,
(x - x0)2 + (y - y0)2 = r2.
 :: 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
semi-axes 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 (x0, y0)
The parametric equations of a translated ellipse with center (x0, y0) and semi-axes a and b,
                                                                x = x0 + a cos t
                                                                y = y0 + b sin t       where,   0 < t < 2p.
To convert the above parametric equations into Cartesian coordinates, we write them as
x - x0 = a cos t
y - y0b 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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