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Parametric Equations |
Parametric
equations definition |
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The parametric equations of a line
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The
parametric equations of a line passing through two points |
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The direction of
motion of a parametric curve |
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Evaluation
of parametric equations for given values of the parameter |
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Sketching parametric
curve |
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Eliminating the
parameter from parametric equations |
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Parametric
and rectangular forms of equations
conversions |
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| 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. |
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| 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. |
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| The parametric equations of a line
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| 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 |
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| the parametric
equations of the line. |
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| To
convert the parametric equations into the Cartesian coordinates solve
given equations for t.
So |
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| by
equating |
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| 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 |
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| 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. |
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| Example: Write
the parametric equations of the line y
= (-1/2)x
+ 3 and sketch its graph. |
| Solution: Since |
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| 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
= 6, y1
= 0, xs
= 2,
and ys
= -1
into the parametric equations of a line |
| x
= x1 + xs · t,
x
= 6 + 2t |
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y
= y1 + ys · t,
y = -t |
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The direction of motion (denoted by red arrows) is given by increasing t. |
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| 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
= -2,
y1
= 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, |
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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 |
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therefore, |
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Functions
contents B
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| Copyright
© 2004 - 2020, Nabla Ltd. All rights reserved. |
