Parametric Equations

When the coordinates of curve are expressed in terms of a third independent variable - parameter - parametric equations are produced.

E.g.

y = 2t; x = t2 - 3

Here, 't' is the parameter.

E.g.1

y = t2 + 2; x = (t-3);
Now, we have a choice - finding the values of x and y manually by substituting the values of t or use algebra to find the Cartesian equation. Let's go for the latter.
t = (x+3) => y =(x + 3)2 + 2
This can easily be sketched: basic quadratic curve translated -3 in the x-axis and 2 in the y axis.

Quadratic Curve

E.g.2

y = t; x = (t - 3)2;
t = √x+3 => y = √x + 3
This can easily be sketched as follows:

Parabolic Curve

E.g.3

y = sec t; x = cos t
xy = cos t . 1/cos t
xy = 1
y = 1/x This is reciprocal curve.
This can easily be sketched as follows:

Reciprocal Curve

E.g.4

y = 3 sin t; x = 3 cos t
cos t = x/3; sin t = y/3;
x2 / 9 + y2 / 9 = sin2 t + cos2 t = 1
x2 / 9 + y2 / 9 = 1
x2 + y2 = 9
x2 + y2 = 32
This is a circle with radius 3.
This can easily be sketched as follows:

Circle

E.g.5

y = -2 + 3 sin t; x = 3 + 3 cos t
cos t = (x-3)/3; sin t = (y+2)/3;
(x-3)2 / 9 + (y+2)2 / 9 = sin2 t + cos2 t = 1
(x-3)2 / 9 + (y+2)2 / 9 = 1
(x-3)sup>2 + (y+2)2 = 9
(x-3)2 + (y+2)2 = 32
This is a circle with the centre at (3,-2) and radius 3.
This can easily be sketched as follows:

Circle eccentric

E.g.6

y = 3 sin t; x = 4 cos t
cos t = x/4; sin t = y)/3;
x2/16 + y2 / 9 = sin2 t + cos2 t = 1
x2 / 16 + y+2 / 9 = 1
if x = 0 => y = +3 or -3
if y = 0 => x = +4 or -4
It is an ellipse. This can easily be sketched as follows:

Ellipse

The World of Spirals

The significance of parametric equations can be seen from the following beautiful shapes, which are produced by the manipulation of them in different ways.

Archimedes

Fermat

Hyperbolic

Logarithmic

Circle

Ellipse

Clear

 

Please press clear button, before choosing a new spiral, so that you can see each smoothly.

 

Area under a curve

Suppose we want to find the area under the curve, y = f(t), using parameter, t in the range, a<=x<=b. x = g(t).

Area under the curve

Area = aby dx
Since ∫ y dx = y (dx/dt) dt
Area = a'b'f(t) g'(t) dt
a' and b' are the boundaries in terms of t.

Area of a Circle

Area of a circle

The parametric equations for the circle of radius r are as follows:
y = r sin (t); x = r cos(t); dx/dt = -r sin(t)
Area = 0y (dx/dt) dt
= 0r sin(t). -r sin (t) dt
= 0-r2 sin2(t) dt
= 0-r2 (1- cos(2t))/2 dt
= 0-r2/2 (1- cos(2t)) dt
= -r2/2 [t - sin(2t)/2]0
= -r2/2 [2π]
= -πr2

Area of a circle = πr2



 

 

 

 

Resources at Fingertips

This is a vast collection of tutorials, covering the syllabuses of GCSE, iGCSE, A-level and even at undergraduate level. They are organized according to these specific levels.
The major categories are for core mathematics, statistics, mechanics and trigonometry. Under each category, the tutorials are grouped according to the academic level.
This is also an opportunity to pay tribute to the intellectual giants like Newton, Pythagoras and Leibniz, who came up with lots of concepts in maths that we take for granted today - by using them to serve mankind.

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