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



 

 

Recommended Reading

 

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.

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