Parametric Vs. Implicit Descriptions

There are two ways, in general, of specifying a geometric figure in three (or higher) dimensional space. A shape is implictly defined by one or more equations which are all satisfied by only points on that shape. For example, the one equation x² + y² + z² = 1 implicitly defines a sphere centered at the origin with radius one. You can think of an implict definition as a giving a test for whether or not a point lies on the shape.

Conversely, an explicit or parametric definition generates the points on the shape, in terms of one or more free parameters. For example, the points on a curve in space can be given by a vector valued function sigma(t) of one parameter t. E.g., (cos t, sin t) describes the points on a circle -- you plug in a t value, and you get out a point on the circle. Surfaces, in turn, are given by vector valued functions of two parameters, e.g. S(u,v) = ( 3 u + v, 7 u v, v cos u ) A special case of this is the surface given by the graph of a function f(x,y) which can be represented explicitly as S(x,y ) = ( x, y, f(x, y ) ) .

To convert from an implicit definition to an explict definition, simply declare one or more of your variables to be parameters and then solve for the rest. For example, if I want to convert x² + y² + z² = 1 to an explict definition, I can let x and y be my parameters and then solve to get z = ± sqrt( 1 - x² - y²) My sphere is then explicitly defined by the two functions S₁ = ( x, y, sqrt( 1 - x² - y²) and S₂ = ( x, y, -sqrt( 1 - x² - y²) To be totally explicit, I also have to say what x and y values I'm mapping from to get the two halves of the sphere; in this case I'm letting (x,y) range over the unit disk x² + y² <= 1.

[In Vector Calculus, one must be careful not to confuse disks with their boundary circles and balls with their boundary spheres!]

To convert from an explicit definition to an implicit definition, solve to eliminate your parameter variables. For example, if I wanted to convert x = cos t, y = sin t to an implicit definition, I could solve to get t = cos^-1 x and then let y = sin cos^-1 x This is a strange way to define a circle, but it works.

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