1. POLAR CO-ORDINATE SYSTEM:
In mathematics, the polar coordinate system is a two-dimensional coordinate system in which
each point on a plane is determined by a distance from a fixed point and an angle from a fixed
direction.
The fixed point (analogous to the origin of a (Cartesian system) is called the pole, and
the ray from the pole in the fixed direction is the polar axis. The distance from the pole is called
the radial coordinate or radius, and the angle is the angular coordinate, polar angle, orazimuth.
Uniqueness of polar coordinates:
Adding any number of full turns (360°) to the angular coordinate does not change the
corresponding direction. Also, a negative radial coordinate is best interpreted as the
corresponding positive distance measured in the opposite direction. Therefore, the same point
can be expressed with an infinite number of different polar coordinates (r, φ ± n×360°) or
(−r, φ ± (2n + 1)180°), where n is any integer.[10] Moreover, the pole itself can be expressed as
(0, φ) for any angle φ.
Where a unique representation is needed for any point, it is usual to limit r to non-negative
numbers (r ≥ 0) and φ to the interval [0, 360°) or (−180°, 180°] (in radians, [0, 2π) or (−π, π]). One
must also choose a unique azimuth for the pole, e.g., φ = 0.
2. The polar coordinates (the radial coordinate) and (the angular coordinate, often called
the polar angle) are defined in terms of Cartesian coordinatesby
(1)
(2)
Where the radial is distance from the origin, and is the counterclockwise angle from the x-axis.
In terms of and ,
(3)
(4)
(Here, should be interpreted as the two-argument inverse tangent which takes the
signs of and into account to determine in which quadrant lies.) It follows immediately that
polar coordinates aren't inherently unique; in particular, will be precisely the same
polar point as for any integer . What's more, one often allows negative values of under the
assumption that is plotted identically to
Area with Polar Coordinates
3. We’ll be looking for the shaded area in the sketch above. The formula for finding this area is,
Polar equation of a curve:
Notice that we user in the integral instead of so make sure and substitute
accordingly when doing the integral.
The equation defining an algebraic curve expressed in polar coordinates is known as a polar
equation. In many cases, such an equation can simply be specified by defining r as
a function of φ. The resulting curve then consists of points of the form (r(φ), φ) and can be
regarded as the graph of the polar function r.
Different forms of symmetry can be deduced from the equation of a polar function r. If r(−φ)
= r(φ) the curve will be symmetrical about the horizontal (0°/180°) ray, if r(π − φ) = r(φ) it will be
symmetric about the vertical (90°/270°) ray, and if r(φ − α) = r(φ) it will be rotationally
symmetric by α counterclockwise about the pole.
4. Because of the circular nature of the polar coordinate system, many curves can be described by
a rather simple polar equation, whereas their Cartesian form is much more intricate. Among the
best known of these curves are the polar rose, Archimedean spiral,lemniscate, limaçon,
and cardioid.
For the circle, line, and polar rose below, it is understood that there are no restrictions on the
domain and range of the curve.
Time Derivaties of the unit vector:
To differentiate this relation with respect to time to obtain and we
need expressions for time to obtain and , in exactly the same way we derived ,in the
5. preceding article. During time dt the coordinat directions rotate through the angle d0, and the
unit vector also rotate through the same angle from and to and as shown in
figure below. We note that the vector change d is in the plus 0-direction and that d is in
the r-direction. Because their magnatudes in the limit are equal to unit vector as radius times the
angle d0 in the radians, we can write them as d = d0. If we divide these equations by dO,
we have
If, on the other hand, we divide them by dt. We have d /dt-(d0/dt) and
d /dt= -(d0/dt) , or simply
equ (2)
Velocity:
We have now rady to differentiate r =r with respect to time. Using the role for
diferentiating the product of a scalar and a vector gives
With the substitution of from equation .2/21, the vector expression for the velocity becomes
Where
The r-component of v is merely the rate at which the vector x streches. The 0-component of v is
due to the rotation of r.
6. Accelaration:
We now diferentiate the expression for v to obtain the acceleration
A= Note that the derivative of r will be produced three terms, since all three factor are
variable. Thus,
Subsitution of and from eq.2 and collecting terms give
Where
We can write the 0-component alternatively as
Which can be verified easily by carrying out the differentiation. This form for will be useful
when we treat the angulat momentum of particles.