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Polar Equation of Line

Learn the concept of polar equation in this page with solved examples and diagrams provided for a better understanding. First let's understand what is polar coordinate system.

Two dimensional coordinate system for determining the distance from the fixed point and determining an angle from a fixed direction on a plane. This is called polar coordinate system which can be found on each point. The pole is the fixed point and the ray from that point with the fixed direction is the polar axis. The distance from the point is known as the radius , and the angle is the polar angle. A point in the plane is addressed by the coordinates (r, theta).

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Solve Polar Equation

Below is provided the method to solve polar equation of line:

To find straight line in polar coordinates

The straight line is y = mx + c can be expressed in polar coordinates form as

y = mx + c

r sin θ = m × r cos θ + c

r(sin θ - mcos θ) = c

r = c/(sin theta - m cos theta)

So the straight line equation y = mx + c is represented in polar coordinates by

r = c/ (sin theta - m cos theta)

Solve the polar into Cartesian form:

To convert the polar straight line coordinates (r, θ) into the straight line Cartesian coordinates (x, y) with the help of formula, given below.

Here, by definition,

sin (θ) = y/r

cos (θ) = x/r

Therefore,

x= r cos (θ)

y=r sin (θ)

Polar Equation of Line Examples

Here are few example problems on polar equation of line for a better understanding:

Problem 1: Solve the polar coordinates (5, 90°) into Cartesian coordinates

Solution: Change the polar into Cartesian form

Here, r = 5 and θ = 90°

x = r cos (θ)

y = r sin (θ)

x = 5 cos 90°

= 5 × 0

x = 0

y = 5 sin 90°

= 5 x 1

y = 5

Cartesian form is (0, 5)

Problem 2: What are each of the fourth-roots of expressed in polar form?

Solution: r =  |sqrt(3)+i|

r = "sqrt(sqrt(3^2+1^2))"

r = "sqrt(3+1)"

r = 2

Since cos α = sqrt(3/2) and sin α = 1/2 , α is in the first quadrant and α = 30°. Therefore, since the sine and cosine are periodic,

z = r(cos α + i sin α )

z = 2[cos ( + k. ) + i sin ( + k. )]

and applying the theorem of nth root, the first four-roots of z are

2^(1/4)

where k = 0, 1, 2, and 3

Thus the fourth-roots are

z1 = 2^(1/4)(cos 6^o + i sin 6^0)

z2 = 2^(1/4)(cos 78^o + i sin 78^o)

z3 = 2^(1/4)(cos150^o + isin150^o)

z4 =2^(1/4)(cos 222^o + i sin 222^o)

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