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Convert the rectangle coordinates to polar coordinates where r > 0

and 0 ≀ 𝞱 ≀ 2𝛑
(3sqrt3/2,3/2)

Can I have it explained? I know the process to change coordinates, but not when they change the constraints.

Draw the triangle in standard position. You have both x and y positive, so that's in QI. As always,

tanΞΈ = y/x
so here you have
tanθ = (1/2)/ (3√3/2) = 1/(3√3)
so ΞΈ β‰ˆ 10.89Β° β‰ˆ 0.19
But I suspect a typo, and that you really wanted
tanθ = (1/2)/ (√3/2) = 1/√3
so ΞΈ = Ο€/6

To convert rectangular coordinates to polar coordinates, you can use the following formulas:

r = √(x² + y²)
𝞱 = arctan(y / x)

In this case, the rectangular coordinates are (3sqrt3/2,3/2).

To find the value of r, substitute the values of x and y into the formula:
r = √((3sqrt3/2)² + (3/2)²)
= √(9/4 * 3 + 9/4)
= √(27/4 + 9/4)
= √(36/4)
= √9
= 3

To find the value of 𝞱, substitute the values of x and y into the formula:
𝞱 = arctan((3/2) / (3sqrt3/2))
= arctan((1/√3))
= Ο€/6

Therefore, the polar coordinates are (r, 𝞱) = (3, Ο€/6).
Since the constraints are r > 0 and 0 ≀ 𝞱 ≀ 2𝛑, the given polar coordinates satisfy these constraints.

To convert rectangular coordinates to polar coordinates, you can use the following formulas:

r = √(x^2 + y^2)
𝞱 = arctan(y / x)

In this case, we are given the rectangular coordinates (3√3/2, 3/2) and want to convert them to polar coordinates where r > 0 and 0 ≀ 𝞱 ≀ 2𝛑.

First, let's compute the value of "r":
r = √((3√3/2)^2 + (3/2)^2)
= √(27/4 + 9/4)
= √(36/4)
= √9
= 3

Next, let's compute the value of "𝞱". Since the given point lies in the first quadrant (x > 0 and y > 0), we can use the arctan formula without any adjustments:
𝞱 = arctan((3/2) / (3√3/2))

However, we need to consider the range of "𝞱" in polar coordinates. The given restrictions state that 0 ≀ 𝞱 ≀ 2𝛑. In this case, the value of "𝞱" will be between 0 and 2𝛑, inclusive.

Using the inverse tangent function, arctan, we can find the principal value of "𝞱" and then add or subtract multiples of 2𝛑 to obtain a value within the given range.

𝞱 = arctan((3/2) / (3√3/2))
β‰ˆ arctan(1 / √3)
β‰ˆ 0.5236 (in radians)

The value of "𝞱" in this case is approximately 0.5236 radians. However, this value is less than 2𝛑 but greater than 0. To bring it within the given range, we need to add 2𝛑 to the value of "𝞱":

𝞱 = 0.5236 + 2𝛑
β‰ˆ 0.5236 + 6.2832
β‰ˆ 6.8068 (in radians)

Therefore, the polar coordinates of the given rectangular coordinates (3√3/2, 3/2) in the range r > 0 and 0 ≀ 𝞱 ≀ 2𝛑 are (r, 𝞱) β‰ˆ (3, 6.8068 radians).