5-6 The Cartesian coordinates of a point are given. (i) Find...

$5-6$ The Cartesian coordinates of a point are given.
(i) Find polar coordinates $(r, \theta)$ of the point, where $r>0$ and
$0 \leqslant \theta<2 \pi$ .
(ii) Find polar coordinates $(r, \theta)$ of the point, where $r<0$ and
$0 \leqslant \theta<2 \pi$
$$(\text a)\quad(2,-2)$$

Key Concepts

Quadrant Considerations

When converting from Cartesian to polar coordinates, simply computing arctan(y/x) may lead to an angle that does not reflect the true orientation of the point because arctan only returns values from -?/2 to ?/2. Adjusting the angle by adding ? or 2?, as necessary, ensures that ? falls within the desired interval [0, 2?) and accurately reflects the point's quadrant.

Conversion between Coordinate Systems

This concept involves translating a point's representation between Cartesian (x, y) coordinates and polar (r, ?) coordinates. The conversion employs the formulas r = ?(x² + y²) and ? = arctan(y/x) while accounting for the sign of x and y to correctly identify the location of the point in the correct quadrant. This is fundamental when moving between algebraic and trigonometric descriptions of points in the plane.

Multiple Polar Representations

A set of Cartesian coordinates can correspond to infinitely many polar coordinate pairs because of the periodicity of the angle ? and alternative sign choices for r. Understanding how to represent a point with a positive r (by choosing the proper angle) or with a negative r (by adding ? to the angle when converting from the positive representation) is crucial, especially when the problem explicitly asks for representations with prescribed conditions on r.

Transcript

00:01 To find the polar coordinates for which r is less than or greater than 0, we have the magnitude of r to be equal to 2 squared plus negative 2 squared.

00:17 And this will give us 4 plus 4 is 8.

00:20 So 8 squared is 2 square root of 2.

00:24 So for r equal to r greater than 0, we have implies that.

00:30 R is equal to two square root of two and if r is less than zero then it means it's going to be negative of that we recall that x it's equal to r cost theta so this implies that you have two r two square root of two cost theta to be equal to 2 so let's divide both sides by 3 so this implies that you have course teta to be equal to 1 divided by square root of 2 then at 12 for the negative side so this is for r r greater than 0 so let's say for r less than 0 we have have x to be equal to r cos theta so then this implies that you have negative 2 square root of 2 cos theta to be equal to 2 so if you divide both sides by this you have cause theta to be equal to minus 1 divided by 2 1 divided by square root of 2 so you have the positive a negative.

02:07 Then for y as well, we know that y is equal to our sign theta.

02:18 So this implies that for this part, you have two square root of two, sine theta to be equal to two.

02:29 Divide both sides by this and you have sine theta to be equal to minus one.

02:37 Divided by square root of 2.

02:40 If you take for r less than 0, you have y to be equal to r sine theta.

02:51 So then this implies that you have minus 2 square roots of 2, sine theta to be equal to negative 2...

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