# Level Two Calculus with Polar Coordinates

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So far, everything we've worked with has used the x-y Cartesian coordinate system. As we shall see, however, this is not always the most convenient coordinate system in which to operate. For this reason, we'll begin our exploration of the polar coordinate system here.

A coordinate system is simply a way to specify a location in space. In the Cartesian system, for instance, a point is identified by the coordinates (left(x,y right)), which we use to specify the location of the point as the result of a horizontal and vertical movement of x and y units, respectively, from the origin. The diagram below illustrates this.

One can define a point in two dimensions in other ways as well. It is also possible to find the angle this line makes with the positive x-axis by starting at the origin and moving directly outward until we reach the point. The point's coordinates would then be the radial distance from the origin plus the required rotation from the positive x-axis. The diagram below illustrates this.

This kind of coordinate system is known as polar coordinates.

Based on what has been said so far, you might conclude that r must be an integer greater than zero. We do, however, allow r to take on a negative value. The two points (left(2,fracpi6 right) and (left( - 2,fracpi6 right)) are depicted in a sketch below.

According to this diagram, if r is positive, the point will be in the same quadrant as (theta ). If r is negative, however, the point moves to the quadrant opposite (theta ). Keep in mind that the point described by the coordinates [left( - 2,fracpi 6 right)] is also described by the coordinates [left(2,fracpi 6 right)]. Given the above sketch, the coordinates [left(2,frac7pi 6 right)] tell us to rotate [frac7pi 6] degrees counterclockwise from the positive [[x]]-axis, which would put us on the dashed line.

As a result, there is a substantial distinction between Cartesian and polar coordinates. With Cartesian coordinates, a point has only one possible location. This is not the case when using polar coordinates. The number of possible polar coordinates for a given point is infinite. In this case, the following four coordinates all refer to the same point

left(5,fracpi3 right) = left(5, - frac5pi 3 right) = left( - 5,frac4pi 3 right) = left( - 5,frac2pi 3 right)

A rough diagram of the angles involved in these four coordinate systems follows.

For the second set of coordinates, we did a clockwise rotation to reach the target. Please remember to rotate clockwise. It's a necessary evil at times.

The last two sets of coordinates take advantage of the fact that if we move to the opposite quadrant from the point, we can use a negative r to move back to the point, and that there is a rotation in both directions to reach the angle.

If you spin the system only once, you'll see that these four points always correspond to the same spot. One point can have an infinite number of coordinates if the angle can make an infinite number of full rotations about the axis system. One can use any of the following sets of coordinates to represent the point (left(r,theta) right):

]left(r),right(theta 2pi n),hspace(0 25in}\hspace{0 25in - r, theta - r, 2n - 1 r, pi - r, hspace - 0, 25in, where n is an integer, where i, j, k, l, and m are all boxes. }}\]

As a next step, we'll explain how the coordinate system got its start. A common name for the reference point used in polar coordinates is "pole." We know r = 0 because we're not actually moving away from the pole. The coordinates of the origin/pole are (left(0, theta) right), but we are free to rotate around the system at any angle we like.

After learning the basics of polar coordinates, we can move on to thinking about how to move data back and forth between the two systems. Let's begin with a quick sketch to review the relationship between the two coordinate systems:

The right triangle in the figure above allows us to derive the following equations for transforming polar to Cartesian coordinates.

#### Formulas for Transforming from Polar to Cartesian Coordinates

The formula is [x = rcos hspace]. Zero in y = cos(theta)

Changing to or from Cartesian coordinates is essentially the same It's important to note the following

[beginalign*x2 y2 & = left(rcos theta right)2 & = r2cos 2theta & = r2sin 2theta & = r2left(cos 2theta &

Although this is a very helpful formula to keep in mind, what we really need is an equation for r, so we can simplify by taking the square root of both sides. What this means is that

r = sqrt(x2 + y2)

Because we know that r can take on positive or negative values, we must remember that a plus or minus must be placed in front of the root. Here, we'll follow the standard practice and assume that r is positive.

Theoretically, deriving an equation for theta is nearly as easy as deriving an equation for pi. Let's get started with,

[fracyx = fracrsintheta rcostheta = tantheta]

A product of the inverse tangents of the two sides gives,

If you plug these numbers into a calculator, you'll get: [theta = tan - 1left(fracyxright]

Because inverse tangents only return values in the range [- fracpi2 theta fracpi2], we'll have to proceed with caution. Remember that there's a second angle that could be taken, and that it's defined by (theta pi ).

As a result, the following are the formulas to use when moving from Cartesian to polar coordinates.

#### Conversion Tables from Cartesian to Polar Coordinates

x2 + y2 + hspace1 = [beginalign*r2 45in r = sqrt x2 y2 &= tan - 1left(frac yxright)hspace 0 35in}\mbox\hspace{0 35in theta 2 = theta 1 pi end align *]

I'll explain with an example.

Example 1 Determine how each of the following coordinates would look in the provided framework.
1. Input the Cartesian coordinates for (left( - 4,frac2pi3 right)).
2. How do I express the polar coordinates (left( - 1,-1 right))?
a Input the Cartesian coordinates for (left( - 4,frac2pi 3 right)). Indicator of a Fix

This is a simple conversion to make. To solve, we need only enter the relevant data into the appropriate equations.

[beginalign*x & = - 4cos left(frac2pi 3 right) = - 4left( - frac12 right) = -2 y & = - 4sin left(fracsqrt 3 2 right) = -2sqrt 3 endalign*

As a result, the Cartesian coordinates of this location are (left(2, - 2sqrt 3 right)).

b Enter the polar coordinates for (left( - 1,-1 right)) Prove the Answer

First, let's establish r.

The equation r = sqrt(left2 - sqrt(right2)) = sqrt(2) is written as [sqrt(left2 - sqrt(right2)]

I think it's time we got theta.

Theorem: [theta = tan - 1left(frac- 1 - 1right) = tan - 1left(1 right) = fracpi 4]

No, this isn't the right perspective. This theta value lies in the first quadrant, while the given point lies in the third. As mentioned before, by adding (p) to this, we obtain the correct angle. Therefore, the correct perspective is,

It can be written as [theta = fracpi4 pi = frac5pi4]

So, the point's polar coordinates are (left(sqrt 2,frac5pi4 right)). Keep in mind that the initial (theta ) obtained with a negative r could have been used instead. In polar coordinates, the point would look like this: (left(-sqrt(2),-frac(4),-right).

The aforementioned formulas can also be used to translate equations between coordinate systems.

Example 2 Use the provided coordinates to transform the following into an equation.
1. Input the polar coordinate solution to the equation (2x - 5x3 = 1 xy)
2. The formula for doing so is (r = - 8cos theta), which must be transformed into Cartesian coordinates.
a Polarize the coordinates (2x - 5x3 = 1 xy) Display the Answer

The only thing left to do here is to plug in the appropriate values for x and y using the appropriate formulas. e relative to r and theta (i.e., the Cartesian coordinates) e where the polar coordinates are)

[beginalign*2left(rcos theta right) - 5left(rcos theta right)3 &= 1left(rcos theta right)left(rsin theta right)2rcos theta - 5r

b Calculate the Cartesian coordinates given the expression (r = - 8cos theta ). Please Exhibit the Answer

In comparison to the last one, this one is slightly more difficult. First, we should probably just omit the r and replace it with a straight. However, we cannot simply replace the cosine with a different number to obtain Cartesian coordinates. A direct substitution would be possible if the cosine were followed by an r on the right. So, if it would be more practical to have an r on the right, go ahead and add one; just make sure to add one to the left, too.

For example, [r2 = - 8rcos theta]

Simply by replacing some values, we can get the Cartesian coordinates.

${x^2} {y^2} = - 8x$

Let's finish up the second half of the preceding example before we move on to the next topic.

The second part of the equation is an equation for a fairly well-known graph, although it is not presented in a form that is easily recognizable by most readers. To find it, we'll rearrange the Cartesian coordinate equation a bit.

${x^2} 8x {y^2} = 0$

Square off the x-coordinate now.

Starting at [beginalign*x2 8x 16 y2 & = 16 left(x 4 right)2 y2 & = 16endalign*], the expression reads as follows:

This was, therefore, a circle with a radius of 4 and a center at (left( - 4,0 right)).

Now we'll discuss what we've been building up to all along:

#### Polar Coordinate Graphs that You're Likely to See

Let's name a few of the most typical polar-coordinated graphs. A couple of unusual polar graphs will also be examined.

Lines
Equations in polar coordinates for some lines can be quite straightforward.

1. when (theta = beta),
By transforming the angles into Cartesian coordinates, we can confirm that this is indeed a straight line: [beginalign*theta & = beta tan - 1left(fracyx right) & = beta fracyx & = tan beta y & = left(t

Taking the positive x-axis as a reference, this line passes through the origin at an angle of (beta ). Alternatively stated, it is a line that starts at the origin and has a slope of tan beta.

2. It's straightforward to transform this into Cartesian coordinates, r cos theta = a, giving us x = a. A vertical line, then.
3. r * sin * theta * = b
And similarly, this is a horizontal line since it represents the equation (y = b).
Example 3 Make a single axis graph of (theta = frac3pi 4), (rcos theta = 4) and (rsin theta = - 3).

This one is straightforward, requiring only the creation of a graph, and so here it is.

Circles
Polar coordinates allow us to examine the circle equations.

1. $$r = a$$
This equation states that the distance from the origin must be (a) at all angles. This is, of course, the definition of a circle with center O and radius a.

As such, we have a circle with center O and radius a. Also, this is why we might prefer to use polar coordinates. In contrast to its equivalent in Cartesian coordinates, the equation of a circle with its center at the origin is quite beautiful.

2. (r = 2a cos theta) When we were transforming algebraic expressions into Cartesian coordinates, we examined a concrete instance of this.

A circle with a radius of and a center at (left, a, 0) is shown here. The absolute value bars are needed on the radius because a(a) could be negative (as it was in our example above). However, they must never be applied to the core.

3. Assume that r = 2b sin theta. That was very similar to this one It's a (left| b right|) circle with a (left(0,b right)) center.
4. (r = 2acostheta2bsintheta)
It can be shown that this is a circle with the radius sqrt(a2 + b2 + c) and the center left(a,b) + right(c) by combining the previous two examples and completing the square twice. This, then, is the universal formula for a symmetric circle that is not centered at the origin.
Example 4 Draw a graph with the values (r = 7), (4costheta), and (- 7sintheta) all on the same axis.

The first is a 7-radius circle with its center at zero. The second is a 2 radii circle whose origin is at [left][2,0][right]. The third is an equidistant circle with a radius of (frac72) and a center at (left(0, - frac72 right)). See below for a plot of the three equations:

Observe that a complete graph of r = a requires a range of 0 le theta le 2pi, while the other circles given here only require a range of 0 le theta le pi. If you'd like, you can check this with a quick table of values.

The Heart and the Limbs
The following three examples illustrate this:

1. Heart sounds: (r = a pm a cos theta) and (r = a pm a sin theta) are cardioids. These always include the origin and have a graph that resembles a heart.
2. Both limacons (r = a pm b cos theta and r = a pm b sin theta) have an inner loop if and only if (a b). These will always include the origin and have an internal loop.
3. Limacons with no inner loop are (r = a pm b cos ttheta) and (r = a pm b sin ttheta) when (a > b).
These are not closed-loop structures and do not include the starting point.
Example 5 You can visualize the relationships between (r = 5 - 5sin theta), (r = 7 - 6cos theta), and (r = 2 - 4cos theta) by plotting these three functions.

All of these will plot identically in the interval [0, theta, 2pi]. You'll find a table of values for everything here, along with accompanying graphs.

$$\theta$$ (r = -5 sin -5)theta For example, "(r = 7-6)costheta" (r) = (24)cos(theta). 5 1 6 "(displaystyle "frac"pi"2") 7 2 $$\pi$$ 5 13 -2 ('displaystyle 'frac-3 pi' 2) 10 7 2 $$2\pi$$ 5 1 6

A final step is required before we move on to the next section. An internal loop was present in the previous example's third graph. Sometimes, we'll need to find the value of (theta ) at which the graph is centered at the origin. We can easily determine these by setting the equation equal to zero and solving as follows:

It is written as [0 = 2 4 cos theta hspace]. 25in}\hspace{0 25in hspace Rightarrow 0 25in,,,,costheta = - frac12hspace0 25in}\hspace{0 0.25in RightArrow hSpace 25in}\hspace{0 25intheta = [frac(2pi)3,[frac(4pi)3]
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