• Algebra > Polar Equation

Polar Equations

Introduction

The Cartesian coordinate system is probably the most familiar way of graphing equations. In this system, a coordinate pair \( (x,y) \) is graphed on the x-axis and the y-axis. Usually, we graph functions as \( (x, f(x)) \), where x is the input and y or \( f(x) \) is the output. We also have an origin at (0, 0) where the axes cross.

There is another graphing system called the polar system. In this case, the distance (usually referred to as the radius, r) from the origin, called the pole, is graphed against the angle formed with the polar axis, equivalent to the x-axis. The polar axis is essentially 0° angle. A polar coordinate is written as \( (r; \theta) \), where r is the distance from the pole, and θ is the angle measure from the polar axis. I will use a semicolon in between the coordinates to differentiate between rectangular coordinates.

Polar Coordinates

The angle can be written in radians or degrees. Note that the output is usually written first and the input is written second, meaning when graphing a polar function, r is the result and θ is the input. So, it is reversed compared to the Cartesian system. The graph above shows a polar coordinate and a polar function.

The point P is at (3; 60°). We go 60° from the polar axis and out 3 units from the pole to graph this point. You can see the concentric circles have a radius 1 more than the previous one. This allows us to find the distance from the pole. The radial lines are lines passing through the pole. They are drawn at π/6, π/3, π/2, etc. They help finding the angle to create.

The angles in polar coordinates can be represented as radians or degrees. For simple points, degrees may be used. However, when graphing functions, radians are generally preferred.

Negative Radius

Although my graphs will show negative numbers on the polar axes (refer to Figure 1a), technically, there is no negative radius in the polar system. Instead, when we obtain a negative radius, we go in the opposite direction instead of the usual direction. Point P on the graph is (3; 60°). Point P' is (-3; 60°). From 0°, we went to 60°, which ends up in quadrant I, but the radius is -3, so we go in the opposite direction of where 60° is located from the pole to find -3. Hence (3; 270°) is the same as (-3; 60°). And this point is also the same as (3; -120°). So there are multiple ways of graphing the same point. When polar curves are plotted, this is sometimes what happens. We can not always tell if a point was graphed because of a negative radius.

Graphing Curves

Most functions normally have a range that covers both negative and positive numbers. As a matter of fact, cosine is negative from the interval \( (\frac{\pi}{2}, \pi) \). In the graph of \( r(\theta) = 4\cos\theta \), the bottom half of the circle is graphed with a negative radius. It may not always be apparent where the radius plotted on a curve is negative.

You can see that the interval \( (0, \frac{\pi}{2}) \) is in the 1st quadrant and the blue part is what results. The interval \( (\frac{\pi}{2}, \pi) \) in the 2nd quadrant and gives the bottom part of the circle as in in red. The interval \( (0, \pi) \) completes the whole circle, and the interval \( (\pi, 2\pi) \) retraces the circle again. The circle is traced over and over again as θ continues beyond 2π.

Referring back to Figure 1, the curve in blue is a circle that has the equation \( (4\cos\theta; \theta) \). We usually write polar equations as \( r(\theta) = 4\cos\theta \) or simply \( r = 4\cos\theta \). The center of the circle is at the polar coordinate (2; 0°). The point F can have infinite values, the first of which is (4; 0°). Remember that cosine is cyclic. So when \( \theta = \pi \), \( \cos\pi = -1 \). Hence, at \( \theta = \pi \), instead of the point being at what would normally be -4, it is at 4 since \( 4\cos\pi = -4 \), meaning we go in the opposite (positive) direction instead of the usual (negative) direction. Point F can also be graphed for all of these values: θ is 0, π, 2π, 3π.... nπ, for all integers n.

Because θ repeats as we go around the circle, the points will naturally repeat depending on the function. Most functions that continuously increase toward infinity as θ increases toward infinity, will produce a path that revolves around the pole. Functions that have asymptotes, such as tangent and secant will have graphs that may not have a revolving path.

Conversion to Parametric Equations

Polar equations can be converted to parametric forms (or Cartesian forms). The rectangular point (the Cartesian coordinate) \( P(x, y) \) can be represented as \( P(x,y) = (r\cos \theta, r\sin \theta) \), where \( r = \sqrt{x^{2}+y^{2}} \) equals the distance of P from the origin and θ is \( \arctan\frac{y}{x} \), the angle the point makes with the x-axis.

I mentioned above that \( r = 4\cos\theta \) is a circle. How did we know it is a circle? It looks like a circle but we cannot assume. We can covert it to rectangular form and show it is the equation of a circle.

Let’s substitute \( r = \sqrt{x^{2}+y^{2}} \) and \( x = r\cos\theta \) or \( \frac{x}{\sqrt{x^{2}+y^{2}}} = \cos\theta \) in our equation.

(i) \( r = 4\cos\theta \)

(ii) \( \sqrt{x^{2}+y^{2}} = 4\cdot \frac{x}{\sqrt{x^{2}+y^{2}}} \)

(iii) \( x^{2}+y^{2} = 4x \)

(iv) \( x^{2} - 4x + y^{2} = 0 \)

(v) \( (x-2)^{2} -4 + y^{2} = 0 \) (Completing the square.)

(vi) \( (x-2)^{2} + y^{2} = 4 \)

The equation \( (x-2)^{2} + y^{2} = 4 \) is that of a circle centered at (2, 0) with radius of 2, which is what we expected.

Equations of the form \( r = a\cos\theta \) are circles with radius of a and centered at (a; 0°) or (a, 0). Equations of the form \( r = a\sin\theta \) are circles with radius a and centered at (a; 90°) or (0, a).

Calculus with Polar Curves

The Extrema

Let \(r = r(\theta) \) be defined as a polar equation. Then the expression \( \frac{dr}{d\theta} = r'(\theta) \) translates as the rate of change of r with respect to θ, but it does not interpret as the slope of a tangent at θ. Wherever \( \frac{dr}{d\theta} = 0 \), a relative maximum or relative minimum of r, the distance from the pole, occurs. This means that the distance r for angle θ is either relatively greatest or the least from the pole. Let’s take the example of the circle we used above \( r(\theta) = 4\cos\theta \).

It is apparent the maximum distance from the pole is at F and the minimum is at the pole. Let’s calculate this with the derivative.

(i) \( r(\theta) = 4\cos\theta \)

(ii) \( \frac{dr}{d\theta} = -4\sin\theta \)

(iii) \( -4\sin\theta = 0 \)

(iv) \( \sin\theta = 0 \)

(v) \( \theta = 0, \, \pi, \, 2\pi, \, 3\pi, \, 4\pi... \)

Any angle that gives sine of 0 is a solution, so all the solutions are 0 and nπ for all integers n. Now, we need to find the radius at these points by substituting these values into \( r(\theta) \).

(i) \(4\cos(0) = 4 \)

(ii) \(4\cos(\pi) = -4 \)

(iii) \(4\cos(2\pi) = 4 \)

It seems -4 is not a solution, but this is in fact a solution because the angle π is in the 2nd quadrant and the radius is negative, so the point lies in the opposite direction of where the angle is. Practically, the only unique solution we get is (4; 0°).

Technically, the point (0; 0°) is not a minimum because the radius changes signs from positive to negative at this point and the values are still decreasing. This can be seen by graphing \( 4\cos\theta\) curve on a Cartesian plane.

Unlike a maximum, a polar graph with a minimum is difficult to spot compared to a rectangular graph. For example, a line has no minimum or maximum on a rectangular graph. However, it has a minimum on a polar graph because we are looking at the minimum distance from the pole. See example below in polar equation of a line.

The Slope of Tangents

To find \( \frac{dy}{dx} \), which is what we normally think of a slope on the Cartesian plane, convert the equation to a parametric equation first:

Here, the slope \( \frac{dy}{dx} \) can also be written as \( \frac{dy}{dx} = \frac{r'(\theta)\tan\theta + r(\theta)}{r'(\theta) - r(\theta)\tan\theta} \).

The acute angle λ formed by the tangent to a polar curve and the polar axis is given by the formula: \( \lambda = \theta + \arctan\frac{r(\theta)}{r'(\theta)} = \arctan\frac{dy}{dx} \).

The angle \( \phi \) formed between the line drawn from the pole to the point of tangency and the tangent line is given by the formula: \( \phi = \arctan\frac{r(\theta)}{r'(\theta)} \).

Example

For the polar circle \( r(\theta) = 4\cos\theta \), the derivative is \( r'(\theta) = \frac{-4(\sin\theta)(\sin\theta) + 4(\cos\theta)(\cos\theta)}{-4(\sin\theta)(\cos\theta) - 4(\cos\theta)(\sin\theta)} \) or \( r'(\theta) = \frac{4\sin^{2}\theta - 4\cos^{2}\theta}{4(\sin\theta)(\cos\theta) + 4(\cos\theta)(\sin\theta)} \).

Using the double-angle identities, the derivative is \( r'(\theta) = \frac{-4\cos^{2}(2\theta) }{4\sin^{2}(2\theta)} = -\cot^{2}(2\theta) \).

Now, we find the solutions to \( \cot(2\theta) = 0 \). The solutions to this equations are \( \pm \frac{(2n-1)}{4}\pi \), for n is an integer. The graph below shows the extrema at I and J, the topmost and bottommost points on the circle. The polar coordinates of the extrema are J = \( (2\sqrt{2}; \frac{\pi}{2}) \) and I = \( (-2\sqrt{2}; \frac{3\pi}{2}) \). Note the negative radius of I. This means the angle \( \frac{3\pi}{4} \) lies in the 2nd quadrant, but the radius has a negative value, so the point is plotted in the 4th quadrant.

Since we know the center of the circle is at (2,0), we expect the extrema to be here. Applying the Pythagorean identity confirms the extrema at I and J.

Area

In polar equations, we can find the area swept out by the angle rotation. For example, in our curve \( r = 4\cos\theta \), we can calculate the area swept out from 0 to π, which would be the complete circle.

The integral reflects the area of a sector of a circle. The area of a circular sector whose angle measure is θ is given by \( A = \frac{1}{2}r^{2}\theta \). Let’s apply this on the curve \( r = 4\cos\theta \) to find the area between 0 and π. (Changing the variable to t to make it easier.)

(i) \( \frac{1}{2}\int_{0}^{\pi}(4\cos(t))^{2} \, dt \)

(ii) \( 8\int_{0}^{\pi}\cos^{2}(t) \, dt \)

This integral is \( \int \cos^{2}t \, dt = \frac{1}{2}(t + (\sin t)(\cos t)) + C\).

(iii) \( 8\left[\frac{1}{2}\left(t + \sin(t)\cos(t)\right)\right]_{0}^{\pi} \)

(iv) \( 4\left[\pi + \sin(\pi)\cos(\pi)\right] - 4\left[0 + \sin(0)\cos(0)\right] \)

(v) \( 4\left[\pi\right] - 4\left[0\right] = 4\pi \)

This is the area for a circle with radius 2. Now, let’s do the same from 0 to π/2 for the area of half the circle, starting at step (iii).

(iii) \( 8\left[\frac{1}{2}\left(t + \sin(t)\cos(t)\right)\right]_{0}^{\pi/2} \)

(iv) \( 4\left[\pi/2 + \sin(\pi/2)\cos(\pi/2)\right] - 4\left[0 + \sin(0)\cos(0)\right] \)

(v) \( 4\left[\pi/2\right] - 4\left[0\right] = 2\pi \)

This also checks.

Arclength

Continuing with our example circle, let’s find the arclength from 0 to π.

(i) \( \int_{0}^{\pi} \sqrt{(4\cos(t))^{2} + (-4\sin(t))^{2}} \, dt \)

(ii) \( 4\int_{0}^{\pi} \sqrt{\cos^{2}(t) + \sin^{2}(t)} \, dt \)

(iii) \( 4\int_{0}^{\pi} 1 \, dt \) (Did you expect it to reduce to this?)

(iv) \( 4[t]_{0}^{\pi} \)

(v) \( 4[\pi - 0] = 4\pi \)

The circumference of a circle (\( C = 2r\pi\)) with radius 2 is \( 2 \cdot 2\cdot\pi = 4\pi \).

Special Polar Equations

This section will highlight special polar equations and give equations of familiar Cartesian equations in their polar form.

The Circle

The simplest rectangular equation is \( y = a \), where for any x, y is always a units from the x-axis. This is a straight line that is a locus of all points equidistant from the x-axis for any x.

The simplest polar equation is \( r = a \), where, at any θ, the point is equidistant from the pole. Hence, it is a circle with a radius of a.

The derivative is 0 since it is a constant. That means there is no change in the distance from the pole at any θ. It has no minimums or maximums.

The tangent slope can be found by following calculation we learned above.

(i) \( \frac{dy}{dx} = \frac{r'(\theta)\sin\theta + r(\theta)\cos\theta}{r'(\theta)\cos\theta - r(\theta)\sin\theta} = \frac{0 + 4\cos\theta}{0-4\sin\theta} \)

(ii) \( \frac{dy}{dx} = -\cot\theta \)

At θ = 0, we expect undefined slope and \( \cot 0 \) is undefined. At \( \theta = \pi/4 \), we expect the slope to be -1. And \( \cot \frac{\pi}{4} = 1 \), so this also checks.

We encountered another equation of a circle already: \( r = 4\cos\theta \). The equations \( r = a\cos\theta \) and \( r = a\sin\theta \) are circles with radius of a. The difference is in their center, which you can see in the graphs below.

Equations of circles at centers other than the pole or polar axes are too complicated.

The Spiral

The next simplest rectangular equation is \( y = x \). As x increases, y also increases. When we convert this to polar form, the same happens but the line wraps around the plane instead. Meaning, the distance from the pole increases in a spiral. Figure 5 shows the spiral \(r = \theta \).

The derivative of this function is \( r'(\theta) = 1 \), which means the distance continuously increases from the pole at a fixed rate. There are no maximums or minimums, just like a line \( y = x \) has no maximum or minimum.

Let’s find the slope.

(i) \( \frac{dy}{dx} = \frac{r'(\theta)\sin\theta + r(\theta)\cos\theta}{r'(\theta)\cos\theta - r(\theta)\sin\theta} = \frac{\sin\theta + \theta\cos\theta}{\cos\theta - \theta\sin\theta} \)

To find the points where slope is 0, we would have to find solutions to the equation \( \sin\theta + \theta\cos\theta = 0 \), which does not have any algebraic solutions for all points. But there are some special points.

The slope of the curve at \( (\frac{\pi}{2}; \frac{\pi}{2}) \) is \( \frac{\sin\frac{\pi}{2} + \frac{\pi}{2}\cos\frac{\pi}{2}}{\cos\frac{\pi}{2} - \frac{\pi}{2}\sin\frac{\pi}{2}} = \frac{1+0}{0-(\frac{\pi}{2})(1)} = -\frac{2}{\pi}\).

The slope of the curve at \( (\frac{3\pi}{2}; \frac{3\pi}{2}) \) is \( \frac{\sin\frac{3\pi}{2} + \frac{3\pi}{2}\cos\frac{3\pi}{2}}{\cos\frac{3\pi}{2} - \frac{3\pi}{2}\sin\frac{3\pi}{2}} = \frac{-1+0}{0-(\frac{3\pi}{2})(-1)} = -\frac{2}{3\pi}\).

The slopes at these special points seems to be the negative inverse. These two tangents have been plotted in Figure 5. The slope also tends to 0 as the spiral progresses.

The other interesting points occur at π, 2π, 3π, etc.

The slope of the curve at \( (\pi; \pi) \) is \( \frac{\sin\pi + \pi\cos\pi}{\cos\pi - \pi\sin\pi} = \frac{0+(\pi)(-1)}{-1-(\pi)(0)} = \pi \).

The slope of the curve at \( (2\pi; 2\pi) \) is \( \frac{\sin (2\pi) + 2\pi\cos (2\pi)}{\cos (2\pi) - 2\pi\sin (2\pi)} = \frac{0+(2\pi)(1)}{1-(2\pi)(0)} = 2\pi \).

The slopes at these points is the same as θ and the radius at θ. The slope tends to infinity as spiral progresses.

Other Powers

If we graph the powers of θ, \( r = \theta^{n} \), we get similar spirals. The difference will be in how far apart the spiral revolutions are. Generally, for n > 1, the revolutions get further and further apart. For 0 < n < 1, the revolutions get closer and closer. The difference can be profound. For example, compare the graphs of \( r = \theta \) in red to \( r = \theta^{2} \) in green to \( r = \sqrt{\theta} \) in blue. The blue graph spirals get closer and closer and there are many more that can be seen. The other two spirals go off the graph quickly.

Inward Spirals

Rectangular functions of the form \( y = ax^{n} \) spiral outward from 0 to ∞ when plotted on a polar plot. Rectangular functions of the form \( y = \frac{a}{x^{n}} \) spiral inward toward the pole.

At \( \theta = 0 \), the radius is infinite. Therefore, the point is off the plot. As \( \theta \) increases, the radius value gets smaller and tends toward the pole. Figure 6c is the zoomed out version of Figure 6b. It may seem the graph is asymptotic toward the positive x-axis. It actually approaches x-axis toward positive infinity at a very slow rate.

We can calculate the area swept by rays in this curve starting with an angle other than 0. For example, the area from \( \frac{\pi}{2} \) to \( \pi \) is given by \( A = \frac{1}{2}\int_{\pi/2}^{\pi} \left(\frac{1}{\theta}\right)^{2} \, d\theta \) = \( \frac{1}{2}\left[ -\frac{1}{\theta} \right ]_{\pi/2}^{\pi} = \frac{1}{2\pi} \approx 0.15915 \). The area is quite small. If you zoom in, the area seems to be equal to or smaller than a quarter circle with radius of 1/2, which is \( \frac{\pi}{16} \approx 0.19635 \), so this makes sense.

The Line

Most rectangular functions that have a domain of (-∞, ∞) can be graphed on a polar system. The resulting curves will trace a path around the pole indefinitely as θ tends to infinity. If the function has a range of (-∞, ∞), such as \( y = x \), the result is a spiral. If the function has a limited range, like sine and cosine, the radius will never get large enough for the graph to ‘go off the chart’.

Lines, on the other hand, do not cross the axes except twice at most. Hence, plotting a line requires functions that are asymptotic. The equation for a line is \( y = mx + b \). We can convert this to polar form by substituting \( x = r\cos\theta \) and \( y = r\sin\theta \).

(i) \( y = mx + b \)

(ii) \( r\sin\theta = mr\cos\theta + b \)

(iii) \( r(\sin\theta- m\cos\theta) = b \)

(iv) \( r = \frac{b}{\sin\theta- m\cos\theta} \)

Equation (iv) is a line with slope m and ‘y-intercept’ at b. Technically, b is the distance from the pole when converted to polar form. The line equation is also written generally as \( r = \frac{d}{a\sin\theta + b\cos\theta} \), where d is the perpendicular distance of the line from the pole and a and b are some constants. You can see that since sine and cosine are in the denominator, the rectangular curve will have asymptotes.

Horizontal and Vertical Lines

In the polar line equation \( r = \frac{d}{a\sin\theta + b\cos\theta} \), if either a or b equals 0, then the equation reduces to a horizontal or vertical line. Therefore, equation of the form \( r = \frac{d}{\sin\theta} \) is a line \( y = d \) and equation of the form \( r = \frac{d}{\cos\theta} \) is a line \( x = d \).

Line Through the Pole

When d is 0, the line equation degenerates. Therefore, an equation of a line that passes throught the pole is \( \theta = \alpha \), for some angle \( \alpha \). Here, \( \alpha \) can also represent the slope of the line.

Example

The line \( y = \frac{1}{2}x + 2 \) is graphed below in polar form \( r = \frac{2}{\sin\theta - \frac{1}{2}\cos\theta} \), shown in orange and red. The graph below also shows the function \( f(x) = \frac{2}{\sin x - \frac{1}{2}\cos x} \) in rectangular form, shown in blue.

The polar line is shown in 2 colors to illustrate that polar graphs are not always plotted continuously. We usually think of a line being traced from -∞ to ∞ without lifting a pencil. However, although we can let θ go backward in the negative direction, we usually start a polar graph from θ = 0 to ∞.

Since we start the polar graph at θ = 0, we look at the rectangular graph and see that at 0, the value is -4; this point is G₂ in yellow. This corresponds to point H₂ in yellow on the polar line. As x increases, the values get smaller and go toward -∞ and are asymptotic to the line \( x = \tan^{-1}(\frac{1}{2}) \). This corresponds to the polar line radius going toward infinity in the opposite direction (the left). So the line traced toward the left from H₂ shown in orange is the result of the small interval of the function \( f(x) = \frac{2}{\sin x - \frac{1}{2}\cos x} \) from \( (0, \tan^{-1}(\frac{1}{2})) \).

After hopping over the asymptote, we go from -∞ to ∞. So the red portion of the line is traced from (∞,∞) back to (-4, 0) ((4; 180°) in polar form), the point where we started. Point L₂ is a relative minimum of the function. This represents the minimum radius distance from the pole, which corresponds to the point N₂ on the polar line. Point O₂ at \( (4; \pi) \) corresponds to H₂. The interval that plots the whole polar line is (0, π). The rest of the function just repeats the process. One difference is that after θ = π, the orange line traced has a positive radius, whereas when we started with θ = 0, it was traced with a negative radius.

Conic Sections

I have a page dedicated to the conic sections in polar form here: The General Polar Equation of Conic Sections. Browse through this page for an in-depth discussion on conics. Here, I’ll just cover them briefly.

The general polar equation of a conic section is \( r(\theta) = \frac{ed}{1-e\sin\theta} \), where e is the eccentricity and d is the distance of the directrix from the focus. This one equation covers all the conic sections. If e = 0, it is a circle, and reduces to the form \( r = a \). If e = 1, it is a parabola. If e > 1, it is a hyperbola. If \( 0 \lt e \lt 1 \), then it is an ellipse.

Alternate Equation of a Parabola

The conic with the simplest equation is the parabola because either x or y is squared but not both. We can take the rectangular equation of a parabole \( y = ax^{2} \) and convert it to the polar form.

(i) \( y = ax^{2} \)

(ii) \( r\sin\theta = ar^{2}\cos^{2}\theta \)

(iii) \( \sin\theta = ar\cos^{2}\theta \)

(iv) \( r = \frac{\sin\theta}{a\cos^{2}\theta} \)

(v) \( r = \frac{1}{a}\tan\theta \sec\theta \)

The graph below shows a polar parabola and the same equation in rectangular form. The reader can determine what interval covers the whole parabola and the starting points.

It may see in Figure 8 that the point on the parabola at the pole is the minimum. But that is not the case. The polar parabola has no minimum when we look at the rectangular graph of the equation. The parabola is graphed from (0, 0) to (+∞, +∞), then jumps to (-∞, -∞) back to (0, 0). Hence, there is no minimum.

Consider the area swept by a ray on the parabola. The area bound by the ray and the parabola can we determined by integration. We will let the interval be general but start from 0 so we can determine the area swept by any angle.

(i) \( A = \frac{1}{2}\int (\frac{1}{a}\tan\theta \sec\theta)^{2} \, d\theta = \frac{1}{2a^{2}}\int \tan^{2}\theta \sec^{2}\theta \, d\theta \)

If we let \( u = \tan\theta \), then \( du = \sec^{2} \theta \, d\theta \). That was a convenient substitution.

(ii) \( A = \frac{1}{2a^{2}}\int u^{2} \, du \) (Making the u-substitution)

(iii) \( A = \frac{1}{6a^{2}}u^3 \)

(iv) \( A = \frac{1}{6a^{2}}\tan^{3}\theta \) (Back substitution)

Therefore, the area swept by angle θ is \( \frac{1}{6a^{2}}\tan^{3}\theta \).

For example, the area swept by an angle of 45° is \( \frac{1}{6a^{2}}\tan^{3}\frac{\pi}{4} = (\frac{1}{6a^{2}})(1) = \frac{1}{6a^{2}} \). If a = 1/2, then the area is 2/3. We can confirm this using integration in rectangular form.

When the angle is 45°, the point on the parabola is (2,2). The triangle created by the x-axis and the y-value has an area of 2 by simple geometry. The area under the parabola is \( \int_{0}^{2} \frac{1}{2}x^{2} \, dx = \frac{8}{6} = \frac{4}{3} \). The bound area is then \( \frac{6}{3} - \frac{4}{3} = \frac{2}{3} \), which is what we got above.

Odd-Numbered Rose Petals

Curves of the form \( r = \sin(a\theta) \) and \( r = \cos(a\theta) \) form petals. When a is odd, the number of petals equal a. When a is even, the number of petals is 2a. Figure 9 shows the rose curve \( r = \sin(3\theta) \). Naturally, the petals are all congruent. The maximum distance from the pole is 1. The cosine petals are similar except in the orientation. Because cos(0°) = 1, the starting point in the cosine petals is (1, 0°).

Let’s calculate the area enclosed in 1 petal. The roots of sine are 0, π, 2π, etc. But for \( \sin (3\theta)\), the roots are a third. The first petal is traced by the interval (0, π/3). The sine curve is shown in red to show the roots. Using the area integral \( A = \frac{1}{2}\int_{a}^{b}r(\theta)^{2}\,d\theta \):

(i) \( A = \frac{1}{2}\int_{0}^{\pi/3} (\sin(3\theta))^{2}\,d\theta \)

We need to use the identity \( \sin^{2} x = \frac{1}{2} - \frac{1}{2}\cos(2x) \).

(ii) \( A = \frac{1}{2}\int_{0}^{\pi/3} \frac{1}{2} - \frac{1}{2}\cos(6\theta)\,d\theta \)

(iii) \( A = \frac{1}{2}\left[ \frac{1}{2}\theta - \frac{1}{12}\sin(6\theta) \right]_{0}^{\pi/3} \)

(iv) \( A = \frac{1}{2}\left[ (\frac{1}{2})(\frac{\pi}{3}) - \frac{1}{12}\sin(2\pi) \right] - \frac{1}{2}\left[ (\frac{1}{2})(0) - (\frac{1}{12})(0) \right] \)

(v) \( A = \frac{\pi}{12} \approx 0.262 \)

The area of all 3 petals is equal to \( A = \frac{\pi}{4} \).

Let’s compare this area to the area under the sine curve from 0 to π/3 since the petal fits nicely inside the curve.

(i) \( A = \int_{0}^{\pi/3} (\sin(3\theta))\,d\theta \)

(ii) \( A = -\frac{1}{3}\left[ \cos(3\pi) \right]_{0}^{\pi/3} \)

(iii) \( A = -\frac{1}{3}\left[ \cos(\pi) - \cos(0) \right] \)

(iv) \( A = -\frac{1}{3}\left[ -1 - 1 \right] \)

(v) \( A = \frac{2}{3} \)

The area under the sine curve is about 0.667 and the area enclosed by the petal is about 0.262.

Even-Numbered Rose Petals

An even-numbered petals rose occurs with even coeffients. In Figure 10, the curve starts at P₁ and goes to O. Then to P₂ and back to O. Then, to P₃ to O to P₄ to O to P₁. The red portion shows what is traced first so you can see how the rest is traced. For the sine petal in Figure 10a, the blue shows how it is traced (but does not start at the blue.)

The petals above are for a = 2. Let’s calculate the area enclosed by the cosine petal this time. This time, we will integrate from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \), since these are the roots and the interval traces the first petal.

(i) \( A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} (\cos(2\theta))^{2}\,d\theta \)

We need to use the identity \( \cos^{2} x = \frac{1}{2} + \frac{1}{2}\cos(2x) \).

(ii) \( A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1}{2} + \frac{1}{2}\cos(4\theta)\,d\theta \)

(iii) \( A = \frac{1}{4}\int_{-\pi/4}^{\pi/4} 1 + \cos(4\theta)\,d\theta \)

(iv) \( A = \frac{1}{4}\left[ \theta + \frac{1}{4}\sin(4\theta) \right]_{-\pi/4}^{\pi/4} \)

(v) \( A = \frac{1}{4}\left[ \frac{\pi}{4} + \frac{1}{4}\sin(\pi) \right] - \frac{1}{4}\left[ -(\frac{\pi}{4}) + (\frac{1}{4})\sin(\pi) \right] \)

(vi) \( A = \frac{1}{4}\left[ \frac{\pi}{4} + 0 \right] - \frac{1}{4}\left[ -(\frac{\pi}{4}) + 0 \right] \)

(vii) \( A = \frac{\pi}{8} \approx 0.393 \)

The area of all 4 petals is equal to \( A = \frac{\pi}{2} \).

General Area of A Petal

We can generalize the formula of 1 petal or any number of petals in the equation \( r = \cos(n\theta) \) or \( r = \sin(n\theta) \). Since sine and cosine functions are equivalent, only shifted, the areas of either would be the same. We will use the sine petal since its interval is easy to calculate. For \( r = \sin(n\theta) \), the interval of intergration for one petal is \( (0, \frac{\pi}{n}) \).

(i) \( A = \frac{1}{2}\int_{0}^{\pi/n} (\sin(n\theta))^{2}\,d\theta \)

We need to use the identity \( \sin^{2} x = \frac{1}{2} - \frac{1}{2}\cos(2x) \).

(ii) \( A = \frac{1}{2}\int_{0}^{\pi/n} \frac{1}{2} - \frac{1}{2}\cos(2n\theta)\,d\theta \)

(iii) \( A = \frac{1}{4}\int_{0}^{\pi/n} 1 - \cos(2n\theta)\,d\theta \)

(iv) \( A = \frac{1}{4}\left[ \theta - \frac{1}{2n}\sin(2n\theta) \right]_{0}^{\pi/n} \)

(v) \( A = \frac{1}{4}\left[ \frac{\pi}{n} - \frac{1}{2n}\sin(2\pi) \right] - \frac{1}{4}\left[ 0 - \frac{1}{2n}\sin(0) \right] \)

(vi) \( A = \frac{1}{4}\left[ \frac{\pi}{n} - 0 \right] - \frac{1}{4}\left[ 0 - 0 \right] \)

(vii) \( A = \frac{\pi}{4n} \)

The area of 1 petal for \( r = \sin(n\theta) \) is equal to \( A = \frac{\pi}{4n} \). If n is even, then area of all petals is \( A = (2n)(\frac{\pi}{4n}) = \frac{\pi}{2} \), and if n is odd, the area of all petal is \( A = (n)(\frac{\pi}{4n}) = \frac{\pi}{4} \).

More Petals

Curves of the form \( r = a + \cos(n\theta) \) also form petals for n > 1. If n = 1, it is a cardioid.

The difference in the rose petals in the previous section and these is that these petals have cusps that seem to be joined at the pole instead of a continuous smooth transition from one petal to another. You can see in Figure 12 how the curve is traced from one petal to another. The path starts at (2; 0°) and follows the blue path from there. The rest of the path is similar to this path. You can also trace a point on the graph below to see what path the point follows.

Another difference is that even n does not result in twice the petals. The curve always has n number of petals.

We can find the area of 1 petal for a general curve for any n. The interval of integration is \( (-\frac{\pi}{n},\frac{\pi}{n}) \). Because of symmetry, let’s just integrate from 0 to \( \frac{\pi}{n} \) and double it.

(i) \( A = 2 \cdot \frac{1}{2}\int_{0}^{\pi/n} (a + \cos(n\theta))^{2}\,d\theta \)

(ii) \( A = \int_{0}^{\pi/n} a^{2} + 2a\cos(n\theta) + \cos^{2}(n\theta) \,d\theta \)

We need to use the identity \( \cos^{2} x = \frac{1}{2} + \frac{1}{2}\cos(2x) \).

(iii) \( A = \int_{0}^{\pi/n} a^{2} + 2a\cos(n\theta) + \frac{1}{2} + \frac{1}{2}\cos(2n\theta) \,d\theta \)

(iv) \( A = \left[ (a^{2} + \frac{1}{2})\theta + \frac{2a}{n}\sin(n\theta) + \frac{1}{4n}\sin(2n\theta) \right]_{0}^{\pi/n} \)

(v) \( A = \left[ (a^{2} + \frac{1}{2})(\frac{\pi}{n}) + \frac{2a}{n}\sin(\pi) + \frac{1}{4n}\sin(2\pi) \right] - \left[ 0 + \frac{2a}{n}\sin(0) + \frac{1}{4n}\sin(0) \right] \)

(vi) \( A = (a^{2} + \frac{1}{2})(\frac{\pi}{n}) \)

For an n number or petals, the area is \( A = (a^{2} + \frac{1}{2})\pi \).

Anirdesh.com Last Updated: July 20 2025, 14:30:08.