Question

Find the following average values. The average distance between points within the cardioid \(r=1+\cos \theta\) and the origin

Short answer

Solution: Following the 7 steps outlined above, we calculate the average distance by dividing the total distance by the total area: Average distance \(= \frac{D}{A}\) First, solve the integral for the area \(A\): $$A = \int_{0}^{2\pi} \frac{1}{2}(1+\cos \theta)^2 d\theta = \frac{3\pi}{2}$$ Next, solve the integral for the total distance \(D\): $$D = \int_{0}^{2\pi} \frac{1}{3}(1+\cos \theta)^3 d\theta = 2\pi$$ Now, calculate the average distance: $$\text{Average distance} = \frac{D}{A} = \frac{2\pi}{\frac{3\pi}{2}} = \frac{4}{3}$$ Thus, the average distance between points within the cardioid and the origin is \(\frac{4}{3}\).

Step-by-step solution

Find the distance function in polar coordinates

Distances in polar coordinates can be determined using the Pythagorean theorem. For the origin, the distance function would simply be the polar coordinate \(r\). Thus, the distance function for the given cardioid is \(r=1+\cos \theta\).

Find the area element in polar coordinates

The area element in polar coordinates is given by \(dA = r d\theta dr\). We will use this to determine the integral bounds and integral.

Determine the integral bounds

The given cardioid \(r=1+\cos \theta\) traces a complete cycle from \(\theta=0\) to \(\theta=2\pi\). The bounds for \(r\) will be between \(r=0\) and \(r=1+\cos \theta\). Therefore, the double integral trivially goes from \(0\) to \(2\pi\) for \(\theta\) and \(0\) to \(1+\cos \theta\) for \(r\).

Set up the double integral for the total area of the cardioid

The total area of the cardioid can be found by integrating the area element over the bounds determined in step 3. This is written as: $$A = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r dr d\theta$$

Set up the double integral for the average distance

Multiplying the distance function by the area element and integrating with respect to the bounds will give us the total distance to points inside the cardioid. Thus: $$D = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} (1+\cos \theta) r dr d\theta$$

Solve the integrals

First, find the area integral: $$A = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r dr d\theta = \int_{0}^{2\pi} \frac{1}{2}(1+\cos \theta)^2 d\theta$$ Now, we solve the double integral for the total distance: $$D = \int_{0}^{2\pi} \int_{0}^{1+\cos \theta} (1+\cos \theta) r dr d\theta = \int_{0}^{2\pi} \frac{1}{3}(1+\cos \theta)^3 d\theta$$

Calculate the average distance

Finally, we find the average distance by dividing the total distance by the total area: $$\text{Average distance} = \frac{D}{A}$$ To find the final answer, you can solve the two integrals from step 6 and use the formula in step 7. This will result in the average distance between points within the cardioid and the origin.

Key concepts

Polar Coordinates

Polar coordinates offer an alternative to the Cartesian coordinate system, where instead of relying on x and y coordinates, we use a radius and an angle. This system is extremely useful when dealing with circular or radial patterns, such as in the geometry of a cardioid. In polar coordinates, a point in the plane is determined by:

• Radius (r): The distance from the origin to the point.
• Angle (θ): Formed between the positive x-axis and a line connecting the origin to the point.

The distance function we use in polar coordinates simplifies the problem of finding distances from the origin. Considering the polar form of a cardioid, such as the one given by the function \(r = 1 + \cos \theta\), simplifies many mathematical operations, like integration, necessary for determining areas, and distances.

Double Integral

Double integrals extend the concept of integration to calculate areas, volumes, and averages over two-dimensional regions. This tool is vital in calculus for summing over a continuous domain, especially in polar coordinates. When solving our problem involving a cardioid, we use a double integral because:

• The inner integral computes over the radius, \(r\), from 0 to \(1 + \cos \theta\).
• The outer integral sums up over the angle \(\theta\), from 0 to \(2\pi\), thus covering the entire cardioid.

By setting up our double integral as:\[\int_{0}^{2\pi} \int_{0}^{1+\cos \theta} r \, dr \, d\theta\]we compute the total area of the cardioid, and similar setups help to determine the total distance to points within it.

Cardioid Geometry

A cardioid is a heart-shaped curve traced by a point on the circumference of a circle as it rolls around another fixed circle of the same radius. In polar coordinates, a typical cardioid equation is \(r = 1 + \cos \theta\). Important aspects of a cardioid involve:

• Its symmetry about the polar axis (the line \(\theta = 0\)).
• The use of trigonometric functions, which offer simplicity in calculations.
• Having features that naturally suggest polar coordinates for integration.

Cardioids are significant because they not only appear in geometry and trigonometry but also physics, where they model various reflective properties and diffraction patterns.

Distance Function

In polar coordinates, the distance from any point to the origin is simply the radius \(r\). This is a direct consequence of how polar coordinates are defined. For a cardioid \(r = 1 + \cos \theta\):

• Each point's distance to the origin is given directly by this formula, reducing computational complexity.
• We use this distance function to set up integrals which help in finding averages, such as the average distance from all points within the cardioid back to the origin.

Using integrals, the entire area can be summed, and this function allows for an elegant computation of entire shapes, leading to useful insights in both theoretical and applied mathematics.