Problem 60 - Thomas Calculus

Question

Replace the Cartesian equations in Exercises \(49-62\) by equivalent polar equations. $$ (x-5)^{2}+y^{2}=25 $$

Step-by-Step Solution

Answer

The polar equation is \(r = 10\cos\theta\).

Step 1: Recognize the Cartesian Equation

The given Cartesian equation represents a circle centered at the point \((5,0)\) with a radius of 5. The equation is \((x-5)^2 + y^2 = 25\).

Step 2: Substitute Polar Coordinates

In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). So, substitute these into the equation: \((r\cos\theta - 5)^2 + (r\sin\theta)^2 = 25\).

Step 3: Expand and Simplify the Expression

Expand \((r\cos\theta - 5)^2\) to get \(r^2\cos^2\theta - 10r\cos\theta + 25\). Combine like terms: \(r^2\cos^2\theta + r^2\sin^2\theta - 10r\cos\theta + 25 = 25\).

Step 4: Utilize the Pythagorean Identity

Recognize that \(r^2\cos^2\theta + r^2\sin^2\theta = r^2(\cos^2\theta + \sin^2\theta)\). Since \(\cos^2\theta + \sin^2\theta = 1\), this simplifies to \(r^2\). Hence, the equation becomes \(r^2 - 10r\cos\theta + 25 = 25\).

Step 5: Simplify the Final Equation

Subtract 25 from both sides to get: \(r^2 - 10r\cos\theta = 0\). Factor out \(r\): \(r(r - 10\cos\theta) = 0\).

Step 6: Solve for the Polar Equation

The factorization gives two possible equations: \(r = 0\) or \(r = 10\cos\theta\). Since \(r = 0\) represents the origin and is not part of our circle, the polar equation is \(r = 10\cos\theta\).

Key Concepts

Cartesian Equations

In mathematics, Cartesian equations are a way to express geometric shapes using coordinates on a plane. These equations use the familiar x-y coordinate system, often referred to as the Cartesian coordinate system, named after René Descartes.
Cartesian equations are prevalent because they allow a wide range of expressions; lines, circles, and more complex curves can all be represented in this system. A typical Cartesian equation for a circle is i.e., often in the form i.e. i.e., \((x-h)^2 + (y-k)^2 = r^2\), which represents a circle with center at the point i.e. i.e., \((h, k)\) and radius i.e., \(r\). The equation given in the exercise, \((x-5)^2 + y^2 = 25\), is such an example. Here, the circle's center is i.e., \((5, 0)\) and its radius is 5. This simple form makes it straightforward to understand where the shape is located and its size.

Conversion to Polar

Polar coordinates are another way to describe the position of points in a plane, using a radius and an angle. This system can sometimes simplify equations that are more cumbersome in Cartesian coordinates.
The relationship between Cartesian and polar coordinates is straightforward:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)

Here, \(r\) stands for the distance from the origin to the point, and \(\theta\) is the angle formed with the positive x-axis. To convert a Cartesian equation like \((x-5)^2 + y^2 = 25\) into a polar equation, replace \(x\) and \(y\) by their polar forms. The equation becomes \((r\cos\theta - 5)^2 + (r\sin\theta)^2 = 25\). This substitution is the first crucial step in the conversion process.

Circle Equations

Understanding circle equations helps in converting between coordinate systems. In the Cartesian form, the equation \((x-h)^2 + (y-k)^2 = r^2\) clearly represents a circle. In polar form, however, circle equations might not be as intuitive.
Through substitution and simplification, we arrive at the equivalent polar equation for our exercise: \(r = 10\cos\theta\). This tells us that as you move around the circle, the radius changes in a cosine wave pattern relative to the angle \(\theta\). The lines in polar coordinates could express a circle centered along one of the axes or radiating from the origin depending on the equation structure. This showcases the versatility of polar coordinates in representing shapes.

Trigonometric Identities

Trigonometric identities are fundamental tools in converting and simplifying equations between Cartesian and polar forms. A particularly important identity used in this process is the Pythagorean identity: \(\cos^2\theta + \sin^2\theta = 1\). This equation is essential when dealing with polar coordinates because it significantly reduces complexity in equations like \(r^2\cos^2\theta + r^2\sin^2\theta = r^2\).
When working with trigonometric identities, it's important to remember they allow for transformation and simplification of expressions, helping to illuminate the relationships between different forms. For example, recognizing the Pythagorean identity helps transform \(r^2 - 10r\cos\theta + 25 = 25\) into a simpler form. It's an excellent example of how identities smooth the bridge between Cartesian and polar views.