Problem 27 - Calculus

Question

(a) Multiply \(r=\cos \theta+\sin \theta\) by \(r\) to convert into an \(x y\) equation. (b) Rewrite the equation as \(\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=R^{2}\) to find the radius \(R\). (c) Draw the graph.

Step-by-Step Solution

Answer

The equation is \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\), with a circle center at \((\frac{1}{2}, \frac{1}{2})\) and radius \(R = \sqrt{\frac{1}{2}}\).

Step 1: Express in Polar Coordinates

Recall that polar coordinates can be expressed in terms of Cartesian coordinates using the equations \(x = r\cos\theta\) and \(y = r\sin\theta\). In the given problem \(r = \cos\theta + \sin\theta\). Express this as \(r = \frac{x}{r} + \frac{y}{r}\), which can be rearranged as \(r^2 = x + y\).

Step 2: Substitute and Simplify to Cartesian Form

Substitute the expression derived from polar coordinates into the equation: \(r^2 = r (\cos\theta + \sin\theta)\), which results in \(r(x+r) = x + y\). Simplify this to \(x^2 + y^2 = x + y\).

Step 3: Complete the Square

To transform \(x^2 + y^2 = x + y\) into the form \((x-a)^2 + (y-b)^2 = R^2\), complete the square for both \(x\) and \(y\). This gives \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\).

Step 4: Identify the Center and Radius

The standard form equation \((x - a)^2 + (y - b)^2 = R^2\) gives the center at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and \(R^2 = \frac{1}{2}\), thus the radius \(R = \sqrt{\frac{1}{2}}\).

Step 5: Draw the Graph

Using the center \((\frac{1}{2}, \frac{1}{2})\) and radius \(R = \sqrt{\frac{1}{2}}\), draw the circle on a coordinate plane. The center is at the point \((\frac{1}{2}, \frac{1}{2})\) and the radius is approximately \(0.707\).

Key Concepts

Polar Coordinates

Polar coordinates are a system used to locate a point in a plane using a distance from a fixed point and an angle from a reference direction.

In polar coordinates, a point is represented by \(r, \theta\), where \(r\) is the distance from the origin, and \(\theta\) is the angle measured counterclockwise from the positive x-axis.

This system is an alternative to the traditional Cartesian coordinates (x, y).

• Use \(x = r \, \cos \, \theta\) to convert the horizontal component from polar to Cartesian.
• Use \(y = r \, \sin \, \theta\) to convert the vertical component.
• The equations allow for transforming between systems depending on the context or problem requirements.
• In complex-plane applications, polar coordinates can simplify calculations involving rotations and angles.

Completing the Square

Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial. This method is essential in solving quadratic equations and geometrically interpreting them as circles or parabolas.

To complete the square for an expression like \(x^2 + bx\), you:

• Add and subtract \(\left(\frac{b}{2}\right)^2\) inside the expression.
• Factor it into the form \(\left(x + \frac{b}{2}\right)^2\).

This step is crucial in rewriting equations to determine geometric properties such as the center and radius of a circle, as shown in the original problem.
In our example, we adapted \(x^2 + y^2 = x + y\) to:

• Rewriting \(x^2 - x + y^2 - y\) by adding and subtracting \(\frac{1}{4}\) to both x and y terms.
• Consequently, we expressed each component as squares: \(\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2}\).

Equations in Mathematics

Equations are fundamental tools in mathematics used to describe relationships between variables.
They often represent lines, curves, shapes, or systems in different fields ranging from algebra to calculus and beyond.
A basic form of an equation is \(ax + by = c\), where a and b define the linear relationship of x and y, and c is the constant term.

• Equations can be linear, quadratic, polynomial, or trigonometric, among many other types.
• Graphically, equations describe paths in the coordinate system, such as lines, circles, and ellipses.
• Transformations, like completing the square, help convert equations to identifiable geometric forms.
• In our example, the given equation, via manipulation turns into a circle with specific properties.

Understanding how to manipulate and solve equations is crucial for problem-solving in different mathematics areas.

Radius of a Circle

The radius of a circle is a key geometric property, representing the distance from the center to any point on the circle's edge.
The equation of a circle in the Cartesian plane can be expressed as \(\left(x-a\right)^2 + \left(y-b\right)^2 = R^2\), where \(a\) and \(b\) are the coordinates of the circle's center, and \(R\) is the radius.

• Finding the radius involves identifying \(R^2\) in the equation's right side, then taking the square root.
• In practical terms, the radius is crucial for determining the circle's size and for calculations involving circle properties like circumference and area.
• In our problem, after completing the square, the radius \(R = \sqrt{\frac{1}{2}}\), which approximately equals 0.707.
• This understanding aids in accurately graphing and visualizing circles.

The circle's radius plays a fundamental role in a wide range of geometry and trigonometry problems.