Problem 27
Question
Find parametric equations for the circle \(x^{2}+y^{2}=a^{2}\)
Step-by-Step Solution
Verified Answer
The parametric equations are \(x = a \cos(t)\) and \(y = a \sin(t)\) with \(t\) ranging from \(0\) to \(2\pi\).
1Step 1: Identify the Equation of a Circle
The given equation of the circle is \(x^2 + y^2 = a^2\), which represents a circle centered at the origin with radius \(a\). The goal is to convert this into parametric form.
2Step 2: Understand the Parametric Representation of a Circle
A circle with radius \(a\) centered at the origin can be represented parametrically using trigonometric functions. The parameters for a circle are typically based on the angle \(t\) swept from the positive x-axis.
3Step 3: Write the Parametric Equations
For a circle of radius \(a\), the parametric equations are derived using sine and cosine functions: \( x = a \cos(t) \) and \( y = a \sin(t) \), where \(t\) is the parameter that varies from \(0\) to \(2\pi\).
4Step 4: Verify the Parametric Form
Substitute both parametric equations into the original circle equation: \(x = a \cos(t)\) and \(y = a \sin(t)\). Then \(x^2 + y^2 = (a \cos(t))^2 + (a \sin(t))^2 = a^2 \cos^2(t) + a^2 \sin^2(t)\). Factor out \(a^2\), giving \(a^2(\cos^2(t) + \sin^2(t)) = a^2\), which verifies the identity \(\cos^2(t) + \sin^2(t) = 1\). Therefore, the parametric equations satisfy the original circle equation.
Key Concepts
Equation of a CircleParametric FormTrigonometric FunctionsCircle Centered at Origin
Equation of a Circle
At the heart of this exercise is the simple yet elegant equation of a circle. The standard form of a circle's equation is given by \(x^2 + y^2 = a^2\), which describes a circle with a radius \(a\) centered at the origin, i.e., the point (0,0) on the Cartesian coordinate system.
Understanding this equation helps us to see that every point \(x, y\) on the circle is equidistant from the origin, with the distance being the radius \(a\). This equation is fundamental to many applications in mathematics, including geometry and trigonometry.
It's crucial to realize that changing \(a\) changes the size of the circle; specifically, increasing \(a\) makes the circle larger, while decreasing \(a\) makes it smaller. This equation is foundational for converting into parametric form, a key objective of this exercise.
Understanding this equation helps us to see that every point \(x, y\) on the circle is equidistant from the origin, with the distance being the radius \(a\). This equation is fundamental to many applications in mathematics, including geometry and trigonometry.
It's crucial to realize that changing \(a\) changes the size of the circle; specifically, increasing \(a\) makes the circle larger, while decreasing \(a\) makes it smaller. This equation is foundational for converting into parametric form, a key objective of this exercise.
Parametric Form
Converting the standard circle equation into a parametric form involves expressing the coordinates \(x\) and \(y\) as functions of a parameter, often denoted as \(t\). This parameter usually represents an angle, allowing a continuous representation of the circle.
In this exercise, we find the parametric equations for the circle by applying trigonometric functions:
In this exercise, we find the parametric equations for the circle by applying trigonometric functions:
- The parametric equation for x is \(x = a \, \cos(t)\)
- The parametric equation for y is \(y = a \, \sin(t)\)
Trigonometric Functions
Trigonometric functions \(\cos(t)\) and \(\sin(t)\) are deeply rooted in the parametrization of the circle. They inherently provide a natural way to express the circular path in mathematics.
Here's why they fit perfectly in understanding circles:
Here's why they fit perfectly in understanding circles:
- Both functions oscillate between -1 and 1, which enables them to perfectly map onto the unit circle where the radius is 1.
- Cosine corresponds to the x-coordinate, and sine corresponds to the y-coordinate of a point on the circle.
- The identity \(\cos^2(t) + \sin^2(t) = 1\) is crucial as it verifies that these functions maintain the constant radius in the parametric form of the circle equation.
Circle Centered at Origin
A circle centered at the origin is a particularly straightforward type of circle to work with because of its symmetry and simplicity.
This clarity arises because the center is at the point (0,0), removing any "shift" in the equation. Modern applications love such symmetry since they reduce complexity. It can simplify mathematical analysis and calculations significantly.
When dealing with parametric equations, this origin-centered circle simplifies computation. It ensures the transformation using \(a\, \cos(t)\) and \(a\, \sin(t)\) directly and purely reflects the radius without additional terms shifting it horizontally or vertically.
This specialization highlights the elegance of mathematical design, allowing mathematicians to focus on the properties and implications of circular motion and path without distractions from more complex translations.
This clarity arises because the center is at the point (0,0), removing any "shift" in the equation. Modern applications love such symmetry since they reduce complexity. It can simplify mathematical analysis and calculations significantly.
When dealing with parametric equations, this origin-centered circle simplifies computation. It ensures the transformation using \(a\, \cos(t)\) and \(a\, \sin(t)\) directly and purely reflects the radius without additional terms shifting it horizontally or vertically.
This specialization highlights the elegance of mathematical design, allowing mathematicians to focus on the properties and implications of circular motion and path without distractions from more complex translations.
Other exercises in this chapter
Problem 26
Use a graphing device to graph the hyperbola. \(\frac{x^{2}}{100}-\frac{y^{2}}{64}=1\)
View solution Problem 26
Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Focus \(F\left(0,-\frac{1}{2}\right)\)
View solution Problem 27
Use a graphing device to graph the ellipse. $$ 6 x^{2}+y^{2}=36 $$
View solution Problem 27
(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$2 x^{2}-4 x y+2 y^{2}-5 x-5=0$$
View solution