Problem 9 - Precalculus: Functions and Graphs

Question

Find an equation in \(x\) and \(y\) whose graph contains the points on the curve \(C\). Sketch the graph of \(C\), and indicate the orientation. $$x=4 \cos t+1, \quad y=3 \sin t, \quad 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

The equation is \(\left(\frac{x-1}{4}\right)^2 + \frac{y^2}{9} = 1\). The curve is an ellipse with center (1, 0).

Step 1: Understand the Parametric Equations

The given parametric equations are \(x(t) = 4 \cos t + 1\) and \(y(t) = 3 \sin t\). These describe a set of points \((x, y)\) on a curve \(C\) as the parameter \(t\) varies from 0 to \(2\pi\).

Step 2: Identify the Type of Curve

The equations \(x=4 \cos t + 1\) and \(y=3 \sin t\) suggest that the curve is an ellipse. This is because they resemble the general parametric form of an ellipse, \(x = a \cos t + h\), \(y = b \sin t + k\), where \((h, k)\) is the center, and \(a\) and \(b\) are the semi-major and semi-minor axes respectively.

Step 3: Express in Standard Form

To convert to the Cartesian (standard) form of an ellipse, eliminate the parameter \(t\). From \(x(t) = 4\cos t + 1\), solve for \(\cos t\), and from \(y(t)=3\sin t\), solve for \(\sin t\). These give \(\cos t = \frac{x-1}{4}\) and \(\sin t = \frac{y}{3}\).

Step 4: Use the Pythagorean Identity

Apply the identity \(\cos^2 t + \sin^2 t = 1\) to obtain:\[\left(\frac{x-1}{4}\right)^2 + \left(\frac{y}{3}\right)^2 = 1\]. This is the equation of the ellipse in standard form.

Step 5: Sketch the Graph and Consider Orientation

The ellipse is centered at \((1, 0)\), with semi-major axis length 4 along the x-direction and semi-minor axis length 3 along the y-direction. As \(t\) increases from 0 to \(2\pi\), \((x, y)\) traverses the ellipse in a counterclockwise orientation.

Key Concepts

Ellipse

An ellipse is a smooth, closed curve that resembles an elongated circle. It is defined as the set of points for which the sum of the distances to two fixed points, called foci, is constant. In simpler terms, imagine stretching a circle along one of its axes. This is how an ellipse is formed.
Here are some key aspects of ellipses:

• The longest diameter is the major axis, while the shortest is the minor axis.
• The points where these axes intersect are known as the center of the ellipse.
• In mathematics, ellipses can be expressed in parametric equations, which is useful for generating points of the curve dynamically.

In the given exercise, we have standard parametric equations for an ellipse: \[ x = a \cos t + h \]\[ y = b \sin t + k \]For our problem, the ellipse is centered at \((1, 0)\), with 4 as the semi-major axis length along the x-axis, and 3 as the semi-minor axis length along the y-axis.

Cartesian form

The Cartesian form of an equation is a standard way to represent curves using algebraic equations in terms of x and y. Transforming parametric equations into Cartesian form provides a more general understanding and can be easier to graph on a standard coordinate system.
For our ellipse, transforming from parametric to Cartesian involves eliminating the parameter \(t\). This was done using trigonometric identities:

• We start with the equations: \(x = 4 \cos t + 1\) and \(y = 3 \sin t\).
• Isolating \(\cos t\) and \(\sin t\) gives: \(\cos t = \frac{x-1}{4}\) and \(\sin t = \frac{y}{3}\).
• Applying the identity \(\cos^2 t + \sin^2 t = 1\) helps transform these into the Cartesian form: \[ \left(\frac{x-1}{4}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \]

This results in an easy-to-read algebraic equation that maps out all points \((x, y)\) on the ellipse.

Pythagorean identity

The Pythagorean identity is a fundamental trigonometric principle stating that for any angle \(t\), \[ \cos^2 t + \sin^2 t = 1 \]This identity is essential when dealing with circles and ellipses, as it relates the trigonometric functions of an angle directly to the geometry of these shapes.
By using this identity, we can connect trigonometric relationships to algebraic expressions. In the exercise, the Pythagorean identity allows us to manipulate parametric forms and convert them to Cartesian forms easily. This is achieved by substituting the expressions for \(\cos t\) and \(\sin t\) into the identity, effectively eliminating the parameter \(t\), and providing a neat algebraic description of the ellipse.
Key points:

• \(\cos^2 t + \sin^2 t = 1\) is valid for any angle \(t\), which makes it universally applicable.
• This simplifies expressing curves such as ellipses in Cartesian form.
• It serves as a bridge between algebra and trigonometry, linking different mathematical domains.

Understanding this identity can simplify many problems involving trigonometric curves.