Question

Create a vector-valued function whose graph matches the given description. An ellipse, centered at (3,-2) with horizontal major axis of length 6 and minor axis of length \(4,\) traced once clockwise on \([0,2 \pi]\)

Short answer

\(\mathbf{r}(t) = \langle 3 + 3\cos(t), -2 - 2\sin(t) \rangle\) traces the ellipse clockwise.

Step-by-step solution

Understanding the Problem

We need to create a vector-valued function that describes an ellipse centered at \((3, -2)\) with a horizontal major axis and a minor axis. The major axis is 6 units in length, and the minor axis is 4 units long. The ellipse should be traced clockwise over the interval \([0, 2\pi]\).

Derive the Parametric Equations of an Ellipse

For an ellipse centered at \((h, k)\), with a major axis of length \(2a\) and a minor axis of length \(2b\), the standard parametric equations are \(x(t) = h + a\cos(t)\) and \(y(t) = k + b\sin(t)\). Here, \(a = 3\) and \(b = 2\).

Adjust for Clockwise Tracing

To trace the ellipse clockwise, we replace \(t\) with \(-t\) in the sine function, obtaining: \(x(t) = 3 + 3\cos(t)\) and \(y(t) = -2 - 2\sin(t)\).

Write the Vector-Valued Function

Combining the x and y components, the vector-valued function is \(\mathbf{r}(t) = \langle 3 + 3\cos(t), -2 - 2\sin(t) \rangle\) for \(t \in [0, 2\pi]\).

Key concepts

Ellipse

An ellipse is a geometric shape that looks like a stretched circle. Imagine a circle being pulled from two sides, and you get an ellipse.
The ellipse has two axes, a major and a minor axis.

• The **major axis** is the longest diameter of the ellipse.
• The **minor axis** is the shortest diameter.

In the given problem, the major axis of the ellipse measures 6 units, and the minor axis is 4 units.
This tells us how wide and tall the ellipse is when placed on a coordinate plane. The center of the ellipse is at the point (3, -2).
This means the entire shape is shifted 3 units right and 2 units down from the origin.

Parametric Equations

Parametric equations allow us to describe curves like ellipses using a parameter, usually denoted as \( t \).
For an ellipse centered at \((h, k)\), the general parametric form is:

• \(x(t) = h + a\cos(t)\)
• \(y(t) = k + b\sin(t)\)

The parameters \(a\) and \(b\) represent half the lengths of the major and minor axes, respectively.
In our problem, \(a = 3\) (half of the horizontal major axis) and \(b = 2\) (half of the vertical minor axis).These equations parametrize the ellipse, meaning they define the ellipse's location for each value of \(t\).
As \(t\) changes from \(0\) to \(2\pi\), the equations take us around the entire shape.

Clockwise Tracing

Clockwise tracing of a path or shape takes you in a direction around it like the hands of a clock, starting from the top.
To make the ellipse trace clockwise, we change the direction of one component of the motion.In mathematical terms, this is done by adjusting the sign of the trigonometric function responsible for the vertical movement.
Instead of \(y(t) = k + b \sin(t)\), we use \(y(t) = k - b \sin(t)\).
This effectively reverses the vertical motion, making \(y\) decrease when it would normally increase.The trick swaps the direction in which points move along the path of the ellipse from counterclockwise to clockwise.
So in our solution, we've used \(-2 - 2\sin(t)\) to make the ellipse trace clockwise.

Vector Function

A vector function describes a curve by combining its horizontal and vertical components into a single mathematical expression.
For any point on our ellipse, this is shown as a vector \(\mathbf{r}(t)\).In our exercise, the vector function is written as:\[ \mathbf{r}(t) = \langle 3 + 3\cos(t), -2 - 2\sin(t) \rangle \] What does this mean?

• \(3 + 3\cos(t)\) gives the x-coordinate (horizontal position).
• \(-2 - 2\sin(t)\) gives the y-coordinate (vertical position).

By using this format, we capture both directions of movement at once.
This vector includes the position (3, -2) as its center and makes sure the ellipse traces along the correct path defined by \(t\).