Question

In Exercises \(5-22,\) a parametrization is given for a curve. (a) Graph the curve. What are the initial and terminal points, if any? Indicate the direction in which the curve is traced. (b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized curve? $$x=4 \cos t, \quad y=2 \sin t, \quad 0 \leq t \leq 2 \pi$$

Short answer

a) The graph of the parametric equations is an ellipse with initial and terminal point at (4, 0). It's traced once counterclockwise. b) The cartesian equation of the curve is \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), and the parametrized curve traces out the entire ellipse.

Step-by-step solution

Part (a)

First, the given parametric equations are \(x=4 \cos t\) and \(y=2 \sin t\), where \(0 \leq t \leq 2\pi\). This describes a sinusoidal motion where the x-coordinate oscillates between -4 and 4, and the y-coordinate moves between -2 and 2. The initial point is \((x(0), y(0)) = (4, 0)\), and the terminal point is \((x(2\pi), y(2\pi)) = (4, 0)\). The curve is traced as t ranges from 0 to \(2\pi\).

Part (b)

To find the Cartesian equation of the curve, eliminate \(t\) from the given parametric equations. From the identity \(\sin^2(t) + cos^2(t) = 1\), obtain the equation \(\frac{x^2}{16} + \frac{y^2}{4} = 1\). This is the equation of an ellipse which the given parametric curve lies on. Since \(t\) ranges from \(0\) to \(2 \pi\), the parametric curve traces out the full ellipse.

Key concepts

Graphing Parametric Equations

Graphing parametric equations involves plotting points on a coordinate system where each point is determined by a pair of equations, each expressing a separate variable (typically the x and y coordinates) in terms of a third variable, commonly known as the parameter. Typically, this parameter is denoted by the letter 't'.

In the exercise given, the parametric equations are
\(x=4 \cos t\), \(y=2 \sin t\)
with the parameter 't' ranging from \(0\) to \(2\pi\). Here's how one would sketch the graph of such equations:

• Assign a range of values to 't', from \(0\) to \(2\pi\), as specified in the question.
• Calculate the corresponding x and y values for each t value to plot coordinate points.
• Plot these points on a graph, and connect them smoothly to illustrate the motion as 't' increases.
• The initial point is where 't' equals zero, and the terminal point is where 't' equals \(2\pi\).
• The direction in which the curve is traced is determined by the increasing values of 't'.

The resulting graph reveals the curve's trajectory as the parameter 't' changes within the given range.

Converting Parametric to Cartesian Equations

Converting parametric to Cartesian equations is a process of elimination, where the goal is to remove the parameter and describe the relationship between 'x' and 'y' directly.

In the provided exercise, we have the parametric equations \(x=4 \cos t\) and \(y=2 \sin t\). To convert these to a Cartesian equation, we utilize trigonometric identities, such as \(\sin^2(t) + \cos^2(t) = 1\).

• Express \(\cos t\) and \(\sin t\) in terms of 'x' and 'y' by rearranging the given parametric equations as \(\cos t = \frac{x}{4}\) and \(\sin t = \frac{y}{2}\).
• Square both equations to get \(\cos^2 t = \frac{x^2}{16}\) and \(\sin^2 t = \frac{y^2}{4}\).
• Add the squared equations. The left side harnesses the Pythagorean identity, becoming just '1', while the right side structures the Cartesian ellipse equation \(\frac{x^2}{16} + \frac{y^2}{4} = 1\).

The resulting equation represents an ellipse in Cartesian form and no longer depends on the parameter 't'.

Ellipse Equation from Parametric Form

The ellipse equation derived from parametric form can provide insights into the shape and orientation of the curve. When given a set of parametric equations that defines an ellipse, the goal is to express it in Cartesian form.

In this case, the equations \(x=4 \cos t\) and \(y=2 \sin t\) represent an ellipse in parametric form. The constants preceding the trigonometric functions determine the length of the semi-axes of the ellipse.

• The coefficient '4' in front of \(\cos t\) indicates that the ellipse extends 4 units along the x-axis. This defines the semi-major axis of the ellipse.
• Similarly, the coefficient '2' before \(\sin t\) describes a semi-minor axis that is 2 units long along the y-axis.

After converting to Cartesian form (as shown in the previous section), we obtain the standard ellipse equation \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), where the denominators (16 and 4) are the squares of the lengths of the semi-major and semi-minor axes, respectively.

Analyzing Parametric Curves

Analyzing parametric curves encompasses understanding their shape, behavior, and other geometric properties based on the given parametric equations.

For the example \(x=4 \cos t\), \(y=2 \sin t\), we know the curve represents an ellipse. We can further analyze the curve by considering the following:

• The angle parameter 't' correlates to the angle of rotation from the positive x-axis to the radius vector of the point (x,y) on the ellipse.
• Since 't' ranges from \(0\) to \(2\pi\), the curve completes a full oscillation around the ellipse, indicating that it traces the entire ellipse once.
• The speed at which the curve is traced depends on the derivative of the parametric equations with respect to 't'. This can show us how fast the point is moving along the curve at any instant.
• Any symmetry properties of the curve, such as reflectional symmetry over the x-axis or y-axis, can be determined by examining the parametric equations and how they change with 't'.

By considering these aspects, students can gain a stronger intuition about how parametric equations define curves and how the curves behave when traced out over their respective intervals.