AOS5 Topic 6: Parametric Equations

Definition:

Using parametric equations, one can express the relationship between two or more variables in terms of a third variable, which is usually referred to as a parameter. These formulae translate a point's coordinates into terms of a parameter.

For example, consider the equations:

\[ x = f(t) \] \[ y = g(t) \]

Here, \(x\) and \(y\) are functions of the parameter \(t\). The parameter \(t\) can represent time, angle, or any other varying quantity. The set of equations \(x = f(t)\) and \(y = g(t)\) defines a curve in the Cartesian plane as the parameter \(t\) varies.

·           vector equations and parametric equations of curves in two or three dimensions involving a parameter (and the corresponding Cartesian equation in the two-dimensional case)

Created with GeoGebra, by SKKY-Matrix-Lab, Link

Types of parametric equations:

Here are the some types of parametric equations:

Linear Parametric Equations:

\[ x = at + c \]

\[ y = bt + d \]

Where \( a \), \( b \), \( c \), and \( d \) are constants.

Circular Parametric Equations (Unit circle):

\[ x = \cos(t) \]

\[ y = \sin(t) \]

Where \( t \) ranges from \( 0 \) to \( 2\pi \).

Ellipse Parametric Equations:

\[ x = a \cos(t) \]

\[ y = b \sin(t) \]

Where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively.

Hyperbola Parametric Equations:

\[ x = a \sec(t) \]

\[ y = b \tan(t) \]

Where \( a \) and \( b \) are constants.

Parabola Parametric Equations (Vertical):

\[ x = at^2 \]

\[ y = 2at \]

Where \( a \) is a constant.

Parabola Parametric Equations (Horizontal):

\[ x = 2at \]

\[ y = at^2 \]

Where \( a \) is a constant.

Example 1

A circle is defined by the parametric equations:
\( x = 2 + 3\cos(\theta) \) and \( y = 1 + 3\sin(\theta) \) for \( \theta \in [0, 2\pi] \)
Find the Cartesian equation of the circle, and state the domain and range of this relation.

Solution
Domain: The range of the function with the rule \( x = 2 + 3 \cos \theta \) is \([-1, 5]\). Hence the domain of the corresponding Cartesian relation is \([-1, 5]\).
Range: The range of the function with the rule \( y = 1 + 3 \sin \theta \) is \([-2, 4]\). Hence the range of the corresponding Cartesian relation is \([-2, 4]\).

Cartesian equation:
Rewrite the parametric equations as:
\[ \frac{x - 2}{3} = \cos \theta \quad \text{and} \quad \frac{y - 1}{3} = \sin \theta \]
Square both sides of each of these equations and add:
\[ \left(\frac{x - 2}{3}\right)^2 + \left(\frac{y - 1}{3}\right)^2 = \cos^2 \theta + \sin^2 \theta = 1 \] \[ \left(\frac{x - 2}{3}\right)^2 + \left(\frac{y - 1}{3}\right)^2 = 1 \] \[ (x - 2)^2 + (y - 1)^2 = 9 \]

Example 2

Find the Cartesian equation of the curve with parametric equations:
\[ x = 3 + 3 \sin t \] \[ y = 2 - 2 \cos t \]
for \( t \in \mathbb{R} \).

Solution
We can rearrange the two equations as:
\[ \frac{x - 3}{3} = \sin t \quad \text{and} \quad \frac{2 - y}{2} = \cos t \]
Now, square both sides of each equation and add:
\[ \left(\frac{x - 3}{3}\right)^2 + \left(\frac{2 - y}{2}\right)^2 = \sin^2 t + \cos^2 t = 1 \]
Since \( (2 - y)^2 = (y - 2)^2 \), this equation can be written more neatly as:
\[ \left(\frac{x - 3}{3}\right)^2 + \left(\frac{y - 2}{2}\right)^2 = 1 \]
This is the equation of an ellipse with center at \( (3, 2) \) and axis intercepts at \( (3, 0) \) and \( (0, 2) \).

Example 3

Find the Cartesian equation of the curve with parametric equations:
\[ x = 3 \sec t \] \[ y = 4 \tan t \]
for \( t \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \).
Describe the curve.

Solution:
Rearrange the two equations:
\[ \frac{x}{3} = \sec t \quad \text{and} \quad \frac{y}{4} = \tan t \]
Square both sides of each equation and subtract:
\[ \frac{x^2}{9} - \frac{y^2}{16} = \sec^2 t - \tan^2 t = 1 \]
The Cartesian equation of the curve is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \).
The range of the function with rule \( x = 3 \sec t \) for \( t \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \) is \((-\infty, -3]\). Hence the domain for the graph is \((-\infty, -3]\).
The curve is the left branch of a hyperbola centered at the origin with x-axis intercept at \((-3, 0)\). The equations of the asymptotes are \( y = \frac{4x}{3} \) and \( y = -\frac{4x}{3} \).

Example 4

A curve is defined parametrically by the equations:
\[ x = at^2 \] \[ y = 2at \]
for \( t \in \mathbb{R} \) where \( a \) is a positive constant. Find:
a) The Cartesian equation of the curve
b) The equation of the line passing through the points where \( t = 1 \) and \( t = -2 \)
c) The length of the chord joining the points where \( t = 1 \) and \( t = -2 \).

Solution
a) The second equation gives \( t = \frac{y}{2a} \). Substitute this into the first equation:
\[ x = at^2 = a \left( \frac{y}{2a} \right)^2 = a \left( \frac{y}{4a^2} \right) = \frac{y^2}{4a} \]
This can be written as \( y^2=4ax \).

b) At \( t = 1 \), \( x = a \) and \( y = 2a \). This is the point \((a, 2a)\).
At \( t = -2 \), \( x = 4a \) and \( y = -4a \). This is the point \((4a, -4a)\).
The gradient of the line is:
\[ m = \frac{2a - (-4a)}{a - 4a} = \frac{6a}{-3a} = -2 \]
Therefore, the equation of the line is:
\[ y - 2a = -2(x - a) \]
which simplifies to \( y = -2x + 4a \).
c) The chord joining \((a, 2a)\) and \((4a, -4a)\) has length:
\[ \sqrt{(a - 4a)^2 + (2a - (-4a))^2} = \sqrt{9a^2 + 36a^2} = \sqrt{45a^2} = 3\sqrt{5}a \quad \text{(since } a > 0 \text{)} \]

Example 5

A circle has center (1, 3) and radius 2. If parametric equations for this circle are:
\[ x = a + b \cos(2\pi t) \] \[ y = c + d \sin(2\pi t) \]
where \( a \), \( b \), \( c \), and \( d \) are positive constants, state the values of \( a \), \( b \), \( c \), and \( d \).

Solution:
Given that the center of the circle is at (1, 3) and the radius is 2, we can use the parametric equations of a circle to find the values of \( a \), \( b \), \( c \), and \( d \).
For a circle with center \((a, c)\) and radius \(r\), the parametric equations are:
\[ x = a + r \cos(2\pi t) \] \[ y = c + r \sin(2\pi t) \]

Comparing these equations with the given equations \( x = a + b \cos(2\pi t) \) and \( y = c + d \sin(2\pi t) \), we can identify the values of \( a \), \( b \), \( c \), and \( d \):
\[ a = 1 \] \[ b = 2 \] \[ c = 3 \] \[ d = 2 \]
Therefore, the values of \( a \), \( b \), \( c \), and \( d \) are \( 1 \), \( 2 \), \( 3 \), and \( 2 \) respectively.

Example 6

An ellipse has x-axis intercepts (-4, 0) and (4, 0) and y-axis intercepts (0, 3) and (0, -3). State a possible pair of parametric equations for this ellipse.

Solution:
Given that the x-axis intercepts of the ellipse are (-4, 0) and (4, 0), the length of the major axis (the x-axis) is the distance between these two points, which is 8 units. However, the length of the major axis is represented by \(2a\), where \(a\) is the semi-major axis. Therefore, \(2a = 8\) which implies \(a = 4\).
Similarly, given that the y-axis intercepts of the ellipse are (0, 3) and (0, -3), the length of the minor axis (the y-axis) is the distance between these two points, which is 6 units. However, the length of the minor axis is represented by \(2b\), where \(b\) is the semi-minor axis. Therefore, \(2b = 6\) which implies \(b = 3\).
Now, we have determined the values of \(a\) and \(b\), which are the semi-major and semi-minor axes, respectively. These values are crucial for defining the parametric equations of the ellipse.
The parametric equations for an ellipse centered at the origin (0, 0) are typically given by:
\[ x = a \cos(t) \]
\[ y = b \sin(t) \]
where \(t\) ranges from \(0\) to \(2\pi\).

Substituting the values \(a = 4\) and \(b = 3\) into these equations, we get:
\[ x = 4 \cos(t) \]
\[ y = 3 \sin(t) \]
These equations represent the ellipse with x-axis intercepts (-4, 0) and (4, 0) and y-axis intercepts (0, 3) and (0, -3).

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

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