A parametric function consists of a pair of functions, given by f(t) = (x(t),\, y(t)), where x(t) and y(t) are functions of an independent variable t, also known as the parameter. In the context of particle motion, the parameter t often represents time, while the dependent variables x and y indicate the particle's position in the plane.

The graph of a parametric function traces the path of the particle as it moves through the plane over time. Each point on the graph corresponds to a specific position of the particle at a particular time, represented by the coordinates (x(t),\, y(t)).

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To analyze the motion of the particle, we can examine the individual parametric equations x(t) and y(t). These equations describe the horizontal and vertical components of the particle's motion, respectively. By studying these equations, we gain valuable insights into the behavior of the particle and its trajectory in the plane. The maximum and minimum values of the functions x(t) and y(t) correspond to the horizontal and vertical extrema of the particle's motion, respectively. These extrema represent the farthest points the particle reaches in the horizontal and vertical directions. By identifying these extrema, we can better understand the overall shape and characteristics of the particle's path.

Particle motion

The movement of a particle in space, which can be modeled using parametric functions in a two-dimensional plane.

Horizontal extrema

The farthest points a particle reaches in the horizontal direction, corresponding to the maximum and minimum values of the function x(t).

Vertical extrema

The farthest points a particle reaches in the vertical direction, corresponding to the maximum and minimum values of the function y(t).

Examples

Example 1

A particle moves in the plane following the given parametric equations: x(t) = 4t - 3 and y(t) = t^{2} + 1, for t in the domain [0,\, 3].

Find the parametric function f(t) that represents the motion of the particle.

Worked Solution

Create a strategy

Substitute the given x(t) and y(t) functions into the general form: f(t) = (x(t),\, y(t)), and then state the domain of the parameter.

Apply the idea

\displaystyle f(t)	\displaystyle =	\displaystyle (x(t),\, y(t))	Write the general form
\displaystyle f(t)	\displaystyle =	\displaystyle (4t-3,\,t^{2}+1)	Substitute x(t) = 4t - 3 and t^{2}+1

Specify the domain of the parameter t. We are given the domain of t as [0,\, 3]. This domain restricts the motion of the particle to the time interval between t = 0 and t = 3, which allows us to analyze the behavior of the particle during this specific period.

t	x(t)	y(t)
0	-3	1
0.75	0	1.5625
1.5	3	3.25
2.25	6	6.0625
3	9	10

So, the parametric function representing the motion of the particle for t in the domain [0,\, 3] is f(t) = (4t - 3,\, t^2 + 1). By finding this function, we can now study the particle's motion and analyze its behavior in the plane during the specified time interval.

Reflect and check

In this problem, we were given the parametric equations for x(t) and y(t) that describe the motion of a particle in the plane. By substituting the given equations into the general form of a parametric function, we were able to find the function f(t) that represents the particle's motion. By specifying the domain of the parameter t, we restrict the motion of the particle to a particular time interval, which could be useful for analyzing its behavior during that period.

Example 2

A particle is moving along a trajectory described by the following parametric equations: \\x(t) = 3\cos(t) and y(t) = 4\sin(t).

Determine the position of the particle at t = \dfrac{\pi}{2}.

Worked Solution

Create a strategy

Substitute t = \dfrac{\pi}{2} to the given parametric equations.

Apply the idea

\displaystyle x\left(\dfrac{\pi}{2}\right)	\displaystyle =	\displaystyle 3\cos\left(\dfrac{\pi}{2}\right)	Substitute t = \dfrac{\pi}{2}
\displaystyle x\left(\dfrac{\pi}{2}\right)	\displaystyle =	\displaystyle 3 \times 0	Evaluate \cos\left(\dfrac{\pi}{2}\right)
\displaystyle \pi x	\displaystyle =	\displaystyle 0	Multiply both sides by 2
\displaystyle x	\displaystyle =	\displaystyle 0	Divide both sides by \pi

\displaystyle y	\displaystyle =	\displaystyle 4\sin\left(\dfrac{\pi}{2}\right)	Substitute t = \dfrac{\pi}{2}
\displaystyle y	\displaystyle =	\displaystyle 4	Evaluate

So, the position of the particle at t = \dfrac{\pi}{2} is (0,\, 4).

Graph the path of the particle for t in the interval [0,\, 2\pi].

Worked Solution

Create a strategy

We need to solve for \cos t and \sin t, then substitute it to Pythagorean identity: \cos^{2} t + \sin^{2} t = 1. And then plot the graph the equation.

Apply the idea

Solve for \cos t and \sin t:

\displaystyle x	\displaystyle =	\displaystyle 3\cos t	Write the parametric equation
\displaystyle \dfrac{x}{3}	\displaystyle =	\displaystyle \cos t	Divide both sides by 3

\displaystyle y	\displaystyle =	\displaystyle 4\sin t	Write the parametric equation
\displaystyle \dfrac{y}{4}	\displaystyle =	\displaystyle \sin t	Divide both sides by 4

Substitute the values to Pythagorean identity:

\displaystyle \cos^{2}t + \sin^{2} t	\displaystyle =	\displaystyle 1	Write the Pythagorean identity
\displaystyle \left(\dfrac{x}{3}\right)^{2}+ \left(\dfrac{y}{4}\right)^{2}	\displaystyle =	\displaystyle 1	Substitute \cos t = \dfrac{x}{3} and \sin t = \dfrac{y}{4}
\displaystyle \dfrac{x^2}{9} + \dfrac{y^2}{16}	\displaystyle =	\displaystyle 1	Evaluate

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As we can see from the graph, the particle follows an elliptical path through the plane as time progresses. By examining the graph, we can gain insights into the overall behavior of the particle's motion and its trajectory.

Idea summary

A parametric function consists of a pair of functions, given by f(t) = (x(t),\, y(t)), where x(t) and y(t) are functions of an independent variable t, also known as the parameter.

Identifying horizontal and vertical extrema

We will explore how to identify the horizontal and vertical extrema of a particle's motion using the functions x(t) and y(t). Extrema are the maximum and minimum values that represent the furthest points reached by the particle in the horizontal and vertical directions.

To determine the horizontal extrema, we focus on the function x(t) which represents the horizontal position of the particle at time t. The maximum value of x(t) corresponds to the furthest point reached by the particle in the positive x-direction, while the minimum value represents the furthest point in the negative x-direction.

Similarly, to find the vertical extrema, we examine the function y(t) which represents the vertical position of the particle. The maximum value of y(t) indicates the highest point reached by the particle, while the minimum value represents the lowest point.

Alternatively, for certain types of functions, we can use graphical analysis to identify the extrema. Graphical analysis is a method used to visually identify the extrema of a particle's motion in the plane by plotting the parametric equations x(t) and y(t) on a coordinate system.

Examples

Example 3

Consider a particle moving in the xy-plane, with its position at time t given by the functions \\x(t) = t^{2} - 2t and y(t) = t + 1. Find the horizontal and vertical extrema of the particle's motion.

Worked Solution

Create a strategy

To find the horizontal and vertical extrema for the given parametric equations x(t) = 3t - 1 and y(t) = 2t + 1, we will first determine the appropriate domain for t. Once we have the domain, we will analyze the behavior of x(t) and y(t) within that domain to identify the maximum and minimum values, which correspond to the horizontal and vertical extrema.

Apply the idea

Determining Horizontal Extrema:

To find the maximum and minimum values of x(t), we need to locate the vertex of the quadratic function. The vertex occurs at the value of t that minimizes or maximizes x(t). We can use the formula t = \dfrac{-b}{2a}, where a and b are the coefficients of the quadratic function x(t).
In this case, a = 1 and b = -2. Applying the formula, we find t = \dfrac{-(-2)}{(2 \times 1)} = 1. Therefore, the vertex of x(t) occurs at t = 1.
To determine the corresponding value of x(t) at the vertex, we substitute t = 1 into the expression for x(t). x(1) = (1)^{2} - 2(1) = -1. Hence, the vertex of x(t) is (1,\, -1).
Thus, the maximum value of x(t) is at t = 1, and the minimum value does not exist in this example.

Determining Vertical Extrema:

To find the vertical extrema, we examine the function y(t) = t + 1. As it represents a linear function, there are no maximum or minimum values. The particle's vertical motion is unrestricted, continuously increasing without reaching a maximum or minimum point.

Example 4

A particle moves in the xy-plane with its position given by the functions x(t) = 2 \sin(t) and \\y(t) = \cos(t) for t in the interval [0,\, 2\pi]. Find the horizontal and vertical extrema of the particle's motion.

Worked Solution

Create a strategy

To determine the horizontal and vertical extrema, we will analyze the functions x(t) = 2\sin(t) and y(t) = \cos(t) over the given interval [0,\, 2\pi]. We need to identify the maximum and minimum values of each function.

Apply the idea

Determining Horizontal Extrema:
To find the horizontal extrema, we need to analyze the function x(t) = 2\sin(t). Since \sin(t) oscillates between -1 and 1, the maximum and minimum values of x(t) occur when \sin(t) reaches its extreme values.

Maximum Value of x(t):

\displaystyle x(t)	\displaystyle =	\displaystyle 2\sin(t)	Write the function
	\displaystyle =	\displaystyle 2\sin\left(\dfrac{\pi}{2}\right)	Substitute t = \dfrac{\pi}{2}
	\displaystyle =	\displaystyle 2 \times 1	Simplify \sin(t) = 1
	\displaystyle =	\displaystyle 2	Evaluate

Minimum Value of x(t):

\displaystyle x(t)	\displaystyle =	\displaystyle 2\sin(t)	Write the function
	\displaystyle =	\displaystyle 2\sin\left(\dfrac{3\pi}{2}\right)	Substitute t = \dfrac{3\pi}{2}
	\displaystyle =	\displaystyle 2 \times -1	Simplify \sin(t) = -1
	\displaystyle =	\displaystyle -2	Evaluate

Determining Vertical Extrema:
To find the vertical extrema, we need to analyze the function y(t) = \cos(t). Similar to \sin(t),\, \cos(t) also oscillates between -1 and 1.
Maximum Value of y(t):
\displaystyle y(t) \displaystyle = \displaystyle \cos(t) Write the function
\displaystyle = \displaystyle \cos(0) Substitute t = 0
\displaystyle = \displaystyle 1 Evaluate
Minimum Value of y(t):
\displaystyle y(t) \displaystyle = \displaystyle \cos(t) Write the function
\displaystyle = \displaystyle \cos(\pi) Substitute t = \pi
\displaystyle = \displaystyle -1 Evaluate

Idea summary

These maximum and minimum values correspond to the horizontal and vertical extrema of the particle's motion, representing the farthest points the particle reaches in the horizontal and vertical directions. By identifying these extrema, we can better understand the overall shape and characteristics of the particle's path.

Finding intercepts of parametric planar motion functions

In this section, we focus on finding intercepts of parametric functions modeling particle motion in the plane. The x-intercept of a parametric function occurs when the particle's path crosses the x-axis (y(t) = 0), and the y-intercept occurs when the path crosses the y-axis (x(t) = 0). These intercepts provide important information about the particle's motion and can help us better understand its trajectory.

Examples

Example 5

Given the parametric equations x(t) = t^{2} - 4t and y(t) = 2t - 4, find the x-intercept and y-intercept of the particle's path.

Worked Solution

Apply the idea

Step 1: Set y(t) equal to 0 and solve for t to find the x-intercept.

\displaystyle 2t-4	\displaystyle =	\displaystyle 0	Set y(t) equal to 0
\displaystyle 2t	\displaystyle =	\displaystyle 4	Add 4 to both sides
\displaystyle t	\displaystyle =	\displaystyle 2	Divide both sides by 2

Step 2: Set x(t) equal to 0 and solve for t to find the y-intercept.

\displaystyle t_{1,\,2}	\displaystyle =	\displaystyle \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}	Write the quadratic formula
\displaystyle t_{1,\,2}	\displaystyle =	\displaystyle \dfrac{-(-4) \pm \sqrt{(-4)^{2}-4\times 1 \times 0}}{2\times 1}	Substitute a=1,\, b=-4,\, and c=0
\displaystyle t_{1,\,2}	\displaystyle =	\displaystyle \dfrac{-(-4) \pm 4}{2\times 1}	Evaluate the operations inside the \sqrt{}

Separate the solutions:

t_{1}= \dfrac{-(-4) + 4}{2\times 1} \,\,\,\,,\, t_{2} = \dfrac{-(-4) - 4}{2\times 1}

t_{1} = 4 \,\,\,\,,\, t_{2} = 0

Step 3: Substitute the t-values found in step 1 and step 2 back into the corresponding parametric equations to find thex and y coordinates of the intercepts.

\displaystyle x(t)	\displaystyle =	\displaystyle t^{2}-4t	Write the parametric equation
\displaystyle x(t)	\displaystyle =	\displaystyle 2^{2}-4 (2)	Substitute t=2
\displaystyle x(t)	\displaystyle =	\displaystyle -4	Evaluate

So, the x-intercept is (-4,\, 0).

\displaystyle y(t)	\displaystyle =	\displaystyle 2t−4	Write the parametric equation
\displaystyle y(t)	\displaystyle =	\displaystyle 2(0)-4	Substitute t=0
\displaystyle y(t)	\displaystyle =	\displaystyle -4	Evaluate

\displaystyle y(t)	\displaystyle =	\displaystyle 2t-4	Write the parametric equation
\displaystyle y(t)	\displaystyle =	\displaystyle 2(4)-4	Substitute t=4
\displaystyle y(t)	\displaystyle =	\displaystyle 4	Evaluate

So, the y-intercepts are (0,\, -4) and (0,\, 4).

Example 6

Find the x-intercept and y-intercept of the particle's path given by the parametric equations \\x(t) = 3t - 1 and y(t) = t^{3} - t.

Worked Solution

Apply the idea

Step 1: Set y(t) equal to 0 and solve for t to find the x-intercept.

\displaystyle t^{3}-t	\displaystyle =	\displaystyle 0	Write the parametric equations
\displaystyle t\left(t+1\right)\left(t-1\right)	\displaystyle =	\displaystyle 0	Factor out the equation

Use the zero factor principle: If ab=0 then a=0 or b=0.

\displaystyle t+1	\displaystyle =	\displaystyle 0	Write the first factor
\displaystyle t	\displaystyle =	\displaystyle -1	Subtract 1 from both sides

\displaystyle t-1	\displaystyle =	\displaystyle 0	Write the second factor
\displaystyle t	\displaystyle =	\displaystyle 1	Add 1 to both sides

The solutions are t=0,\,t=-1 and t=1.

Step 2: Set x(t) equal to 0 and solve for t to find the y-intercept.

\displaystyle 3t-1	\displaystyle =	\displaystyle 0	Write the parametric equation
\displaystyle 3t	\displaystyle =	\displaystyle 1	Add 1 to both sides
\displaystyle t	\displaystyle =	\displaystyle \dfrac{1}{3}	Divide both sides by 3

Step 3: Substitute the t-values found in step 1 and step 2 back into the corresponding parametric equations to find the x and y coordinates of the intercepts.

For solutions: t=0,\,t=-1 and t=1.

\displaystyle x(t)	\displaystyle =	\displaystyle 3(0)-1	Substitute t=0
	\displaystyle =	\displaystyle -1	Evaluate

\displaystyle x(t)	\displaystyle =	\displaystyle 3(-1)-1	Substitute t=-1
	\displaystyle =	\displaystyle -4	Evaluate

\displaystyle x(t)	\displaystyle =	\displaystyle 3(1)-1	Substitute t=1
	\displaystyle =	\displaystyle 2	Evaluate

So, the x-intercepts are (-1,\,0),\,(-4,\,0),\, and (2,\,0).

For solution: t=\dfrac{1}{3}.

\displaystyle y(t)	\displaystyle =	\displaystyle \left(\dfrac{1}{3}\right)^{3}-\dfrac{1}{3}	Substitute t=\dfrac{1}{3}
	\displaystyle =	\displaystyle -\dfrac{8}{27}	Evaluate

So, the y-intercept is \left(0,\,-\dfrac{8}{27}\right).

Idea summary

To find the x-intercepts, we set the y(t) function equal to zero and solve for the parameter t. We substitute the values of t back into the x(t) function to find the corresponding x-values.

To find the y-intercepts, we set the x(t) function equal to zero and solve for t. Then, we substitute these t-values into the y(t) function to find the corresponding y-values.

By finding the intercepts, we gain a better understanding of the particle's trajectory and how it interacts with the coordinate axes.

Outcomes

4.2.A

Identify key characteristics of a parametric planar motion function that are related to position.

4.2 Parametric functions modeling planar motion

Introduction

Ideas

Introduction to parametric functions for planar motion

Examples

Example 1

Example 2

Identifying horizontal and vertical extrema

Examples

Example 3

Example 4

Finding intercepts of parametric planar motion functions

Examples

Example 5

Example 6

Outcomes

4.2.A

What is Mathspace

About Mathspace

\displaystyle y(t)	\displaystyle =	\displaystyle \cos(t)	Write the function
	\displaystyle =	\displaystyle \cos(0)	Substitute t = 0
	\displaystyle =	\displaystyle 1	Evaluate