Every function we have studied so far has had one input and one output. Usually our variable for the input is [math]x[/math] and our variable for the output is [math]y[/math]. To indicate that our output variable is [i]dependent[/i] on our choice of input variable we write [math]y=f\left(x\right)[/math], literally read "[math]y[/math] is a function of [math]x[/math]".[br][br]In this unit we will introduce a new type of function, one that has one input, usually named [math]t[/math] and two outputs, usually named [math]x[/math] and [math]y[/math]. Our two output variables are [i]dependent [/i]on our choice of input variable. We will often write [math]x\left(t\right)[/math] and [math]y\left(t\right)[/math] indicating that [math]x[/math] and [math]y[/math] are each functions of [math]t[/math]. Later when you take my Multivariable Calculus class I will teach you to write [math]c\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)[/math] indicating that the function [math]c[/math] inputs a single number [math]t[/math] and outputs a two-dimensional point [math]\left(x\left(t\right),y\left(t\right)\right)[/math].[br][br]We refer to the functions [math]x\left(t\right)[/math] and [math]y\left(t\right)[/math] as [b][color=#ff0000]parametric equations[/color][/b], and the variable [math]t[/math] is called the [b][color=#ff0000]parameter[/color][/b]. In the GeoGebra file below you can practice plotting the output of such a function by plugging in various values for the parameter.
What point is plotted when [math]t=0[/math]?
[math]\left(-1,3\right)[/math]
What do the arrows represent?
The direction of movement as the parameter [math]t[/math] increases.
The resulting picture is the graph of a function of the form [math]y=f\left(x\right)[/math]. What is the equation and the domain of that function?[br][Hint: Start with the equation for [math]x\left(t\right)[/math] and rearrange to have an equation for [math]t[/math]. Then perform a substitution in the equation for [math]y\left(t\right)[/math] to obtain an equation relating only [math]x[/math] and [math]y[/math]).
[math]y=3-4\left(\frac{1}{3}\left(x+1\right)\right)[/math][br]or more simply: [math]y=-\frac{4}{3}x+\frac{5}{3}[/math][br]The domain is [math]x\in\left[-16,14\right][/math]
The parametric equations [math]x\left(t\right)=3t-1[/math] and [math]y\left(t\right)=3-4t[/math] ended up plotting the graph of a line segment. Describe how to find the slope of this line segment?
One method is to find the equation for the line segment as we did in the last question (this technique is called [b][color=#ff0000]eliminating the parameter[/color][/b]). The slope is then the coefficient of [math]x[/math]. That is:[br][math]m=-\frac{4}{3}[/math][br][br]However if all we need is the slope we can skip the step of eliminating the parameter and read the slope directly from the parametric equations. Remember that slope is change in [math]y[/math] over the change in [math]x[/math]. The change in [math]y[/math] is the coefficient of the parameter in the [math]y\left(t\right)[/math] equation and the change in [math]x[/math] is similarly the coefficient of the parameter in the [math]x\left(t\right)[/math] equation.[br]Hence the slope is:[br][math]m=\frac{\text{change in y}}{\text{change in x}}=\frac{-4}{3}=-\frac{4}{3}[/math]