In Exercises 67-70, let c(t)=(t^2-9, t^2-8 t) (see Eigure 19). FIGURE CANT COPY FIGURE 19 Plot o

step 1 All right, we're given this parametric curve and this picture of it. And you can look in the boo

In Exercises 67-70, let c(t)=(t^2-9, t^2-8 t) (see Eigure...

In Exercises 67-70, let $c(t)=\left(t^2-9, t^2-8 t\right)$ (see Eigure 19). FIGURE CANT COPY FIGURE 19 Plot of $c(t)=\left(t^2-9, t^2-8 t\right)$. Draw an arrow indicating the direction of motion, and determine the interval(s) of $t_{\text {-values corresponding to the }}$ portion(s) of the curve in each of the four quadrants. Let $c(t)=\left(t^2-9, t^2-8 t\right)$ (see Eigure 19).

In Exercises 67-70, let $c(t)=\left(t^2-9, t^2-8 t\right)$ (see Eigure 19).
FIGURE CANT COPY
FIGURE 19 Plot of $c(t)=\left(t^2-9, t^2-8 t\right)$.
Draw an arrow indicating the direction of motion, and determine the interval(s) of $t_{\text {-values corresponding to the }}$ portion(s) of the curve in each of the four quadrants.
Let $c(t)=\left(t^2-9, t^2-8 t\right)$ (see Eigure 19).

Key Concepts

Analysis of Quadrants

Determining which portions of a parametric curve lie in specific quadrants involves analyzing the signs of the x- and y-coordinates as functions of the parameter. This requires solving inequalities to find the intervals of t that correspond to the positive or negative values of x and y, which classify the location of the curve within the four quadrants of the Cartesian plane.

Direction of Motion

The direction of motion on a parametric curve is determined by the derivatives of the coordinate functions with respect to the parameter. By analyzing these derivatives, one can understand how the point moves along the curve as the parameter increases, which is crucial for drawing arrows indicating orientation and for determining how the curve traverses the coordinate plane.

Parametric Equations

Parametric equations describe a set of related quantities as functions of one or more parameters, typically denoted as t. They allow one to represent curves by expressing the x- and y-coordinates in terms of this parameter, which can provide insights into the curve’s structure and motion that may not be immediately evident from a standard Cartesian form.

Transcript

00:01 All right, we're given this parametric curve and this picture of it.

00:05 And you can look in the book if you want a better picture.

00:08 And it wants us to draw an arrow indicating the direction of motion and find the t values for the curves in each quadrant.

00:16 Like what t value gets you in the first quadrant and what gets you in the second quadrant, etc.

00:23 Okay, so let's go ahead and find out what t values give you these points that are on the x and y axes.

00:31 First list you x equals 0.

00:34 When x equals 0, then t squared minus 9 equals 0, so t is plus and minus 3.

00:41 And it's allowing t to be plus and minus 3, so it's not time, it's just a parameter.

00:48 And then y equals 0, t times t minus 8 equals 0.

00:56 So at t equals 0 and 8.

01:01 Okay, so let's just make us a little bit of a table here.

01:07 When t is zero, what's x and what's y? well, when t is zero, x is minus nine and y is zero, because that's where we got it.

01:22 When t is three, x is zero, and y is three times minus five, so minus 15.

01:34 And when t is eight, y is zero, and x is 64 minus 9, so 55.

01:44 Oh, i forgot to do minus 3.

01:45 I should have done that first.

01:47 When t is negative 3, then x is 0, and y is negative 3 times negative 11, so 33.

02:01 Okay, so first we're at minus 9, 0, which is minus 9 on the x, this one.

02:08 We'll call this 1.

02:13 And then oh we better better not call it one let's start at minus 3 when t is minus 3 we're at 0 3 0 on the x 3 and the y so right here that's where t is minus 3 and then when t is 0 we're at negative 9 0 okay so that's here t equals 0 and then when we're at 3 we're at 0 on the x minus 15 on the y so that's this one.

02:55 That's t equals 3.

02:57 And then when we're at t equals 8, we're here.

03:04 All right.

03:05 So you can see that the direction is going this way.

03:13 All right.

03:14 And then you can see that from t equals minus 3 to t equals 0.

03:20 You're in quadrant 2.

03:24 So minus, oops, round bracket, minus 3 to 0...