Suppose we want to graph the circle with equation (x-2)² + (y-1)² = 4. This is easy to do by hand, but not so easy with a graphing calculator, because the circle is not the graph of a function.

Exercises:

1.    How do you know that the circle is not the graph of a function?

2.    Solve the equation of the circle for y, either by hand or using the calculator's solve command. Graph the resulting solution(s). Does your picture look like a circle? Can you make it look like a circle by adjusting the WINDOW parameters?
 

An alternative approach to graphing a circle, and in fact an alternative approach to thinking about what graphs represent, is to picture the circle as the path followed by a very small moving object. At each time t we'll compute the x-coordinate and the y-coordinate of the moving object. Suppose it starts (at t=0) at the point (4,1) and travels counterclockwise around the circle, completing one revolution at time t=2p (seconds, minutes, whatever).
 

3.    Find an equation for the x-coordinate of the object when 0 <= t <= 2.

4.    Find an equation for the y-coordinate of the object when 0 <= t <= 2.

5.     Graph these parametric equations for the circle on your calculator in PARAMETRIC mode. (Make sure Angle is set to RADIAN on the MODE screen while you're there changing Graph mode to PARAMETRIC.)

6.    Change tmax to 4 on the WINDOW screen. Does the resulting graph look different?

7.    To see the difference, go to the Graph Formats menu (F1,9) and change Leading Cursor to ON. Now ReGraph (F4). What is the difference between the old graph and the new one?

8.    If you also change Graph Order to SIMUL (in the Graph Formats menu), you can distinguish between places where two parametric graphs intersect and places where the two moving objects would collide. See problem 32 on page 55 of Stewart.

Unfortunately, the Intersection calculator, F5,5, is inactive in PARAMETRIC mode, so you'll have to find intersection points for problem 32 by hand.
The Solve command on the HOME screen can help you find collision points, but using it to find ordinary intersection points is tricky.
    For collision points, the command you want is solve(xt1(t)=xt2(t) and yt1(t)=yt2(t),t).
    If there are any solutions, the @n1 you get stands for an arbitrary integer; there are infinitely many times for each collision point.
    For ordinary intersection points, you want to allow different times for different curves.
    The command is solve(xt1(t1)=xt2(t2) and yt1(t1) = yt2(t2), {t1,t2}), but it's slow and unreliable.

9.    Compare the graphs of x=t, y=t²;  x=t², y=t; and  x=t², y=-t;  for -5 < t < 5.

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