Parametric Curves

A Presentation for Pre-Calculus Students

By Paulo Tan

 

 

Lets take a close look at the parametric equation x=cos (t) and y=sin(t) where t goes from 0 to 2pi.  You see that the variables x and y will depend on t; and each t value gives us a coordinate point. 

 

So lets think about what kind of graph this parametric equation will produce.

We know that sine and cosine are periodic functions.  Hence, lets construct a table to get a sense of what kind of values and trends this equation gives us.  The table below describes the values of x and y as t increases. 

 

 

CASE 1 x=cos(t), y=sin(t)

t

t

x

y

0

0.00

1

0

pi/4

0.79

0.71

0.71

pi/3

1.05

0.50

0.87

pi/2

1.57

0.00

1.00

3pi/4

2.36

-0.71

0.71

5pi/6

2.62

-0.87

0.50

pi

3.14

-1.00

0.00

7pi/6

3.66

-0.87

-0.50

5pi/4

3.93

-0.71

-0.71

3pi/2

4.71

0.00

-1.00

11pi/6

5.76

0.86

-0.50

2pi

6.28

1.00

0.00

 

 

These values should not surprise you since you are familiar with trigonometric functions.  It also should be no surprise that the graph of this parametric function is that of a unit circle (fig1).

 

Fig1.

 

Now letÕs think about what kind of graph would be produced if the parametric equation is x=cos(3t) and y=sin(t).  Again, lets start with a table of values for this case. 

 

 

CASE 2

x=cos(3t)

y=sin(t)

 

 

 

 

 

 

t

t

3t

x

y

0

0.00

0.00

1.00

0.00

pi/4

0.79

2.36

-0.71

0.71

pi/3

1.05

3.14

-1.00

0.87

pi/2

1.57

4.71

0.00

1.00

3pi/4

2.36

7.07

0.71

0.71

5pi/6

2.62

7.85

0.00

0.50

pi

3.14

9.42

-1.00

0.00

7pi/6

3.66

10.99

-0.01

-0.50

5pi/4

3.93

11.78

0.70

-0.71

3pi/2

4.71

14.13

0.01

-1.00

11pi/6

5.76

17.27

-0.01

-0.50

2pi

6.28

18.84

1.00

0.00

 

Remember that the period of cos(3t) is now 2pi/3.  Thus, the x values  complete three cycles, while the y values are still on a single cycle.  Recall that in a single cycle, the cosine function has two common values since cos(x)=cos(-x).  The exceptions are when x=pi and 2pi.  We would then expect that each cos(3t) value will have up to six different t values. 

 

Think about the intercepts.  In Case 1, we have two x-intercepts and two y-intercepts.  Lets do some computation to see if we get six y-intercepts.

 

We know that cos(t)=0 when t is pi/2 or 3pi/2. 

 

Thus cos(3t)=0 when 3t=pi/3 or 3t=3pi/2. 

So t=pi/2 and pi/6.  Remember that cos(x)=-cos(x).

Thus, we have t=pi/2, -pi/2, pi/6, and -pi/6.  Since (-pi/2)=3pi/2,

and (-pi/6)=11pi/6.

 

We also need to consider the additional period of 2pi/3 in cos(3t).  Since cos(x)=cos(x+2pi), then cos(t)=cos(t+2pi/3).

So use t=pi/2 since the sum of pi/2 and 2pi/3 is 7pi/6 which is still within the domain of t.  Similarly use t=pi/6 where the sum of pi/6 and 2pi/3 is 5pi/6. 

 

Hence cos(3t)=0 for t=pi/2, pi/6, 7pi/6, 11pi/6, 5pi/6, and 3pi/2. 

 

 

So we know the graph in Case 2 has six y-intercepts and two x-intercepts. 

 

A table of values for the y-intercepts shows:

 

 

CASE 2:

 

cos(3t)

sin(t)

t

t

3t

x

y

pi/6

0.52

1.57

0.00

0.50

pi/2

1.57

4.71

0.00

1.00

5pi/6

2.62

7.85

0.00

0.50

7pi/6

3.66

10.99

-0.01

-0.50

3pi/2

4.71

14.13

0.01

-1.00

11pi/6

5.76

17.27

-0.01

-0.50

 

Although there are six different t values for the y-intercept, there are only four unique y-intercepts since sin(7pi/6)=sin(11pi/6) and sin(pi/6)=sin(5pi/6).

 

 

Lets investigate other values such as when x=.5. 

 

         We know that cos(t)=.5 when t is pi/3 or 5pi/3.

 

                  Thus cos(3t)=.5 when 3t=pi/3 or 3t=5pi/3. 

                  So t=pi/9 and 5pi/9.  Again cos(x)=- cos (x).

                  Thus, we have t=pi/9, (-pi/9), 5pi/9, (-5pi/9) or

                  t= pi/9, 17pi/9, 5pi/9, and 13pi/9

 

We also need to consider the additional period of 2pi/3 in cos(3t).  Since cos(x)=cos(x+2pi), then cos(t)=cos(t+2pi/3).

So use t=pi/9 since the sum of pi/9 and 2pi/3 is 7pi/9 which is still within the domain of t.  Similarly use t=5pi/9 where the sum of 5pi/9 and 2pi/3 is 11pi/3. 

 

Hence cos(3t)=.5 for t=pi/9, 5pi/9, 7pi/9, 11pi/9, 13pi/9, and 17pi/9. 

 

Just as we expected the calculation above shows six t values for each cos(3t) value. 

 

The table of values for cos(3t)=.5 shows six distinct points. 

 

 

CASE 2

 

cos(3t)

sin(t)

t

t

3t

x

y

pi/9

0.35

1.05

0.50

0.34

5pi/9

1.74

5.23

0.50

0.98

7pi/9

2.44

7.33

0.50

0.64

11pi/9

3.84

11.51

0.49

-0.64

13pi/9

4.54

13.61

0.51

-0.98

17pi/9

5.93

17.79

0.49

-0.34

 

 

Lets investigate the value of x=1. 

 

         We know that cos(t)=1 when t is 2pi.

 

                  Thus cos(3t)=1 when 3t=2pi. 

                  So t=2pi/3.  Again cos(x)= -cos (x).

                  Thus, we have t=2pi/3 and (-2pi/3)

                  t= 2pi/3 and 4pi/3

 

We also need to consider the additional period of 2pi/3 in cos(3t).  Since cos(x)=cos(x+2pi), then cos(t)=cos(t+2pi/3).

So use t=2pi/3 since the sum of 2pi/3 and 2pi/3 is 4pi/3 which is still within the domain of t.  Similarly use t=4pi/3 where the sum of 4pi/3 and 2pi/3 is 2pi.

 

Hence cos(3t)=1 for t=2pi/3, 4pi/3, and 2pi.  We knew that there is only one solution for t for cos(t)=1.  So for cos(3t) should result in three solutions. 

 

 

CASE 2

 

cos(3t)

sin(t)

t

t

3t

x

y

2pi/3

2.09

6.27

1.00

0.87

4pi/3

4.19

12.56

1.00

-0.86

2pi

6.28

18.84

1.00

0.00

 

 

Here is a summary of our findings so far for the parametric equation in Case 2:

 

a)    cos(3t) has a period of 2pi/3, thus we should have at most six values of t for each cos(3t) value.

 

b)   there are six y-intercept values, however, only four of those are distinct.

 

c)    there is only one solution for cos (t)=1 or cos(t)=-1, thus for cos(3t)=1 or cos(3t)=-1 we should expect only three solution for each of those cases. 

 

This gives us a good starting point before plotting the graph of x=cos(3t) and y=sin(t). We could also continue calculating table of values for different cos(3t) values if needed.  Lets now look at the graph of the equation for Case 2 (fig2).

 

Fig2.

 

What would you expect for a parametric equation: x=cos(5t) and y=sin(t) for t between 0 and 2pi?  At most 10 different solutions for each cos(5t)? 

Lets graph and see (fig3):

 

Case 3: x=cos(5t) and y=sin(t)

 

Fig3.

 

Thus, we could expect similar results for x=cos(7t) and y=sin(t) (fig4).

 

Case 4: x=cos(7t) and y=sin(t)

 

Fig4.

 

What about when the roles are switched?  Take for instance x=cos(t) and y=sin(3t) for t values from 0 to 2pi (fig5).

 

Case5: x=cos(t) and y=sin(3t)

 

Fig5.

 

We could see that Case 5 is very similar to Case 3.  Remember that sin(x) is an even function, thus symmetric about the y-axis.  When the odd multiplier is on the sine function (i.e. sin(nt)) we should expect n number of y values for each t value.  A good way to think about this is to use the horizontal line test to see how many times the graph intersects the horizontal line.  Where as when the odd multiplier is on the cosine function (i.e. cos(nt)) we should expect n number of x values for each t value.  Thus the intersection of a vertical line and the graph provides a good test for this case. 

 

Notice that in Case5, the graph shows only three y values at -1 and 1.  A similar computation to the cosine computation in the previous cases could be done to verify why. 

 

We should expect similar results for the equation x=cos(t) and y=sin(5t) (fig6).

 

Case 6: x=cos(t) and y=sin(5t)

 

Fig6.

 

 

What about the parametric equation x=cos(2t) and y = sin(t) for t values between 0 and 2pi (fig7)?

 

Case 7: x=cos(2t) and y=sin(t)

 

Fig7.

 

The graph of Case 7 looks very different from the previous cases.  It looks like the equation is producing two x values for each t value instead of four as we might expect.  The exception is when x=1. 

 

Lets investigate other values such as when x=.5. 

 

         We know that cos(t)=.5 when t is pi/3 or 5pi/3.

 

                  Thus cos(2t)=.5 when 2t=pi/3 or 2t=5pi/3. 

                  So t=pi/6 and 5pi/6.  Again cos(x)=- cos (x).

                  Thus, we have t=pi/6, (-pi/6), 5pi/6, (-5pi/6) or

                  t= pi/6, 5pi/6, 7pi/6, 11pi/6. 

 

                  The corresponding y-values are: .5, .5. -.5, and -.5.

 

We see that there are four values of x for each t value.  However, only two of the corresponding y-values are distinct.  The graph indicates similar results for other x values. 

 

 

CASE 7

 

cos(2t)

sin(t)

t

t

2t

x

y

0

0.00

0.00

1.00

0.00

pi/4

0.79

1.57

0.00

0.71

pi/3

1.05

2.09

-0.50

0.87

pi/2

1.57

3.14

-1.00

1.00

3pi/4

2.36

4.71

0.00

0.71

5pi/6

2.62

5.23

0.50

0.50

pi

3.14

6.28

1.00

0.00

7pi/6

3.66

7.33

0.50

-0.50

5pi/4

3.93

7.85

0.00

-0.71

3pi/2

4.71

9.42

-1.00

-1.00

11pi/6

5.76

11.51

0.49

-0.50

2pi

6.28

12.56

1.00

0.00

 

 

For cos(2t)=1 when t=pi and t=2pi.  The corresponding y-values would be sin(pi)=0=sin(2pi). 

 

On the other hand cos(2t)=-1 when t=pi/2 and 3pi/2.  The corresponding y-values are sin(pi/2)=1 and sin(3pi/2)=-1.  Thus, the graph of Case 7 does not ÒcloseÓ like in the previous cases. 

 

Lets look at the next even number (fig8):

 

Case 8: x=cos(4t) and y=sin(t) 

 

Fig8.

 

We notice a similar graph to Case 7, but it ÒopensÓ to the opposite side.  As expected, most of the graph shows four x values for each t value.  The exception is once again at x=1 and -1. 

 

Some computation should clarify these exceptions:

 

cos(4t)=1 when t=pi/2, pi, 3p/2, and 2pi.  The corresponding y values are 1, 0, -1, 0. 

 

cos(4t)=-1 when t=pi/4, 3pi/4, 5pi/5, and 7pi/4.  The corresponding y values: root 2/2 and –root2/2.

 

Thus, the computation shows that Case 8 has three distinct y values at x=1 and two distinct y values at x=-1. 

 

We begin to see a pattern as the even number increases (fig9 and fig10).

 

Case9: x=cos(6t) and y=sin(t)

 

Fig9.

 

Case10: x=cos(8t) and y=sin(t)

 

Fig10.

 

Now lets look when the even numbers multiplies t in the sine function (fig11).

 

Case 11: x=cos(t) and y=sin(2t)

 

Fig11.

 

The even multiplier on the sine function does not seem to have the same outcome as the even multiplier on the cosine function.  We do see up to four distinct solutions for each t value.  The exceptions are once again at y values -1, 1, and 0. 

 

sin(t)=1 when t=pi/2.  So sin(2t)=1 for only two t values, t=pi/4, 5pi/4.  The corresponding x values are root 2/2 and –root2/2.

 

sin(t)=0 when t=pi, and 2pi.  So sin(2t) =0 when t=pi, pi/2, 3pi/2, and 2pi.  The corresponding x values are -1, 0, 0, and 1.  So there are three distinct points when sin(2t)=0.

 

We see similar results for higher even multipliers in the sine function (fig12 and fig 13).

 

Case 12: x= cos(t) and y=sin(4t)

 

Fig12.

 

Case 13: x=cos(t) and y = sin(6t)

 

Fig13.

 

In this presentation we discussed the parametric curves x=cos (nt) and

y= sin (mt)  where m and n are positive integers; and in each of the thirteen cases discussed, either m or n was 1. 

 

For homework:

 

a)    Investigate the cases where m and n are negative integers.  Explain why the graph behaves a certain way. 

 

b)   Investigate the cases where m and n are positive integers but neither m nor n is 1. 

 

c)    Investigate the cases where m and n are the same.  Explain why the graph is the same.

 

 

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