1 Parametric curves

Here we explore the use of a parameter t in the description of curves. We shall see that it has some advantages over the more usual Cartesian description. We start with a simple example.

Example 6

Plot the curve x = 2 cos t y = 3 sin t ⏟ / 0 ≤ t ≤ π 2 ⏟ \

parametric equations of the curve  parameter range

Solution

The approach to sketching the curve is straightforward. We simply give the parameter t various values as it ranges through 0 → π 2 and, for each value of t , calculate corresponding values of ( x , y ) which are then plotted on a Cartesian x y plane. The value of t and the corresponding values of x , y are recorded in the following table:

t 0 π 20 2 π 20 3 π 20 4 π 20 5 π 20 6 π 20 7 π 20 8 π 20 9 π 20 10 π 20
x 2 1.98 1.90 1.78 1.62 1.41 1.18 0.91 0.62 0.31 0
y 0 0.47 0.93 1.36 1.76 2.12 2.43 2.67 2.85 2.96 3
Plotting the ( x , y ) coordinates gives the curve in Figure 16.

Figure 16

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The curve in Figure 16 resembles part of an ellipse. This can be verified by eliminating t from the parametric equations to obtain an expression involving x , y only. If we divide the first parametric equation by 2 and the second by 3, square both and add we obtain

x 2 2 + y 3 2 = cos 2 t + sin 2 t ≡ 1 i.e. x 2 4 + y 2 9 = 1

which we easily recognise as an ellipse whose major-axis is the y -axis. Also, as t ranges from 0 → π 2   x = 2 cos t decreases from 2 → 0 , and y = 3 sin t increases from 0 → 3 . We conclude that the parametric equations x = 2 cos t , y = 3 sin t together with the parametric range 0 ≤ t ≤ π 2 describe that part of the ellipse x 2 4 + y 2 9 = 1 in the positive quadrant. On the curve in Figure 16 we have used an arrow to indicate the direction that we move along the curve as t increases from its initial value 0.

Task!

Plot the curve x = t + 1 y = 2 t 2 − 3 0 ≤ t ≤ 1

Do you recognise this curve as a conic section?

First construct a table of ( x , y ) values as t ranges from 0 → 1 :

t 0 0.25 0.5 0.75 1
x 1 1.25 1.5 1.75 2
y − 3 − 2.88 − 2.5 − 1.88 − 1

Now plot the points on a Cartesian plane:

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Now eliminate the t -variable from x = t + 1 , y = 2 t 2 − 3 to obtain the x y form of the curve:

y = 2 x 2 − 4 x − 1 which is the equation of a parabola.

Example 7

Sketch the curve x = t 2 + 1 y = 2 t 4 − 3 0 ≤ t ≤ 1

Solution

This is very similar to the previous Task (except for t 4 replacing t 2 in the expression for y and t 2 replacing t in the expression for x ). The corresponding table of values is

t 0 0.25 0.5 0.75 1
x 1 1.06 1.25 1.56 2
y − 3 − 2.99 − 2.88 − 2.37 − 1

Figure 17

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We see that this is identical to the curve drawn previously. This is confirmed by eliminating the t -parameter from the expressions defining x , y . Here t 2 = x − 1 so y = 2 ( x − 1 ) 2 − 3 which is the same as obtained in the last Task. The main difference is that particular values of t locate (in general) different ( x , y ) points on the curve for the two parametric representations.

We conclude that a given curve in the x y plane can have many (in fact infinitely many) parametric descriptions.

Task!

Show that the two parametric representations below describe the same curve.

  1. x = cos t y = sin t 0 ≤ t ≤ π 2
  2. x = t y = 1 − t 2 0 ≤ t ≤ 1

Eliminate t from the parametric equations in (1):

x 2 + y 2 = cos 2 t + sin 2 t = 1 Eliminate t from the parametric equations in (2):

y = 1 − x 2 ∴ y 2 = 1 − x 2 or x 2 + y 2 = 1 What do you conclude?

Both parametric descriptions represent (part of) a circle centred at the origin of radius 1.