41: Parametric Equations “Teach A Level Maths” Vol. 2: A2 Core Modules.

41: Parametric 41: Parametric Equations Equations

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

Parametric Equations

sin3,cos3 yx

The Cartesian equation of a curve in a plane is an equation linking x and y.

Some of these equations can be written in a way that is easier to differentiate by using 2 equations, one giving x and one giving y, both in terms of a 3rd variable, the parameter.

Letters commonly used for parameters are s, t and . ( is often used if the parameter is an angle. )

e.gs. tytx 4,2 2


Converting between Cartesian and Parametric formsWe use parametric equations because they are simpler, so we only convert to Cartesian if asked to do so !e.g. 1 Change the following to a Cartesian

equation and sketch its graph:tytx 4,2 2

Solution: We need to eliminate the parameter t.

We substitute for t from the easier equation:

ty 44

yt

Subst. in :2 2tx

2

42

y

x8

2yx


The Cartesian equation is 8

2yx

We usually write this as xy 82

Either, we can sketch using a graphical calculator with

xy 8and entering the graph in 2 parts.Or, we can notice that the equation is quadratic with x and y swapped over from the more usual form.


The sketch is

The curve is called a parabola.

xy 82

tytx 4,2 2 Also, the parametric equationsshow that as t increases, x increases faster than y.


e.g. 2 Change the following to a Cartesian equation: sin3,cos3 yx

Solution: We need to eliminate the parameter .

BUT appears in 2 forms: as andso, we need a link between these 2 forms.

cos sin

Which trig identity links and ?

cos sin

ANS: 1sincos 22 To eliminate we substitute into this

expression.


cos3

x

9sin

22 y

9

cos2

2 x

sin3

y

1sincos 22 199

22

yx

922 yxMultiply by 9:

becomes

sin3,cos3 yxSo,

N.B. = not

We have a circle, centre (0, 0), radius 3.


Since we recognise the circle in Cartesian form, it’s easy to sketch.However, if we couldn’t eliminate the parameter or didn’t recognise the curve having done it, we can sketch from the parametric form.


Solution:Let’s see how to do it without eliminating the parameter.We can easily spot the min and max values of x and y:

22 x and 33 y

( It doesn’t matter that we don’t know which angle is measuring. )For both and the min is 1 and the max is 1, so

cos sin

sin3,cos2 yx

e.g. Sketch the curve with equations

It’s also easy to get the other coordinate at each of these 4 key values e.g. 002 yx


sin3,cos2 yx 22 x and

33 y

We could draw up a table of values finding x and y for values of but this is usually very inefficient. Try to just pick out significant features.

x

x

x90

x

0



x

33 y

This tells us what happens to x and y.

90Think what happens to and as increases from 0 to .

cos sin

We could draw up a table of values finding x and y for values of but this is usually very inefficient. Try to just pick out significant features.

x

x

x90

x

0



x

Symmetry now completes the diagram.

33 y

This tells us what happens to x and y.

90Think what happens to and as increases from 0 to .

cos sin

x

x

x90

x

0



33 y

Symmetry now completes the diagram.

x

x

x90

x

0


sin3,cos2 yx

So, we have the parametric equations of an ellipse ( which we met in Cartesian form in Transformations ).

The origin is at the centre of the ellipse.

x

x

x

xO

x


You can use a graphical calculator to sketch curves given in parametric form. However, you will have to use the setup menu before you enter the equations.

You will also have to be careful about the range of values of the parameter and of x and y. If you don’t get the right scales you may not see the whole graph or the graph can be distorted and, for example, a circle can look like an ellipse.By the time you’ve fiddled around it may have been better to sketch without the calculator!


The following equations give curves you need to recognise:

sin,cos ryrx

atyatx 2,2

)(sin,cos babyax

a circle, radius r, centre the origin.

a parabola, passing through the origin, with the x-axis as an axis of symmetry.

an ellipse with centre at the origin, passing through the points (a, 0), (a, 0), (0, b), (0, b).


To write the ellipse in Cartesian form we use the same trig identity as we used for the circle.

)(sin,cos babyax So, for

use 1sincos 22

12

2

2

2

b

y

a

x

122

b

y

a

x

The equation is usually left in this form.

Parametric EquationsThere are other parametric equations you might be asked to convert to Cartesian equations. For example, those like the ones in the following exercise.Exercise

tan2,sec4 yx

tytx

3,3

( Use a trig identity )

1.

2.

Sketch both curves using either parametric or Cartesian equations. ( Use a graphical calculator if you like ).


Solution:

tan2,sec4 yx1.

Use 22 sectan1 22

421

xy

1641

22 xy

We usually write this in a form similar to the ellipse:

1416

22

yx

Notice the minus sign. The curve is a hyperbola.


tan2,sec4 yxSketch:

1416

22

yx

or

A hyperbola

Asymptotes


tytx

3,3

( Eliminate t by substitution. )

2.

Solution: 3

3x

ttx

t

y3

Subs. in

xy

9

9 xy

3

3x

y

The curve is a rectangular hyperbola.

xx 33 33


tytx

3,3 9xySketch

:or

A rectangular hyperbola.

Asymptotes


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


sin3,cos3 yx

The Cartesian equation of a curve in a plane is an equation linking x and y.

Some of these equations can be written in a way that is easier to differentiate by using 2 equations, one giving x and one giving y, both in terms of a 3rd variable, the parameter.

Letters commonly used for parameters are s, t and . ( is often used if the parameter is an angle. )

e.gs. tytx 4,2 2


Converting between Cartesian and Parametric formsWe use parametric equations because they are simpler, so we only convert to Cartesian if asked to do so !e.g. 1 Change the following to a Cartesian

equation and sketch its graph:tytx 4,2 2

Solution: We need to eliminate the parameter t.Substitution is the easiest way.

ty 44

yt

Subst. in :2 2tx

2

42

y

x8

2yx


The Cartesian equation is 8

2yx

We usually write this as xy 82

Either, we can sketch using a graphical calculator with

xy 8and entering the graph in 2 parts.Or, we can notice that the equation is quadratic with x and y swapped over from the more usual form.


e.g. 2 Change the following to a Cartesian equation: sin3,cos3 yx

Solution: We need to eliminate the parameter .

BUT appears in 2 forms: as andso, we need a link between these 2 forms.

cos sin

To eliminate we substitute into the expression.

1sincos 22


cos3

x

9sin

22 y

9

cos2

2 x

sin3

y

1sincos 22 199

22

yx

922 yxMultiply by 9:

becomes

sin3,cos3 yxSo,

N.B. = not

We have a circle, centre (0, 0), radius 3.


The following equations give curves you need to recognise:

sin,cos ryrx

atyatx 2,2

)(sin,cos babyax

a circle, radius r, centre the origin.

a parabola, passing through the origin, with the x-axis an axis of symmetry.

an ellipse with centre at the origin, passing through the points (a, 0), (a, 0), (0, b), (0, b).

41: Parametric Equations “Teach A Level Maths” Vol. 2: A2 Core Modules.

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Transcript of 41: Parametric Equations “Teach A Level Maths” Vol. 2: A2 Core Modules.