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Parametric

Department of MathematicsUniversity of Leicester

What is it?

• A parametric equation is a method of defining a relation using parameters.

• For example, using the equation:

• We can use a free parameter, t, setting:

and

3xy

tx 3ty

What is it?

• We can see that this still satisfies the equation, while defining a relationship between x and y using the free parameter, t.

Why do we use parametric equations• Parameterisations can be used to integrate

and differentiate equations term wise.

• You can describe the motion of a particle using a parameterisation:

• r being placement.

))(),(),(()( tztytxtr

Why do we use parametric equations• Now we can use this to differentiate each

term to find v, the velocity:

))('),('),('()(')( tztytxtrtv

Why do we use parametric equations• Parameters can also be used to make

differential equations simpler to differentiate.

• In the case of implicit differentials, we can change a function of x and y into an equation of just t.

Why do we use parametric equations• Some equations are far easier to describe

in parametric form.

• Example: a circle around the origin

Cartesian form:

Parametric form:

222 ryx

))sin(),cos(()( tatatr

How to get Cartesian from parametric• Getting the Cartesian equation of a

parametric equation is done more by inspection that by a formula.

• There are a few useful methods that can be used, which are explored in the examples.

How to get Cartesian from parametric• Example 1:

• Let:

• So that:

and

)3,()( 2 tttr

2tx ty 3

How to get Cartesian from parametric• Next set t in terms of y:

• Now we can substitute t in to the equation of x to eliminate t.

3

yt

How to get Cartesian from parametric• Substituting in t:

• Which expands to:

2

3

y

x

29 yx

How to get Cartesian from parametric• Example 2:

• Let:

• So that:

and

))sin(3),cos(2()( tttr

)cos(2 tx )sin(3 ty

How to get Cartesian from parametric• To change this we can see that:

• And

)(cos2

22

tx

)(sin3

22

ty

How to get Cartesian from parametric• And as we know that

• We can see that:

1)(cos)(sin 22 tt

123

22

xy

How to get Cartesian from parametric• Which equals:

• This is the Cartesian equation for an ellipse.

149

22

xy

Example

• Example 3: let:

• Be the Cartesian equation of a circle at the point (a,b).

• Change this into parametric form.

222 )()( rbyax

Example

• If we set:

• And:

• Then we can solve this using the fact that:

1cossin 22 tt

trax 222 cos)(

trby 222 sin)(

Example

• From this we can see that:

• So:

• Therefore:

trax 222 cos)(

trax cos

atrx cos

Example

• Similarly:

• So:

• Therefore:

trby 222 sin)(

trby sin

btry sin

Example

• Compiling this, we can see that:

• Which is the parametric equation for a circle at the point (a,b).

btratrtytx sin,cos)(),(

Polar co-ordinates

• Parametric equations can be used to describe curves in polar co-ordinate form:

• For example:

t x

y

r

Polar co-ordinates

• Here we can see, that if we set t as the angle, then we can describe x and y in terms of t:

• Using trigonometry:

• and

trx cos

try sin

Polar co-ordinates

• These can be used to change Cartesian equations to parametric equations:

Polar co-ordinates: example

• Let:

• Be the equation for a circle.

• If we set:

222 ayx

try sintrx cos

Polar co-ordinates: example

• We can see that if we substitute these in, then the equation still holds:

• Therefore we can use:

• As a parameterisation for a circle.

)sin,cos())(),(( trtrtytx

Finding the gradient of a parametric curve

• To find dy/dx we need to use the chain rule:

dx

dt

dt

dy

dx

dy

How to get Cartesian from parametric: example

• Example:

• Let:

and

Then:

and

tx 2 2ty

2dt

dxt

dt

dy2

How to get Cartesian from parametric: example

• Then, using the chain rule:

ttdx

dt

dt

dy

dx

dy

2

12

Extended parametric example

• Let:

• Be the Cartesian equation.

23

2

3

2

ayx

Extended parametric example

• Then to change this into parametric form, we need to find values of x and y that satisfy the equation.

• If we set:

• And:

tax 33 cos

tay 33 sin

Extended parametric example

• Then we have:

• Which expands to:

23

2333

233 )sin()cos( atata

22222 sincos atata

Extended parametric example

• We know that:

• Therefore we can see that our values of x and y satisfy the equation. Therefore:

1sincos 22 tt

)sin,cos())(),(()( 3333 tatatytxtr

Extended parametric example

• Now, as this is the placement of the particle, we can find the velocity of the particle by differentiating each term:

)(sin)cos(3),(cos)sin(3 2323 ttatta

))('),('()(')( tytxtrtv

Extended parametric example

• Next, we can find the gradient of the curve.

• Using the formula:

dx

dt

dt

dy

dx

dy

Extended parametric example

• Using this:

• And:

)(cos)sin(3 23 ttadt

dx

)(sin)cos(3 23 ttadt

dy

Extended parametric example

• Therefore the gradient is:

)sin()(cos3

)(sin)cos(323

23

tta

tta

dx

dy

tt

ttan

)cos(

)sin(

Conclusion

• Parametric equations are about changing equations to just 1 parameter, t.

• Parametric is used to define equations term wise.

• We can use the chain rule to find the gradient of a parametric equation.

Conclusion

• Standard parametric manipulation of polar co-ordinates is:

• x=rcos(t)

• Y=rsin(t)