Parametric
Department of Mathematics
University of Leicester
www.le.ac.uk
What is it?
• A parametric equation is a method of defining a
relation using parameters.
• For example, using the equation:
y  x
• We can use a free parameter, t, setting:
xt
and
y t
3
3
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 ( t )  ( x ( t ), y ( t ), z ( t ))
• r being placement.
Why do we use parametric equations
• Now we can use this to differentiate each term
to find v, the velocity:
v ( t )  r ' ( t )  ( x ' ( t ), y ' ( t ), z ' ( t ))
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:
x  y
2
2
r
2
Parametric form:
r ( t )  ( a cos( t ), a sin( t ))
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:
r ( t )  ( t ,3t )
2
• So that:
x t
2
and
y  3t
How to get Cartesian from parametric
• Next set t in terms of y:
t 
y
3
• Now we can substitute t in to the equation of x
to eliminate t.
How to get Cartesian from parametric
• Substituting in t:
 y
x  
3
• Which expands to:
9x  y
2
2
How to get Cartesian from parametric
• Example 2:
• Let:
r ( t )  ( 2 cos( t ), 3 sin( t ))
• So that:
x  2 cos( t )
and
y  3 sin( t )
How to get Cartesian from parametric
• To change this we can see that:
2
• And
x
2
   cos ( t )
2
2
 y
2
   sin ( t )
3
How to get Cartesian from parametric
• And as we know that
sin ( t )  cos ( t )  1
2
2
• We can see that:
2
2
 y
x
    1
3
2
How to get Cartesian from parametric
• Which equals:
y
2
9

x
2
1
4
• This is the Cartesian equation for an ellipse.
Example
• Example 3: let:
( x  a)  ( y  b)  r
2
2
2
• Be the Cartesian equation of a circle at the
point (a,b).
• Change this into parametric form.
Example
• If we set:
( x  a )  r cos t
2
2
2
• And:
( y  b )  r sin
2
2
2
t
• Then we can solve this using the fact that:
sin
2
t  cos t  1
2
Example
• From this we can see that:
( x  a )  r cos t
2
• So:
2
x  a  r cos t
• Therefore:
x  r cos t  a
2
Example
• Similarly:
( y  b )  r sin
2
• So:
2
2
y  b  r sin t
• Therefore:
y  r sin t  b
t
Example
• Compiling this, we can see that:
 x ( t ), y ( t )    r cos t  a , r sin
t  b
• Which is the parametric equation for a circle at
the point (a,b).
Polar co-ordinates
• Parametric equations can be used to describe
curves in polar co-ordinate form:
• For example:
y
r
t
x
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
x  r cos t
y  r sin t
Polar co-ordinates
• These can be used to change Cartesian
equations to parametric equations:
Polar co-ordinates: example
• Let:
x  y
2
2
 a
• Be the equation for a circle.
• If we set:
y  r sin t
x  r cos t
2
Polar co-ordinates: example
• We can see that if we substitute these in, then
the equation still holds:
• Therefore we can use:
( x ( t ), y ( t ))  ( r cos t , r sin t )
• As a parameterisation for a circle.
Finding the gradient of a parametric
curve
• To find dy/dx we need to use the chain rule:
dy
dx

dy
dt

dt
dx
How to get Cartesian from parametric:
example
• Example:
• Let:
x  2t
Then:
dx
dt
 2
y t
and
and
dy
dt
 2t
2
How to get Cartesian from parametric:
example
• Then, using the chain rule:
dy
dx

dy
dt

dt
dx
 2t 
1
2
t
Extended parametric example
• Let:
2
x
3
2
 y
3
 a
• Be the Cartesian equation.
2
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:
x  a cos t
• And:
y  a sin
3
3
3
3
t
Extended parametric example
• Then we have:
2
3
3
( a cos t )
3
2
 ( a sin
3
 a
3
t)
2
t  a
3
• Which expands to:
a cos t  a sin
2
2
2
2
2
Extended parametric example
• We know that:
cos t  sin
2
2
t 1
• Therefore we can see that our values of x and y
satisfy the equation. Therefore:
r ( t )  ( x ( t ), y ( t ))  ( a cos t , a sin t )
3
3
3
3
Extended parametric example
• Now, as this is the placement of the particle,
we can find the velocity of the particle by
differentiating each term:
v ( t )  r ' ( t )  ( x ' ( t ), y ' ( t ))
  3 a sin( t ) cos ( t ), 3 a cos( t ) sin ( t ) 
3
2
3
2
Extended parametric example
• Next, we can find the gradient of the curve.
• Using the formula:
dy
dx

dy
dt

dt
dx
Extended parametric example
• Using this:
dx
  3 a sin( t ) cos ( t )
3
2
dt
• And:
dy
dt
 3 a cos( t ) sin ( t )
3
2
Extended parametric example
• Therefore the gradient is:
dy
3

dx
2
3 a cos( t ) sin ( t )
3
2
3 a cos ( t ) sin( t )

sin( t )
cos( t )
  tan t
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 coordinates is:
• x=rcos(t)
• Y=rsin(t)