Parametric Differentiation
If  x, y  is a point and this point depends ontime t ,
then both, x and y are a function of t.
let x  t 2 and y  2t , at time t  3
x   3  9
2
y  2  3  6
 the coordinates of the point at t  3 are  9, 6 
An equation for x and y in terms of t is known as
a parametric equation, with t being the parameter
xt
dx
x' 
dt
and
y t
and
dy
y' 
dt
dy
dt  dy  dt  dy
dx dt dx dx
dt
x  t2
y  2t
dx
dy
 2t
2
dt
dt
dy
dy dt
2 1

 
dx dx 2t t
dt
A curve is defined by the parametric equations
1
1
2
2
xt  2 ,
y  t  2 t  0
t
t
dy
Find an expression for
in terms of t , simplify your answer
dx
1
dx
2
2
2
2
3
x  t  2  t t
 2t  2t  2t  3
t
dt
t
1
dy
2
y  t 2  2  t 2  t 2
 2t  2t 3  2t  3
t
dt
t
dy
2
2t  3 t 4  1
dy dt
t 


4
dx
2
t
1
dx
2t  3
dt
t
A curve is defined by the parametric equations
x    sin  ,
y  1  cos 
 0    2 
dy
Find an expression for
in terms of 
dx
x    sin 
y  1  cos 
dx
 1  cos 
d
dy
 sin 
d
dy
dy d
sin 


dx dx 1  cos 
d
A curve is defined by the parametric equations
t
1 t
x
,
y
t  0
1 t
1 t
dy
Find an expression for
in terms of t , simplify your answer
dx
t
dx 1  t   t 1
1
x


2
2
1 t
dt
1  t 
1  t 
1 t
y
1 t
dy 11  t   1  t  1 1  t  1  t
2



2
2
2
dt
1  t 
1  t 
1  t 
2
dy
2
2
2
2
1

t
1

t
1

t

 
 
dy dt 
2
 1 t 




2
 2

2
2
dx
1
dx
1
1

t


1 t 
1 t 


dt 1  t 2
d 2 y d  dy 
  
2
dx
dx  dx 
note: that we can only differentiate
with respect to t
d  dy  d  dy  dt
    
dx  dx  dt  dx  dx
note:
dt
1

dx dx
dt
d y d  dy  dt 

  
2
dx
dt  dx  dx 
2
x  t2
y  2t
dx
dy
 2t
2
dt
dt
dy
dy dt
2 1

 
dx dx 2t t
dt
d y d  1  dt
  
2
dx
dt  t  dx
2
d2y
1 1
1
 2  3
2
dx
t 2t
2t
d 1
1
2
t   t   2

dt
t
dt
1
1


dx dx 2t
dt
Equation of tangents
A curve is defined by the parametric equations
x  t2 1
y  t  t 2  1
Find the equation of the tangent to the curve
at the point with parameter t  2
x  t2 1
y  t  t 2  1  t 3  t
dx
 2t
dt
dy
 3t 2  1
dt
x '  2t
y '  3t 2  1
dy y ' 3t 2  1
 
dx x '
2t
when t  2 x   2   1  5
2
when t  2 y   2    2   10
3
 tangent at  5,10 
at t  2,
dy 3  2   1 13


dx
2  2
4
2
13
mtangent 
4
y  b  m x  a
13
y  10   x  5 
4
4 y  40  13 x  65
13x  4 y  25  0
Stationary
Points
A curve is defined by the parametric equations
xt
y  t 3  3t
Find the coordinate and nature of the stationary points on the curve.
xt
y  t 3  3t
dx
dy
1
 3t 2  3
dt
dt
dy y ' 3t 2  3
 
 3t 2  3
dx x '
1
dy y ' 3t 2  3
 
 3t 2  3
dx x '
1
dy
 0  3t 2  3  0
dx
 t 2 1  0
 t  1
t 1
t  1
x  1  1
x   1  1
y  1  3 1  2
y   1  3  1  2
SP at 1, 2 
SP at  1, 2 
3
3
dy
 3t 2  3
dx
d2y d
dt
2
  3t  3   6t 1  6t
2
dx
dt
dx
d2y
at t  1 6 1  0  1, 2  min
2
dx
d2y
at t  1 6  -1  0   1, 2  max
2
dx