Introduction
• This chapter focuses on Parametric equations
• Parametric equations split a ‘Cartesian’ equation
into an x and y ‘component’
• They are used to model projectiles in Physics
• This in turn allows video game programmers to
create realistic physics engines in their games…
Coordinate Geometry in the (x,y)
plane
You can define x and y coordinates
on a curve using separate functions,
x = f(t) and y = g(t)
t
-3
-2
-1
0
1
2
3
x = 2t
-6
-4
-2
0
2
4
6
y = t2
9
4
1
0
1
4
9
The letter t is used as the primary
application of this is in Physics, with t
representing time.
10
Draw the curve given by the
Parametric Equations:
𝑥 = 2𝑡
𝑦 = 𝑡2
for
-10
10
−3 ≤ 𝑡 2 ≤ 3
Work out the x and y co-ordinates
to plot by substituting the t
values
-10
2A
Coordinate Geometry in the (x,y)
plane
You can define x and y coordinates
on a curve using separate functions,
x = f(t) and y = g(t)
A curve has parametric equations:
𝑥 = 2𝑡 𝑦 = 𝑡 2
Find the Cartesian equation of the
curve
Rewrite t in terms of x, and then
substitute it into the y equation…
𝑥 = 2𝑡
Divide by 2
𝑥
=𝑡
2
𝑦 = 𝑡2
𝑥
𝑦=
2
2
𝑦=
𝑥
4
2
Replace t
with x/2
Simplify
2A
Coordinate Geometry in the (x,y)
plane
You can define x and y coordinates
on a curve using separate functions,
x = f(t) and y = g(t)
A curve has parametric equations:
𝑥=
1
𝑡+1
𝑦 = 𝑡2
Find the Cartesian equation of the
curve
Rewrite t in terms of x, and then
substitute it into the y equation…
𝑥=
1
𝑡+1
Multiply by (t + 1)
𝑥(𝑡 + 1) = 1
Divide by x
1
𝑡+1=
𝑥
Subtract 1
1
𝑡 = −1
𝑥
𝑦 = 𝑡2
1
𝑦=
−1
𝑥
2
Replace t
1 𝑥
𝑦=
−
𝑥 𝑥
2
1 = x/x
1−𝑥
𝑥
2
Put the fractions
together
𝑦=
(1 − 𝑥)2
𝑦=
𝑥2
Simplify
2A
Coordinate Geometry in the (x,y)
plane
You need to be able to use Parametric
equations to solve problems in
coordinate geometry
The diagram shows a sketch of the curve
with Parametric equations:
𝑥 =𝑡−1
A
𝑦 = 4 − 𝑡2
The curve meets the x-axis at the points
A and B. Find their coordinates.
At points A and B, y = 0
 Sub y = 0 in to find t at these points
So t = ±2
 Sub this into the x equation to
find the x-coordinates of A and B
 A and B are (-3,0) and (1,0)
𝑦 = 4 − 𝑡2
0=4−𝑡
2
𝑡2 = 4
𝑡 = ±2
𝑥 =𝑡−1
B
y=0
+ t2
Remember both answers
Sub in t = 2
𝑥 =𝑡−1
𝑥 = (2) − 1
𝑥 = (−2) − 1
𝑥=1
𝑥 = −3
Sub in t = -2
2B
Coordinate Geometry in the (x,y)
plane
You need to be able to use Parametric
equations to solve problems in
coordinate geometry
A curve has Parametric equations:
𝑥 = 𝑎𝑡
3
𝑦 = 𝑎(𝑡 + 8)
where a is a constant. Given that the
curve passes through (2,0), find the value
of a.
We know there is a coordinate
where x = 2 and y = 0, Sub y = 0
into its equation
Since we know that at (2,0), x = 2
and t = -2, we can put these into the
x equation to find a
𝑦 = 𝑎(𝑡 3 + 8)
0 = 𝑎(𝑡 3 + 8)
𝑡3 + 8 = 0
𝑡 3 = −8
𝑡 = −2
y=0
The ‘a’ part cannot be 0 or there would not be
any equations!
Subtract 8
Cube root
So the t value at the coordinate (2,0) is -2
𝑥 = 𝑎𝑡
(2) = 𝑎(−2)
−1 = 𝑎
Sub in x and t values
Divide by -2
𝑥 = 𝑎𝑡
𝑦 = 𝑎(𝑡 3 + 8)
𝑥 = −𝑡
𝑦 = −(𝑡 3 + 8)
The actual equations
with a = -1
2B
Coordinate Geometry in the (x,y)
plane
You need to be able to use Parametric
equations to solve problems in
coordinate geometry
A curve is given Parametrically by:
𝑥 = 𝑡2
𝑦 = 4𝑡
The line x + y + 4 = 0 meets the curve at
point A. Find the coordinates of A.
𝑥 = 𝑡2
𝑥+𝑦+4=0
(𝑡 2 ) + (4𝑡) + 4 = 0
2
𝑡 + 4𝑡 + 4 = 0
(𝑡 + 2)2 = 0
𝑡 = −2
The first thing we need to do is
to find the value of t at
coordinate A
 Sub x and y equations into
the line equation
Replace x and y with their equations
‘Tidy up’
Factorise
Find t
So t = -2 where the curve and line meet (point A)
𝑥 = 𝑡2
2
𝑥 = (−2)
We know an equation for x and
one for y, and we now have the t
value to put into them…
𝑦 = 4𝑡
𝑥=4
Sub t = -2
𝑦 = 4𝑡
𝑦 = 4(−2)
Sub t = -2
𝑦 = −8
The curve and line meet at (4, -8)
2B
Coordinate Geometry in the (x,y)
plane
You need to be able to convert
Parametric equations into Cartesian
equations
A Cartesian equation is just an
equation of a line where the
variables used are x and y only
𝑥 = 𝑠𝑖𝑛𝑡 + 2
𝑥 − 2 = 𝑠𝑖𝑛𝑡
𝑠𝑖𝑛2 𝑡 + 𝑐𝑜𝑠 2 𝑡 ≡ 1
(𝑥 − 2)
A curve has Parametric equations:
𝑥 = 𝑠𝑖𝑛𝑡 + 2 𝑦 = 𝑐𝑜𝑠𝑡 − 3
a) Find the Cartesian equation of the
curve
We need to eliminate t from the
equations
Isolate
sint
2
+ (𝑦 + 3)
2
= 1
𝑦 = 𝑐𝑜𝑠𝑡 − 3
𝑦 + 3 = 𝑐𝑜𝑠𝑡
Isolate
cost
Replace sint and cost
5
(𝑥 − 2)2 + (𝑦 + 3)2 = 1
Centre = (2, -3)
Radius = 1
b) Sketch the curve
The equation is that of a circle
 Think about where the centre
will be, and its radius
-5
5
-5
2C
Coordinate Geometry in the (x,y)
plane
You need to be able to convert
Parametric equations into Cartesian
equations
A Cartesian equation is just an
equation of a line where the
variables used are x and y only
A curve has Parametric equations:
𝑥 = 𝑠𝑖𝑛𝑡
𝑦 = 𝑠𝑖𝑛2𝑡
a) Find the Cartesian equation of the
curve
We need to eliminate t from the
equations
𝑥 = 𝑠𝑖𝑛𝑡
2
2
𝑥 = 𝑠𝑖𝑛 𝑡
Square
𝑦 = 𝑠𝑖𝑛2𝑡
𝑦 = 2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡
𝑦 = 2𝑥𝑐𝑜𝑠𝑡
𝑠𝑖𝑛2 𝑡 + 𝑐𝑜𝑠 2 𝑡 ≡ 1
2
2
𝑐𝑜𝑠 𝑡 = 1 − 𝑠𝑖𝑛 𝑡
𝑐𝑜𝑠 2 𝑡 = 1 − 𝑥 2
𝑐𝑜𝑠𝑡 =
Use the double angle
formula from C3
Replace sint with x
Rearrange
Replace sin2t with x2
Square root
1 − 𝑥2
𝑦 = 2𝑥𝑐𝑜𝑠𝑡
𝑦 = 2𝑥 1 − 𝑥 2
𝑦 2 = 4𝑥 2 (1 − 𝑥 2 )
We can now replace cos t
Another way of writing this (by
squaring the whole of each side)
2C
Coordinate Geometry in the (x,y)
plane
You need to be able to find the area
under a curve when it is given by
Parametric equations
The Area under a curve is given by:
𝑦 𝑑𝑥
By the Chain rule:
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Integrate the equation
with respect to x
Integrate the whole expression with
respect to t
(Remember you don’t have to do
anything with the dt at the end!)
y multiplied by dx/dt (x
differentiated with
respect to t)
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
𝑥 = 5𝑡 2
You need to be able to find the area
under a curve when it is given by
Parametric equations
A curve has Parametric equations:
𝑥 = 5𝑡 2
𝑦 = 𝑡3
Work out:
2
1
𝑑𝑥
𝑦
𝑑𝑡
𝑑𝑡
𝑑𝑥
= 10𝑡
𝑑𝑡
2
𝑦
1
𝑑𝑥
𝑑𝑡
𝑑𝑡
2
Differentiate
𝑦 = 𝑡3
Sub in y and
dx/
dt
𝑡 3 (10𝑡) 𝑑𝑡
1
Multiply
2
10𝑡 4 𝑑𝑡
1
10𝑡 5
=
5
2
= 2𝑡 5
1
2
1
Remember to Integrate
now. A common mistake
is forgetting to!
Sub in the two limits
= 2(2)5 − 2(1)5
= 64 − 2
= 62
Work out the answer!
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
You need to be able to find the area
under a curve when it is given by
Parametric equations
R
The diagram shows a sketch of the
curve with Parametric equations:
𝑥 = 𝑡2
𝑦 = 2𝑡(3 − 𝑡)
The curve meets the x-axis at x = 0
and x = 9. The shaded region is
bounded by the curve and the x-axis.
a)
0
𝑡≥0
Find the value of t when:
i)
x=0
ii) x = 9
b) Find the Area of R
i)
𝑥 = 𝑡2
0=𝑡
0=𝑡
2
x=0
Square root
9
ii)
𝑥 = 𝑡2
9=𝑡
2
±3 = 𝑡
3=𝑡
x=9
Square root
t≥0
Normally when you integrate to find an area, you use the limits of
x, and substitute them into the equation
When integrating using Parametric Equations, you need to use the
limits of t (since we integrate with respect to t, not x)
The limits of t are worked out using the limits of x, as we have
just done
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
𝑦 = 2𝑡(3 − 𝑡)
You need to be able to find the area
under a curve when it is given by
Parametric equations
𝑥 = 𝑡2
R
The diagram shows a sketch of the
curve with Parametric equations:
𝑥 = 𝑡2
𝑦 = 2𝑡(3 − 𝑡)
0
𝑡≥0
3
The curve meets the x-axis at x = 0
and x = 9. The shaded region is
bounded by the curve and the x-axis.
𝑑𝑥
= 2𝑡
𝑑𝑡
𝑦
0
9
𝑑𝑥
𝑑𝑡
𝑑𝑡
Sub in y and
Find the value of t when:
i)
x=0
ii) x = 9
2𝑡 3 − 𝑡 2𝑡 𝑑𝑡
Multiply
3
12𝑡 2 − 4𝑡 3 𝑑𝑡
INTEGRATE!!
(don’t forget!)
0
4 3
=
b) Find the Area of R
 Limits of t are 3 and 0
dx/
dt
3
0
a)
Differentiate
12𝑡 3 4𝑡
−
3
4
= 4(3)3 −(3)4
= 108 − 81
= 27
= 4𝑡 3 − 𝑡 4
3
0
0
Sub in 3 and 0
− 4(0)3 −(0)4
Work out the answer
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
You need to be able to find the area
under a curve when it is given by
Parametric equations
The diagram shows a sketch of the
curve with Parametric equations:
𝑥 = 2𝑡 2
0
8
𝑦 = 𝑡(4 − 𝑡 2 )
Calculate the finite area inside the
loop…
We have the x limits, we need the t
limits
𝑥 = 2𝑡 2
0 = 2𝑡 2
0=𝑡
x=0
Halve and
square root
𝑥 = 2𝑡 2
8 = 2𝑡 2
±2 = 𝑡
x=8
Halve and
square root
Our t values are 0 and ±2
We have 3 t values. We now have to integrate twice,
once using 2 and once using -2
 One gives the area above the x-axis, and the other
the area below
(in this case the areas are equal but only since the
graph is symmetrical)
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
You need to be able to find the area
under a curve when it is given by
Parametric equations
The diagram shows a sketch of the
curve with Parametric equations:
𝑥 = 2𝑡
2
𝑥 = 2𝑡 2
2
𝑦
0
2
𝑦 = 𝑡(4 − 𝑡 )
Calculate the finite area inside the
loop…
The t limits are 0 and ±2
Differentiate
𝑑𝑥
= 4𝑡
𝑑𝑡
2
𝑑𝑥
𝑑𝑡
𝑑𝑡
Sub in y and
𝑦 = 𝑡(4 − 𝑡 2 )
dx/
dt
𝑡(4 − 𝑡 2 ) (4𝑡) 𝑑𝑡
0
Multiply out
2
16𝑡 2 − 4𝑡 4 𝑑𝑡
0
INTEGRATE!!!!
5 2
=
16𝑡 3 4𝑡
−
3
5
3
=
0
8
0
16(2)
4(2)
−
3
5
= 17
Sub in 0 and -2
5
3
−
5
16(0)
4(0)
−
3
5
1
15
At this point we can just double the answer, but
just to show you the other pairs give the same
answer (as the graph was symmetrical)
2D
𝑦
𝑑𝑥
𝑑𝑡
𝑑𝑡
Coordinate Geometry in the (x,y)
plane
You need to be able to find the area
under a curve when it is given by
Parametric equations
The diagram shows a sketch of the
curve with Parametric equations:
𝑥 = 2𝑡
2
𝑥 = 2𝑡 2
0
𝑦
−2
2
𝑦 = 𝑡(4 − 𝑡 )
Calculate the finite area inside the
loop…
The t limits are 0 and ±2
Differentiate
𝑑𝑥
= 4𝑡
𝑑𝑡
0
𝑑𝑥
𝑑𝑡
𝑑𝑡
Sub in y and
𝑦 = 𝑡(4 − 𝑡 2 )
dx/
dt
𝑡(4 − 𝑡 2 ) (4𝑡) 𝑑𝑡
−2
Multiply out
0
16𝑡 2 − 4𝑡 4 𝑑𝑡
−2
INTEGRATE!!!!
5 0
=
16𝑡 3 4𝑡
−
3
5
3
=
0
8
−2
16(0)
4(0)
−
3
5
= 17
Sub in 0 and -2
5
3
−
1
15
2
= 34
15
16(−2)
4(−2)
−
3
5
5
Here you end up
subtracting a negative…
Double
2D
Summary
• We have learnt what Parametric
equations are
• We have seen how to write them as a
Cartesian equation
• We have seen how to calculate areas
beneath Parametric equations