Specialist Maths: Calculus Week 2

Specialist Maths
Calculus
Week 2
Bezier Curves
• These are graphic curves to enable us to
draw curves on computers.
• They were designed by Pierre Bezier in
1962 so he could create computer designs
for creating new car designs for Renault.
Constructing Bezier Curves
• Start the curve at point S and end it at point E.
• Introduce two control points C1 and C2, and a
parameter t where 0≤ t ≤ 1.
C1 C2




E
S
• Points F, G and H are located on SC1, C1C2 and
C2E respectively, such that SF:FC1 = t:1-t,
C1G:GC2 = t:1-t and C2H:HE = t:1-t
C1 C2
t G 1-t 
 
F 1-t t

H

t 1-t


E
S
• Points I and J are located on FG, and GH
respectively, such that FI:IG = t:1-t,
GJ:JH = t:1-t.
C1 C2
t G 1-t 
 
1-t
 1-t t t
F I
t 
J 1-t H
t 1-t


E
S
• Finally P(x(t),y(t)) are the points that describe the
Bezier curve such that IP:PJ = t:1-t
C1 C2
t G 1-t 
 
1-t
 1-t t t
F I t 
t P 1-t 
J 1-t H
t 1-t


E
S
Finding the coordinates of P
Given the start point S( x0 , y0 ), control points C1 ( x1 y1 ) and C 2 ( x2 , y2 ),
end point E( x3 , y3 ), then P x(t ), y (t )  is given by
x(t )  a x t 3  bxt 2  c x t  d x
y (t )  a y t 3  by t 2  c y t  d y
Where and
a x  x3  3 x2  3 x1  x0 a y  y3  3 y2  3 y1  y0
bx  3 x2  6 x1  3 x0 by  3 y2  6 y1  3 y0
c x  3 x1  3 x0 c y  3 y1  3 y0
d x  x0 d y  y0
Using Matricis
Given the start point S( x0 , y0 ), control points C1 ( x1 y1 ) and C 2 ( x2 , y2 ),
end point E( x3 , y3 ), then P x(t ), y (t )  is given by
x(t )  a x t 3  bx t 2  c x t  d x
y (t )  a y t 3  by t 2  c y t  d y
 a x  1  3 3  1  x3   a y  1  3 3  1   y 3 
 b  0 3  6 3   x  b  
0 3  6 3  y 
 x  2   y  2 
 c x  0 0 3  3  x1   c y  0 0 3  3  y1 
         
d
 y   0 0 0 1   y0 
 d x  0 0 0 1   x0 
Note: P is at S when t = 0, and at E when t = 1.
Example 4 (Ex 6C2)
Find the Bezier curve from A(0,1) to B(3,2) with control points
C1 (1,3) and C 2 (2,4),  a x  1  3 3  1  x3   a y  1  3 3  1  y3 
 b  0 3  6 3   x  b    
 y   0 3  6 3   y 2 
x(t )  a x t 3  bx t 2  c x t  d x  x
 c x  0 0
 2 
3  3  x1   c y  0 0 3  3  y1 
         
 d x  0 0 0 1   x0  d y  0 0 0 1   y0 
y (t )  a y t 3  by t 2  c y t  d y
Solution 4
Find the Bezier curve from A(0,1) to B(3,2) with control points
C1 (1,3) and C 2 (2,4),
 a x  1  3 3  1  x3   a y  1  3 3  1  y3 
x(t )  a x t  bx t  cx t  d x
3 2
 b  0 3  6 3   x 
 x  2 
b    
 y   0 3  6 3   y 2 
 c x  0 0 3  3  x1   c y  0 0 3  3  y1 
y (t )  a y t 3  by t 2  c y t  d y   
 d x  0 0 0
 
1   x0 
  
d y  0 0 0
 
1   y0 
Example 5 (Ex 6C2)
For the Bezier curves defined by
x(t )  4t 3  5t 2  2t  3
y (t )  t  2t  4t  5
3 2
Find the coordinates of the starting point and end point.

Solution 5
x(t )  4t 3  5t 2  2t  3
y (t )  t 3  2t 2  4t  5
Find the coordinates of the starting point and end point.
Example 6 (Ex 6C2)
S(-3,5)
x(t )  4t  5t  2t  3
3 2
E(-2,4)
y (t )  t  2t  4t  5
3 2
Find the coordinates of the highest and lowest points.

Solution 6
S(-3,5)
x(t )  4t 3  5t 2  2t  3
E(-2,4)
y (t )  t 3  2t 2  4t  5
Find the coordinates of the highest and lowest points.
Example 7 (Ex 6C2)
For the Bezier curves defined by S(-3,5)
x(t )  4t 3  5t 2  2t  3 E(-2,4)
y (t )  t 3  2t 2  4t  5
Find the coordinates of the most left and right points.
Solution 7
S(-3,5)
x (t )  4t  5t  2t  3
3 2
E(-2,4)
y (t )  t  2t  4t  5
3 2
Find the coordinates of the most left and right points.

Example 8 (Ex 6C2)
x(t )  4t 3  5t 2  2t  3
y (t )  t 3  2t 2  4t  5
Find the velocity vector and the speed of the animation when t  0.3.
Solution 8
x(t )  4t 3  5t 2  2t  3
y (t )  t 3  2t 2  4t  5
Find the velocity vector and the speed of the animation when t  0.3.
Parametric Equations of Tangents to
Curves
The direction vector of the tangent to curve ( x(t), y (t )) at t  a,

is  x' (a), y' (a)
( x(a), y (a ))
The equation of the tangent is given by

x( s )  x(a )  x' (a ) s
 y ( s )  y (a)  y ' (a) s
(x( 0 ),y( 0 ))
Example 9 (Ex 6C3)
Find the parametric equation of the tangent to the curve
x(t )  t 3  t 2  2t  3
y (t )  t 2  6t  5
at the point where t  -2
Solution 9
Find the parametric equation of the tangent to the curve
x(t )  t 3  t 2  2t  3
y (t )  t 2  6t  5
at the point where t  -2
Using Parametric Forms
• Parametric representation enables us to
express the coordinates of any point in
terms of one variable “t”.
• Example P(t2-2,2t+1).
• We need however to be able to convert
from parametric form back to Cartesian
form.
Example 11 (Ex 6E1)
Find the Cartesian equations of the curve with
parametric equations
x(t )  5  t  2t 2
y (t )  8  4t
Solution 11
x(t )  5  t  2t 2
y (t )  8  4t
Example 12 (Ex 6E1)
1
x(t )   t
t
1
y (t )   t
t
Solution 12
Find the Cartesian
equations of the
curve with
parametric
equations
1
x(t )   t
t
1
y (t )   t
t
Tangents and Normals
• If P has coordinates (x(t),y(t)), then
dy y ' (t )

dx x' (t )
• The slope of the tangent at t=a is given by:
y ' (a )
m
x' (a )
Example 13 (Ex 6E2)
• Find the equation to the tangent and the normal to
the curve with paramedic equations when t = -1:
x(t )  t  4
2
y (t )  4t  5
Solution 13
Example 14 (Ex 6E2)
• Find the equation of the tangent to the curve
with parametric equations x = 3t +7 and
y = 8 – 2t2, having slope –4.
Solution 14
This Week
• Text pages 216 to 220, 223 to 226
• Exercise 6C2 Q 1 - 5
• Exercise 6C3 Q1, 2
• Exercise 6E1 Q 1, 2
• Exercise 6E2 Q 1-5

Specialist Maths: Calculus Week 2

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Specialist Maths

Find the coordinates of the starting point and end point.

Find the coordinates of the highest and lowest points.

Find the coordinates of the most left and right points.

The direction vector of the tangent to curve ( x(t), y (t )) at t  a,

The equation of the tangent is given by

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