specialist maths calculus week 2. bezier curves these are graphic curves to enable us to draw curves...
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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.
SE
C1C2
Constructing Bezier Curves• 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
SE
C1C2
t
1-t
t 1-t
t
1-t
F
G
H
Constructing Bezier Curves• Points I and J are located on FG, and GH
respectively, such that FI:IG = t:1-t, GJ:JH = t:1-t.
SE
C1C2
t
1-t
t 1-t
t
1-t
I
Jt
1-t t
1-t
F
G
H
Constructing Bezier Curves• Finally P(x(t),y(t)) are the points that describe the
Bezier curve such that IP:PJ = t:1-t
SE
C1C2
t
1-t
t 1-t
t
1-t
I
Jt
1-t t
1-t
F
G
Ht1-tP
Finding the coordinates of P
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xxxx
dtctbtaty
dtctbtatx
tytxyx
yxyxyx
23
23
33
22211100
)(
)(
bygiven is )(),(P then ),,(Epoint end
),,(C and )(C points control ),,(Spoint start Given the
0
01
012
0123
33
363
33
Where
xd
xxc
xxxb
xxxxa
x
x
x
x
0
01
012
0123
33
363
33
and
yd
yyc
yyyb
yyyya
y
y
y
y
Using Matricis
0
1
2
3
1000
3300
3630
1331
x
x
x
x
d
c
b
a
x
x
x
x
0
1
2
3
1000
3300
3630
1331
y
y
y
y
d
c
b
a
y
y
y
y
yyyy
xxxx
dtctbtaty
dtctbtatx
tytxyx
yxyxyx
23
23
33
22211100
)(
)(
bygiven is )(),(P then ),,(Epoint end
),,(C and )(C points control ),,(Spoint start Given the
Note: P is at S when t = 0, and at E when t = 1.
Example 4 (Ex 6C2)
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xxxx
dtctbtaty
dtctbtatx
23
23
21
)(
)(
),4,2(C and )3,1(C
points control with B(3,2) toA(0,1) from curveBezier theFind
0
1
2
3
1000
3300
3630
1331
x
x
x
x
d
c
b
a
x
x
x
x
0
1
2
3
1000
3300
3630
1331
y
y
y
y
d
c
b
a
y
y
y
y
Solution 4
yyyy
xxxx
dtctbtaty
dtctbtatx
23
23
21
)(
)(
),4,2(C and )3,1(C
points control with B(3,2) toA(0,1) from curveBezier theFind
0
1
2
3
1000
3300
3630
1331
x
x
x
x
d
c
b
a
x
x
x
x
0
1
2
3
1000
3300
3630
1331
y
y
y
y
d
c
b
a
y
y
y
y
Example 5 (Ex 6C2)
point. end andpoint starting theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttx
Solution 5
point. end andpoint starting theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttx
Example 6 (Ex 6C2)
points.lowest andhighest theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttxS(-3,5)
E(-2,4)
Solution 6
points.lowest andhighest theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttxS(-3,5)
E(-2,4)
Example 7 (Ex 6C2)
points.right andleft most theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttxS(-3,5)
E(-2,4)
Solution 7
points.right andleft most theof scoordinate theFind
542)(
3254)(
by defined curvesBezier For the
23
23
tttty
ttttxS(-3,5)
E(-2,4)
Example 8 (Ex 6C2)
.3.0when animation theof speed theandvector velocity theFind
542)(
3254)(
by defined curvesBezier For the
23
23
t
tttty
ttttx
Solution 8
.3.0when animation theof speed theandvector velocity theFind
542)(
3254)(
by defined curvesBezier For the
23
23
t
tttty
ttttx
Parametric Equations of Tangents to Curves
(a)y'(a),x' is
,at ))(,( curve o tangent t theofvector direction The attyx(t)
))(),(( ayax
))),y((x( 00
sayaysy
saxaxsx
)(')()(
)(')()(
bygiven is tangent theofequation The
Example 9 (Ex 6C3)
2 epoint wher at the
56)(
32)(
curve theo tangent t theofequation parametric theFind
2
23
-t
ttty
ttttx
Solution 9
2 epoint wher at the
56)(
32)(
curve theo tangent t theofequation parametric theFind
2
23
-t
ttty
ttttx
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
tty
tttx
48)(
25)( 2
Solution 11Find the Cartesian equations of the curve with
parametric equations
tty
tttx
48)(
25)( 2
Example 12 (Ex 6E1)Find the Cartesian equations of the curve with
parametric equations
tt
ty
tt
tx
1)(
1)(
Solution 12Find the Cartesian
equations of the
curve with parametric
equations
tt
ty
tt
tx
1)(
1)(
Tangents and Normals
• If P has coordinates (x(t),y(t)), then
• The slope of the tangent at t=a is given by:
)('
)('
tx
ty
dx
dy
)('
)('
ax
aym
Example 13 (Ex 6E2)• Find the equation to the tangent and the normal to
the curve with paramedic equations when t = -1:
54)(
4)( 2
tty
ttx
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