Let to < t1 < t2 < t3 and wo, w₁, w2, and w³ be constants. (a) S[w] = [ -t3 the stationary path w(t) of the functional [{"ʻdt w" (t)², w(to) = wo, w(t) = wr, w(tz)=wz, w(ts)=wa, is given by w(t) = co+c₁t + c2t² + c3t³ where co, €1, c2 and c3 satisfy the matrix equation Vc = = W where Wo co 1 to W1 €1 and w = 1 t₁ c = 7 W2 €2 1 t₂ 03 C3 1 t3 You may use the fact that w(t) satisfies the equation OF d OF ď² OF + = 0, მი dt ǝw dt² ǝw" where F(t, w, w', w") = w" (t)². (b) let Pi = (xi, Yi, Zi), i = 0, 3 be four points in three-dimensional space. Let paths x(t), y(t) and z(t), parametrized by t, interpolate Po, … … …, P3 at t = to,……, t3 respectively, so that (x(ti), y(ti), z(ti)) = (xi, Yi, Zi) for i = 0, … … …, 3. A spline is a choice of interpolating functions x(t), y(t) and z(t) such x, y and z are each stationary paths of the functional S[w] given in (a) above. (i) x(t) = €2,0 + €2,1t+Cz,2t² + Cz,3t³, y(t) = Cy,0 + Cy,1t + Cy,2t² + Cy,3t³, z(t) = €2,0 + cz,1t+Cz,2t² + Cz,3t³, VC = x, where (CÃ,0 €₂,0 €2,0\ xo Yo Zo C= Cz,1 Cy,l Cz,1 X1 Y1 21 and X = = Cz,2 Cy,2 Cz,2 Cz3 Cy3 Cz,3/ X2 Y2 Z2 3 Y3 Z3, (ii) For the case to = 0, t₁ = 1, t2 = 2 and t3 3 and = - Po = (0,0,0), P₁ = (1,0,0), P₂ = (1,1,0), and P3 = (1,1,1), find the interpolating functions x(t), y(t) and z(t). You may find the following matrix result useful: 1 1 0 0 0 1 0 0 0 1 1 1 1 1 2 4 8 1 3927 = -11/6 3 -3/2 1/3 1 -5/2 2 -1/2 −1/6 1/2 −1/2 1/6 02@ knew to get stationary point which minimizers is given functional, we consider associated Euler- Lagrange's equation : = = Cy,ot 裴一起()+()= = 0 зст = 0261 x(t) = a + a₁ (t-to) + a₂ (t-to) (t-t₁) + a3 (t-to) (t-t₁) (t-t>) Similarly, y Cx,0 + Cx,, •t + Cx₁₂ •t² + Cx,³t³ Cy,1 .t Cz,o + G₁₁ • t + G₁₂²±² + G₁st³ +Cyz →d (20") =0 ⇒ =0 (integrating wir.to it) اطمه dt = (again integrating) Art + A1 → dn = Aof² + Alt + A₂ → Wits = Aof ³ + AL + ² + Art + A3 (again integracting) For t = to, Xx₁ = x(to)=Cx,0 + Cx,1 to + Cx12 to² + Cx,3 to³ Yo = y(to) = Cyro + Cyll to + Cyr 2 to² + Cy, 3 to ³ 30 = 3 (to) = C₂,0 + C₂,1 •to + C₂₁₂ to ² + C₂13 to ³ Lagain integrating) Let Co = A 3 Cl = Az Cz = > = A and C3 = to where Ao, A1, A2, As are For t = ti (i.e. 1, 2, 3) in general, So we get wit Co + c₁t + c₂t² + c₂t³ arbitrary constants. xi = Cx10 + Cx, ti + Cx, 2 t₁ + Cx,3 ti Yi = Cyro + Cyll ti + Cy,zti + Cysti Zi = G₁o + C₁₁te + G₁zt² + G₁3 ti Now w(to) = Wo ⇒ Co + Cito + C₂ to² + C3 to ³ wlti) = Wo = W₁ = Co + at₁ + cati² + as t₁³ = W₁ wct₂) = W₂ Co + C₁tz +C₂t₂² + c₂+33 = W₂ Wcb3) = W3 ⇒ Co + Cits + C₂tz² + Czt = W3 For X In • matrix form, * * * = V Cx, o Cx,1 655 می کی کی می Co = हु हु हु W₂ Cx,2 -Cx, 3 LW3. for y, = vx, For 31 30 31 on on A or CE 32 Cy1o yz Cy Cy1z Cy,3] = = VG,1 4,2 C310 = √x3 Gis ⇒ VC = W where v= b₁ fox for t +ty of of C = Co and w= Wo G W₁ Cz W₂ LW3. Thus we ' get the stationary functional Co, C., C2,C3 obtained from matrix equation VC=w. path is given is Web) = Co + C₁t + C₂t² + Cst³ where Схіо Сую X Yo 30 vc = v Cxpl Cy₁' G₁ = [VX √x₂ VX3] 81 = X Cx, 2 Cy, 2 C₁z Уг 82 Cx13 Cy.3 48.31 X3 Уз 83
Let to < t1 < t2 < t3 and wo, w₁, w2, and w³ be constants. (a) S[w] = [ -t3 the stationary path w(t) of the functional [{"ʻdt w" (t)², w(to) = wo, w(t) = wr, w(tz)=wz, w(ts)=wa, is given by w(t) = co+c₁t + c2t² + c3t³ where co, €1, c2 and c3 satisfy the matrix equation Vc = = W where Wo co 1 to W1 €1 and w = 1 t₁ c = 7 W2 €2 1 t₂ 03 C3 1 t3 You may use the fact that w(t) satisfies the equation OF d OF ď² OF + = 0, მი dt ǝw dt² ǝw" where F(t, w, w', w") = w" (t)². (b) let Pi = (xi, Yi, Zi), i = 0, 3 be four points in three-dimensional space. Let paths x(t), y(t) and z(t), parametrized by t, interpolate Po, … … …, P3 at t = to,……, t3 respectively, so that (x(ti), y(ti), z(ti)) = (xi, Yi, Zi) for i = 0, … … …, 3. A spline is a choice of interpolating functions x(t), y(t) and z(t) such x, y and z are each stationary paths of the functional S[w] given in (a) above. (i) x(t) = €2,0 + €2,1t+Cz,2t² + Cz,3t³, y(t) = Cy,0 + Cy,1t + Cy,2t² + Cy,3t³, z(t) = €2,0 + cz,1t+Cz,2t² + Cz,3t³, VC = x, where (CÃ,0 €₂,0 €2,0\ xo Yo Zo C= Cz,1 Cy,l Cz,1 X1 Y1 21 and X = = Cz,2 Cy,2 Cz,2 Cz3 Cy3 Cz,3/ X2 Y2 Z2 3 Y3 Z3, (ii) For the case to = 0, t₁ = 1, t2 = 2 and t3 3 and = - Po = (0,0,0), P₁ = (1,0,0), P₂ = (1,1,0), and P3 = (1,1,1), find the interpolating functions x(t), y(t) and z(t). You may find the following matrix result useful: 1 1 0 0 0 1 0 0 0 1 1 1 1 1 2 4 8 1 3927 = -11/6 3 -3/2 1/3 1 -5/2 2 -1/2 −1/6 1/2 −1/2 1/6 02@ knew to get stationary point which minimizers is given functional, we consider associated Euler- Lagrange's equation : = = Cy,ot 裴一起()+()= = 0 зст = 0261 x(t) = a + a₁ (t-to) + a₂ (t-to) (t-t₁) + a3 (t-to) (t-t₁) (t-t>) Similarly, y Cx,0 + Cx,, •t + Cx₁₂ •t² + Cx,³t³ Cy,1 .t Cz,o + G₁₁ • t + G₁₂²±² + G₁st³ +Cyz →d (20") =0 ⇒ =0 (integrating wir.to it) اطمه dt = (again integrating) Art + A1 → dn = Aof² + Alt + A₂ → Wits = Aof ³ + AL + ² + Art + A3 (again integracting) For t = to, Xx₁ = x(to)=Cx,0 + Cx,1 to + Cx12 to² + Cx,3 to³ Yo = y(to) = Cyro + Cyll to + Cyr 2 to² + Cy, 3 to ³ 30 = 3 (to) = C₂,0 + C₂,1 •to + C₂₁₂ to ² + C₂13 to ³ Lagain integrating) Let Co = A 3 Cl = Az Cz = > = A and C3 = to where Ao, A1, A2, As are For t = ti (i.e. 1, 2, 3) in general, So we get wit Co + c₁t + c₂t² + c₂t³ arbitrary constants. xi = Cx10 + Cx, ti + Cx, 2 t₁ + Cx,3 ti Yi = Cyro + Cyll ti + Cy,zti + Cysti Zi = G₁o + C₁₁te + G₁zt² + G₁3 ti Now w(to) = Wo ⇒ Co + Cito + C₂ to² + C3 to ³ wlti) = Wo = W₁ = Co + at₁ + cati² + as t₁³ = W₁ wct₂) = W₂ Co + C₁tz +C₂t₂² + c₂+33 = W₂ Wcb3) = W3 ⇒ Co + Cits + C₂tz² + Czt = W3 For X In • matrix form, * * * = V Cx, o Cx,1 655 می کی کی می Co = हु हु हु W₂ Cx,2 -Cx, 3 LW3. for y, = vx, For 31 30 31 on on A or CE 32 Cy1o yz Cy Cy1z Cy,3] = = VG,1 4,2 C310 = √x3 Gis ⇒ VC = W where v= b₁ fox for t +ty of of C = Co and w= Wo G W₁ Cz W₂ LW3. Thus we ' get the stationary functional Co, C., C2,C3 obtained from matrix equation VC=w. path is given is Web) = Co + C₁t + C₂t² + Cst³ where Схіо Сую X Yo 30 vc = v Cxpl Cy₁' G₁ = [VX √x₂ VX3] 81 = X Cx, 2 Cy, 2 C₁z Уг 82 Cx13 Cy.3 48.31 X3 Уз 83
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 93E
Question
100%
What is part bii?
![Let to < t1 < t2 < t3 and wo, w₁, w2, and w³ be constants.
(a)
S[w] = [
-t3
the stationary path w(t) of the functional
[{"ʻdt w" (t)², w(to) = wo, w(t) = wr, w(tz)=wz, w(ts)=wa,
is given by w(t) = co+c₁t + c2t² + c3t³ where co, €1, c2 and c3
satisfy the matrix equation
Vc =
= W
where
Wo
co
1 to
W1
€1
and w =
1 t₁
c =
7
W2
€2
1 t₂
03
C3
1 t3
You may use the fact that w(t) satisfies the equation
OF
d OF
ď² OF
+
= 0,
მი
dt ǝw
dt² ǝw"
where F(t, w, w', w") = w" (t)².
(b) let Pi
=
(xi, Yi, Zi), i = 0, 3 be four points in three-dimensional
space. Let paths x(t), y(t) and z(t), parametrized by t, interpolate
Po, … … …, P3 at t = to,……, t3 respectively, so that
(x(ti), y(ti), z(ti)) = (xi, Yi, Zi) for i = 0, … … …,
3.
A spline is a choice of interpolating functions x(t), y(t) and z(t)
such x, y and z are each stationary paths of the functional S[w]
given in (a) above.
(i)
x(t) = €2,0 + €2,1t+Cz,2t² + Cz,3t³,
y(t) = Cy,0 + Cy,1t + Cy,2t² + Cy,3t³,
z(t) = €2,0 + cz,1t+Cz,2t² + Cz,3t³,
VC = x,
where
(CÃ,0 €₂,0 €2,0\
xo Yo Zo
C=
Cz,1 Cy,l Cz,1
X1 Y1
21
and X =
=
Cz,2 Cy,2 Cz,2
Cz3 Cy3 Cz,3/
X2 Y2 Z2
3 Y3 Z3,
(ii) For the case to = 0, t₁
= 1, t2 = 2 and t3 3 and
=
-
Po = (0,0,0), P₁ = (1,0,0), P₂ = (1,1,0), and P3 = (1,1,1),
find the interpolating functions x(t), y(t) and z(t).
You may find the following matrix result useful:
1
1 0 0 0
1
0
0
0
1 1 1
1
1 2 4 8
1
3927
=
-11/6 3 -3/2 1/3
1 -5/2 2 -1/2
−1/6 1/2 −1/2 1/6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F3d8317a8-f07a-4666-b0f3-39aac6af3716%2Fqbjpmy8_processed.png&w=3840&q=75)
Transcribed Image Text:Let to < t1 < t2 < t3 and wo, w₁, w2, and w³ be constants.
(a)
S[w] = [
-t3
the stationary path w(t) of the functional
[{"ʻdt w" (t)², w(to) = wo, w(t) = wr, w(tz)=wz, w(ts)=wa,
is given by w(t) = co+c₁t + c2t² + c3t³ where co, €1, c2 and c3
satisfy the matrix equation
Vc =
= W
where
Wo
co
1 to
W1
€1
and w =
1 t₁
c =
7
W2
€2
1 t₂
03
C3
1 t3
You may use the fact that w(t) satisfies the equation
OF
d OF
ď² OF
+
= 0,
მი
dt ǝw
dt² ǝw"
where F(t, w, w', w") = w" (t)².
(b) let Pi
=
(xi, Yi, Zi), i = 0, 3 be four points in three-dimensional
space. Let paths x(t), y(t) and z(t), parametrized by t, interpolate
Po, … … …, P3 at t = to,……, t3 respectively, so that
(x(ti), y(ti), z(ti)) = (xi, Yi, Zi) for i = 0, … … …,
3.
A spline is a choice of interpolating functions x(t), y(t) and z(t)
such x, y and z are each stationary paths of the functional S[w]
given in (a) above.
(i)
x(t) = €2,0 + €2,1t+Cz,2t² + Cz,3t³,
y(t) = Cy,0 + Cy,1t + Cy,2t² + Cy,3t³,
z(t) = €2,0 + cz,1t+Cz,2t² + Cz,3t³,
VC = x,
where
(CÃ,0 €₂,0 €2,0\
xo Yo Zo
C=
Cz,1 Cy,l Cz,1
X1 Y1
21
and X =
=
Cz,2 Cy,2 Cz,2
Cz3 Cy3 Cz,3/
X2 Y2 Z2
3 Y3 Z3,
(ii) For the case to = 0, t₁
= 1, t2 = 2 and t3 3 and
=
-
Po = (0,0,0), P₁ = (1,0,0), P₂ = (1,1,0), and P3 = (1,1,1),
find the interpolating functions x(t), y(t) and z(t).
You may find the following matrix result useful:
1
1 0 0 0
1
0
0
0
1 1 1
1
1 2 4 8
1
3927
=
-11/6 3 -3/2 1/3
1 -5/2 2 -1/2
−1/6 1/2 −1/2 1/6
![02@
knew to get stationary point which minimizers is given functional, we consider associated
Euler- Lagrange's equation :
=
=
Cy,ot
裴一起()+()=
= 0
зст
=
0261
x(t) = a + a₁ (t-to) + a₂ (t-to) (t-t₁) + a3 (t-to) (t-t₁) (t-t>)
Similarly, y
Cx,0 + Cx,, •t + Cx₁₂ •t² + Cx,³t³
Cy,1
.t
Cz,o + G₁₁ • t + G₁₂²±² + G₁st³
+Cyz
→d (20")
=0
⇒
=0
(integrating wir.to it)
اطمه
dt
=
(again integrating)
Art + A1
→ dn = Aof² + Alt + A₂
→ Wits = Aof ³ + AL + ² + Art + A3
(again integracting)
For t = to,
Xx₁ = x(to)=Cx,0 + Cx,1 to + Cx12 to² + Cx,3 to³
Yo = y(to) = Cyro + Cyll to + Cyr 2 to² + Cy, 3 to ³
30 = 3 (to) = C₂,0 + C₂,1 •to + C₂₁₂ to ² + C₂13 to ³
Lagain integrating)
Let
Co = A 3
Cl = Az Cz =
>
= A and C3 = to where Ao, A1, A2, As are
For t = ti
(i.e. 1, 2, 3) in general,
So we get wit
Co + c₁t + c₂t² + c₂t³
arbitrary constants.
xi
=
Cx10 + Cx, ti + Cx, 2 t₁ + Cx,3 ti
Yi
=
Cyro + Cyll ti
+
Cy,zti +
Cysti
Zi
=
G₁o + C₁₁te + G₁zt² + G₁3 ti
Now
w(to) = Wo ⇒ Co + Cito + C₂ to² + C3 to ³
wlti)
= Wo
= W₁ = Co + at₁ + cati² + as t₁³
= W₁
wct₂)
=
W₂ Co + C₁tz +C₂t₂² + c₂+33 = W₂
Wcb3) = W3 ⇒ Co + Cits + C₂tz² + Czt
=
W3
For X
In
• matrix form,
* * *
= V
Cx, o
Cx,1
655
می کی کی می
Co
=
हु हु हु
W₂
Cx,2
-Cx, 3
LW3.
for y,
=
vx,
For 31
30
31
on on A or CE
32
Cy1o
yz
Cy
Cy1z
Cy,3]
=
=
VG,1
4,2
C310
= √x3
Gis
⇒
VC = W
where
v=
b₁
fox for t
+ty of of
C = Co
and w=
Wo
G
W₁
Cz
W₂
LW3.
Thus
we
'
get the stationary
functional
Co, C., C2,C3 obtained from matrix equation VC=w.
path is
given
is
Web) = Co + C₁t + C₂t² + Cst³ where
Схіо Сую
X
Yo
30
vc = v
Cxpl
Cy₁'
G₁ = [VX √x₂ VX3]
81
= X
Cx, 2
Cy, 2
C₁z
Уг
82
Cx13
Cy.3
48.31
X3
Уз
83](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F3d8317a8-f07a-4666-b0f3-39aac6af3716%2Fq2ldb9y_processed.png&w=3840&q=75)
Transcribed Image Text:02@
knew to get stationary point which minimizers is given functional, we consider associated
Euler- Lagrange's equation :
=
=
Cy,ot
裴一起()+()=
= 0
зст
=
0261
x(t) = a + a₁ (t-to) + a₂ (t-to) (t-t₁) + a3 (t-to) (t-t₁) (t-t>)
Similarly, y
Cx,0 + Cx,, •t + Cx₁₂ •t² + Cx,³t³
Cy,1
.t
Cz,o + G₁₁ • t + G₁₂²±² + G₁st³
+Cyz
→d (20")
=0
⇒
=0
(integrating wir.to it)
اطمه
dt
=
(again integrating)
Art + A1
→ dn = Aof² + Alt + A₂
→ Wits = Aof ³ + AL + ² + Art + A3
(again integracting)
For t = to,
Xx₁ = x(to)=Cx,0 + Cx,1 to + Cx12 to² + Cx,3 to³
Yo = y(to) = Cyro + Cyll to + Cyr 2 to² + Cy, 3 to ³
30 = 3 (to) = C₂,0 + C₂,1 •to + C₂₁₂ to ² + C₂13 to ³
Lagain integrating)
Let
Co = A 3
Cl = Az Cz =
>
= A and C3 = to where Ao, A1, A2, As are
For t = ti
(i.e. 1, 2, 3) in general,
So we get wit
Co + c₁t + c₂t² + c₂t³
arbitrary constants.
xi
=
Cx10 + Cx, ti + Cx, 2 t₁ + Cx,3 ti
Yi
=
Cyro + Cyll ti
+
Cy,zti +
Cysti
Zi
=
G₁o + C₁₁te + G₁zt² + G₁3 ti
Now
w(to) = Wo ⇒ Co + Cito + C₂ to² + C3 to ³
wlti)
= Wo
= W₁ = Co + at₁ + cati² + as t₁³
= W₁
wct₂)
=
W₂ Co + C₁tz +C₂t₂² + c₂+33 = W₂
Wcb3) = W3 ⇒ Co + Cits + C₂tz² + Czt
=
W3
For X
In
• matrix form,
* * *
= V
Cx, o
Cx,1
655
می کی کی می
Co
=
हु हु हु
W₂
Cx,2
-Cx, 3
LW3.
for y,
=
vx,
For 31
30
31
on on A or CE
32
Cy1o
yz
Cy
Cy1z
Cy,3]
=
=
VG,1
4,2
C310
= √x3
Gis
⇒
VC = W
where
v=
b₁
fox for t
+ty of of
C = Co
and w=
Wo
G
W₁
Cz
W₂
LW3.
Thus
we
'
get the stationary
functional
Co, C., C2,C3 obtained from matrix equation VC=w.
path is
given
is
Web) = Co + C₁t + C₂t² + Cst³ where
Схіо Сую
X
Yo
30
vc = v
Cxpl
Cy₁'
G₁ = [VX √x₂ VX3]
81
= X
Cx, 2
Cy, 2
C₁z
Уг
82
Cx13
Cy.3
48.31
X3
Уз
83
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