S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola
S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola
Advanced Engineering Mathematics
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Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Can you help me with 8.4.4, I dont know what its asking or how to solve
Transcribed Image Text:S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2
8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that
8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show
4X
4Y
y=
4 Y
4 Y
that this line meets the veil where:
4y
4x'
х%3
y'4
Solve for y as follows:
Z=
y 4
4y'
— Yу' + 4) — 4y'
Y =
у — у'
y'4
х — х'
z- 0
x'- 0
0 4
y'4
y'Y4Y 4y' » 4Y = 4y' - y'Y
х — х'
У — у'
Z
4Y y'(4 Y)
(1)
y'4
У — у"
4Y
х — х'
y=
4 Y'
y'4
Solve for x as follows:
— х(у + 4) — х'(у' + 4) — х'(у — у')
4x
and obtain the equation for x' below:
y'4
Substitute the equation obtained for y' in x =
— х(у' + 4) — х'(у'+4+у—у) %3D 0
— х(у' + 4) — х'(4 + у)
4x'
4Y
X
4Y
x'(4y)
+4
2)
X
(y' 4)
4XY 4X(4 Y)
4XY
+4X
4 - Y
4x'
= 4x'
У — у"
y'4
Z
4 Y
-4
XҮ + 4X — ХY
= x
XҮ + X(4 — Ү)
= x' >
— 2(y' + 4) — -4(у — у')
4 Y
4 Y
4(y' — у)
4X
x' =
(3)
Z=
4 - Y
y'4
The line (1) meets the veil at y = 0 hence (2) (3)
4Y
Hence it is proved that x' =
y =
4-Y
4y
y'4
4-Y
4x'
— х —
;Z=
;y = 0
y'4
x-x'
Suppose
у-у'
у'+4
8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the
image on the veil
= k
x = (k + 1)x'
(Ү— 2)2
X2
у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у'
4
And check that this is the ellipse shown in Figure
z = -4k
8.13
Also suppose
= a
y'+4
y'
4(1 — а)
х — ах',z — ау',4 — ау' + 4a — у' —
;z = 4(1 - a)
а
Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k
Figure 8.13: The parabola
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