Let U be a vector function of position in R3 with continuous second partial deriva- tives. We write (U₁, U2, U3) for the components of U. (a) Show that ¹⁄▼ (Ü · Ü) – Ü × (▼ × Ü) = (Ũ · ▼) Ū, where ((UV)U), = U₁ (b) We define the vector function of position, by setting = V × . If the condition V U = 0 holds true, show that ▼ × ((Ū · ▼) Ú) = (Ū · ▼) Ñ — (Ñ · ▼) Ú. . (Hint: The representation of (UV) U from the first part might be useful.)
Let U be a vector function of position in R3 with continuous second partial deriva- tives. We write (U₁, U2, U3) for the components of U. (a) Show that ¹⁄▼ (Ü · Ü) – Ü × (▼ × Ü) = (Ũ · ▼) Ū, where ((UV)U), = U₁ (b) We define the vector function of position, by setting = V × . If the condition V U = 0 holds true, show that ▼ × ((Ū · ▼) Ú) = (Ū · ▼) Ñ — (Ñ · ▼) Ú. . (Hint: The representation of (UV) U from the first part might be useful.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![2. Let U be a vector function of position in R³ with continuous second partial deriva-
tives. We write (U₁, U2, U3) for the components of U.
(a) Show that
V (U • U) – Ü × (▼ × Ü) = (U · ▼) Ū,
(V
where ((UV) U), = U₁
(b) We define the vector function of position, by setting = V × . If the
condition V U = 0 holds true, show that
▼ × ((Ū · V) Ű) = (Ũ · V) Ñ - (Ñ· ▼) ū.
(Hint: The representation of (UV) U from the first part might be useful.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1e1f7ce-6847-4965-81f8-e638576c5e85%2F1c0bbfbf-5e3f-4438-97e8-df0e1d38d609%2Fcbsykbg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2. Let U be a vector function of position in R³ with continuous second partial deriva-
tives. We write (U₁, U2, U3) for the components of U.
(a) Show that
V (U • U) – Ü × (▼ × Ü) = (U · ▼) Ū,
(V
where ((UV) U), = U₁
(b) We define the vector function of position, by setting = V × . If the
condition V U = 0 holds true, show that
▼ × ((Ū · V) Ű) = (Ũ · V) Ñ - (Ñ· ▼) ū.
(Hint: The representation of (UV) U from the first part might be useful.)
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