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.)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section8.3: Vectors
Problem 60E
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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.)
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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