Engineering Calculus Notes 384

Engineering Calculus Notes 384 - 372 CHAPTER 3. REAL-VALUED...

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Unformatted text preview: 372 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Differentiating with respect to x1 , this gives ∂f = 2a11 x1 + 2a12 x2 + 2a13 x3 ∂x1 = 2(a11 x1 + a12 x2 + a13 x3 ). Note that the quantity in parentheses is exactly the product of the first → → row of A = [Q] with [− ], or equivalently the first entry of A [− ]. For x x → − for the vector convenience, we will abuse notation, and write simply A x → whose coordinate column is A times the coordinate column of − : x → → [A− ] = A [− ] . x x You should check that the other two partials of f are the other coordinates → of A− , so x →→ −− → ∇ f ( u ) = 2A− . u If we also recall that the dot product of two vectors can be written in terms of their coordinate columns as →→ − · − = [− ]T [− ] →→ xy x y → → then the matrix form of f (− ) = Q(− ) becomes x x → →→ f (− ) = − · A− ; x x x we separate out this calculation as Remark 3.9.1. The gradient of a function of the form → →→ f (− ) = − · A− x x x is →→ −− → ∇ f ( x ) = 2A− . x Note that, while our calculation was for a 3 × 3 matrix, the analogous result holds for a 2 × 2 matrix as well. Now, the Lagrange multiplier condition for extrema of f on S 2 becomes → → A− = λ− . u u (3.31) → → Geometrically, this means that A− and − have the same direction (up to u u reversal, or possibly squashing to zero). Such situations come up often in → problems involving matrices; we call a nonzero vector − which satisfies u ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Spring '08 term at University of Florida.

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