Engineering Calculus Notes 248

Engineering Calculus Notes 248 - 236 CHAPTER 3. REAL-VALUED...

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Unformatted text preview: 236 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION → → but if we try to calculate d(0,0) f (− ) for the vector − = (1, m) using v v Equation (3.6), 1 (f (t, mt) − f (0, 0)) t→0 t d(0,0) f (1, m) = lim then, since along the line y = mx the function has a constant value f (t, mt) = m 1 + m2 —which is nonzero if m is—but f (0, 0) = 0, we see that the limit above does not exist: 1 1 (f (t, mt) − f (0, 0)) = lim t→0 t t→0 t lim m 1 + m2 diverges, and the differential cannot be evaluated along the vector →→ − = − + m− if m = 0. In fact, we saw before that this function is not → v ı continuous at the origin, which already contradicts differentiability, by Remark 3.3.2. Another example, this time one which is continuous at the origin, is √ 2xy if (x, y ) = (0, 0), x 2 +y 2 f (x, y ) = . 0 at (0, 0) This function is better understood when expressed in polar coordinates, where it takes the form 2r 2 cos θ sin θ r = 2r cos θ sin θ f (r cos θ, r sin θ ) = = r sin 2θ. From this we see that along the line making angle θ with the x-axis, f (x, y ) is a constant (sin 2θ ) times the distance from the origin: geometrically, the graph of f (x, y ) over this line is itself a line through the origin of slope m = sin 2θ . Along the two coordinate axes, this slope is zero, but for example along the line y = x (θ = π/4), the slope is sin π/2 = 1. So this time, the function defined by Equation (3.6) (without asking about differentiability) exists at the origin, but it is not linear → → (since again it is zero on each of the standard basis elements − and − ). ı ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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