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Unformatted text preview: 236 CHAPTER 3. REALVALUED FUNCTIONS: DIFFERENTIATION
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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
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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 xaxis,
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
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(since again it is zero on each of the standard basis elements − and − ).
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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.
 Fall '08
 ALL
 Calculus

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