Engineering Calculus Notes 248

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

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236 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION but if we try to calculate d (0 , 0) f ( −→ v ) for the vector −→ v = (1 ,m ) using Equation ( 3.6 ), d (0 , 0) f (1 ,m ) = lim t 0 1 t ( f ( t,mt ) f (0 , 0)) then, since along the line y = mx the function has a constant value f ( t,mt ) = m 1 + m 2 —which is nonzero if m is—but f (0 , 0) = 0, we see that the limit above does not exist: lim t 0 1 t ( f ( t,mt ) f (0 , 0)) = lim t 0 1 t parenleftbigg m 1 + m 2 parenrightbigg diverges, and the differential cannot be evaluated along the vector −→ v = −→ ı + m −→ if m negationslash = 0. In fact, we saw before that this function is not 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 f ( x,y ) = 2 xy x 2 + y 2 if ( x,y ) negationslash = (0 , 0) , 0 at (0 , 0) . This function is better understood when expressed in polar coordinates,
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