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Engineering Calculus Notes 234

# Engineering Calculus Notes 234 - sin θ = r 3 cos 2 θ sin...

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222 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION and so lim ( x,y ) −→ 0 x 3 x 2 + y 2 = 0 . How does this play out for our earlier examples? To see what happens in the Frst example above f ( x,y ) = xy x 2 + y 2 , ( x,y ) n = (0 , 0) write it in terms of polar coordinates: f ( r cos θ,r sin θ ) = ( r cos θ )( r sin θ ) r 2 cos 2 θ + r 2 sin 2 θ = r 2 cos θ sin θ r 2 = cos θ sin θ. We see that the limiting behavior of the function very much depends on the behavior of θ ; thus the function diverges as ( x,y ) (0 , 0). This trick is not universally useful. The second example f ( x,y ) = x 2 y x 4 + y 2 is expressed in polar coordinates by f ( r cos θ,r
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Unformatted text preview: sin θ ) = r 3 cos 2 θ sin θ r 4 cos 4 θ + r 2 sin 2 θ where things don’t cancel quite so nicely. We can try to pull out an r 2 factor, to get = r (cos 2 θ sin θ ) r 2 cos 4 θ + sin 2 θ . If sin θ stays bounded away from zero, then this goes to zero as r → (why?), but it is less clear what happens when r and sin θ both tend to zero. Recall that a function is continuous at a point x in its domain if lim x → x f ( x ) = f ( x ) ....
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