Engineering Calculus Notes 268

# Engineering Calculus Notes 268 - to get 2 x dx dy 8 y = 0...

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256 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION (2 , 1) (2 2 , 0) Figure 3.2: The Curve x 2 + 4 y 2 = 8 which we can solve for dy/dx : dy dx = 4 8 = 1 2 . However, the process can break down: at the point (2 2 , 0), substitution into ( 3.13 ) yields 4 2 + 0 dy dx = 0 which has no solutions. Of course, here we can instead di±erentiate ( 3.12 ) treating x as a function of y
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Unformatted text preview: , to get 2 x dx dy + 8 y = 0 . Upon substituting x = 2 √ 2, y = 0, this yields 4 √ 2 dx dy + 0 = 0 which does have a solution, dx dy = 0 . In this case, we can see the reason for our di²culty by explicitly solving the original equation ( 3.12 ) for y in terms of x : near (2 , − 1), y can be...
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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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