test the continuity of the following functions: f(x,y)= {xy/(x^2 + xy + y^2, (x,
ID: 3215048 • Letter: T
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test the continuity of the following functions: f(x,y)= {xy/(x^2 + xy + y^2, (x,y) not equal (0,0) ; 0,(x,y)=(0,0) --------------------------------------------- calculus @T1 3(k)Explanation / Answer
Given f(x,y) = xy(x² - y²)/(x² + y²) = (x³y - xy³)/(x² + y²) if (x,y) ? (0,0), and f(0, 0) = 0: a) By using the Quotient Rule (for (x,y) ? (0,0)): ?f/?x = [(3x²y - y³)(x² + y²) - (x³y - xy³)(2x)] / (x² + y²)² = (x4y + 4x²y² - y5) / (x² + y²)². Similarly, ?f/?y = (x5 - xy4 - 4x²y²) / (x² + y²)². Next, we use the definition of a partial derivative to compute them at (0, 0). ?f(0,0)/?x = lim(h-->0) [f(h, 0) - f(0, 0)] / h = lim(h-->0) (0 - 0) / h = 0. ?f(0,0)/?y = lim(k-->0) [f(0, k) - f(0, 0)] / k = lim(k-->0) (0 - 0) / k = 0. --------------------- Now, we check whether the first order partials are continuous at (0, 0). This comes down to checking whether lim((x,y)-->(0,0)) (x4y + 4x²y² - y5) / (x² + y²)² = 0, and lim((x,y)-->(0,0)) (x5 - xy4 - 4x²y²) / (x² + y²)² = 0. Letting y = x in the first limit before letting x --> 0, we obtain lim((x,y)-->(0,0)) (x4x + 4x²x² - x5) / (x² + x²)² = lim((x,y)-->(0,0)) 4x4 / (4x4) = 1. Thus, ?f/?x (and similarly ?f/?y) is discontinuous at the origin. --------------------------------------… b) Again, use the definition of the partial derivative. ?²f(0,0)/?y?x = lim(k-->0) [?f(0, k)/?x - ?f(0, 0)/?x] / k = = lim(k-->0) (-k - 0) / k = -1. ?²f(0,0)/?x?y = lim(h-->0) [?f(h, 0)/?y - ?f(0, 0)/?y] / k = = lim(h-->0) (h - 0) / h = 1. Note that the mixed partials are not equal at (0, 0). -------------------- c) The result in (b) is allowed to happen, because all of the second partial derivatives are NOT continuous at the origin. One may check that away from the origin, we have ?²f/?x?y = (x6 + 9x4y² - 9x²y4 - y6) / (x² + y²)³. Letting x = 0 and y--> 0, we get lim(y-->0) (-y6) / (y²)³ = -1. But, letting y = 0 and x--> 0, we get lim(x-->0) (x6) / (x²)³ = 1. Thus, ?²f/?x?y is discontinuous at the origin. So, Clairaut's Theorem does not apply. ------------------------------------- d) Suppose that fxy(a,b) ? fyx(a,b) for some (a,b) in R². Without loss of generality, suppose that fxy(a,b) > fyx(a,b). (The other case is done identically.) Since both fxy and fyx are continuous at (a,b), we can find a square S around the point so that fxy - fyx > 0 for all points in S. However, then we have ??s (fxy - fyx) dA > 0. This, however, contradicts Fubini's Theorem. With the square described by [-c, c] x [-c, c]: ??s fxy(x, y) dA = ?(-c to c) ?(-c to c) fxy(x, y) dy dx = ?(-c to c) [fx(x, c) - f(x, -c)] dx = [f(c, c) - f(c, -c)] - [f(-c, c) - f(-c, -c)] = f(c, c) - f(c, -c) - f(-c, c) + f(-c, -c). Similarly, ??s fyx(x, y) dA = f(c, c) - f(c, -c) - f(-c, c) + f(-c, -c). Hence, we must have fxy = fyx. ---------------------- I hope this helps!
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