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Problem 1 Let f (x, y) = xy(x^2 − y^2 )/(x^2 + y^2 ), f (0, 0) = 0. (a) if (x, y) 6 = (0, 0) compute fx, fy. (b) find f (x, 0), f (0, y). (c) find fx(0, 0), fy (0, 0) (d) show that fx(0, y) = −y when y 6 = 0, fy (x, 0) = x when x 6 = 0. (e) show that fyx(0, 0) = 1, fxy (0, 0) = − 1. The partial derivative are not equal!! Why? Problem 2 Let f (x, y) = y^3 /(x^2 + y^2 ), f (0, 0) = 0. (a) Compute fx, fy , fx(0, 0), fy (0, 0). (b) Show that for any θ the directional derivative d dr f^ (r^ cos^ θ, r^ sin^ θ)|r= exists. (c) Show that the directional derivatives are not all given by dotting the direction vector with the gradient vector. (Why does this not contradict the chain rule?)
Solution of Problem 1. (a) fx = y^
(x (^2) − y 2 ) x^2 + y^2 + 2^
x^2 y x^2 + y^2 −^2
x^2 y (x^2 − y^2 ) (x^2 + y^2 )^2
fy = x^
(x (^2) − y 2 ) x^2 + y^2 −^2
xy^2 x^2 + y^2 −^2
xy^2 (x^2 − y^2 ) (x^2 + y^2 )^2 (b) f (x, 0) = 0, f (0, y) = 0. (c) fx(0, 0) = lim h→ 0 f^ (0 +^ h,^ 0) h^ −^ f^ (0,^ 0)= lim h→ 00 − h 0 = 0 fy (0, 0) = lim h→ 0 f^ (0,^ 0 +^ h h)^ −^ f^ (0,^ 0)= lim h→ 00 − h 0 = 0
(d) Just set x = 0 in the expression for fx computed in (a), you find fx(0, y) = −y^3 /y^2 = −y for y 6 = 0. Similarly if you set y = 0 in fy you find fy = x^3 /x^2 = x. (e) fyx(0, 0) = lim h→ 0 fy^ (0 +^ h,^ 0) h^ −^ fy^ (0,^ 0)= lim h→ 0 h^ − h 0 = 1 and fxy (0, 0) = lim h→ 0 fx(0,^ 0 +^ h h)^ −^ fx(0,^ 0)= lim h→ 0 −h h^ − 0 = − 1. The mixed partial derivatives are different because fx and fy are not continuous at (0, 0). The graph of f has a ”crinkle”. Solution of Problem 2. (a) fx = − 2 y (^3) x (x^2 + y^2 )^2 and fy = 3 y 2 x^2 + y^2 −^2
y^4 (x^2 + y^2 )^2 Moreover f (x, 0) = 0 for x 6 = 0, f (0, y) = y, for y 6 = 0. Therefore fx(0, 0) = lim h→ 0 f^ (0 +^ h,^ 0) h^ −^ f^ (0,^ 0)= lim h→ 00 − h 0 = 0 and fy (0, 0) = lim h→ 0 f^ (0,^ 0 +^ h h)^ −^ f^ (0,^ 0)= lim h→ 0 h^ − h 0 = 1
(b) Normally, to show that the directional derivative exists in any direction, one can to show that f is differen- tiable at (0, 0) (and use Theorem 3 pag 978). This is done checking that fx and fy are continuous at (0, 0). However ,in this problem, the function is not differentiable (therefore fx, fy need not be continuous), but the directional derivatives in any direction exist and one can compute them. Since f (r cos θ, r sin θ) = r sin^3 θ then (^) d dr f^ (r^ cos^ θ, r^ sin^ θ)|r=0^ = sin
(^3) θ.
