265-B2 Math Solutions to Homework #7: Sec. 15.3 #5: Set r = x2 + y 2 . Then sin r sin(x2 + y 2 ) 1 = lim lim = . r0 3r 3 (x,y)(0,0) 3x2 + 3y 2 2 x y Sec. 15.3 #17: f (x, y) = 4 x + y2 a.) Along every line through the origin y = mx, x2 (mx) mx3 mx x2 y = 4 = 4 = 2 . So along y = mx, 4 + y2 2 x x + (mx) x + m2 x2 x + m2 mx m(0) x2 y = lim 2 = 2 = 0. lim x0 x + m2 0 + m2 (x,y)(0,0) x4 + y 2 b.) Along the parabola y = x2 , x2 (x2 ) x4 1 x2 y = 4 = 4 = . So along y = x2 , 4 + y2 2 )2 4 x x + (x x +x 2 x2 y 1 1 lim = lim = . x0 2 2 (x,y)(0,0) x4 + y 2 c.) Since two different paths through the origin produce two different limits, x2 y we conclude that lim f (x, y) = lim does not exist. 4 + y2 (x,y)(0,0) (x,y)(0,0) x 2 x - 4y 2 if x = 2y Sec. 15.3 #25: f (x, y) = . x - 2y g(x) if x = 2y We want f to be continuous on all of R2 . For every (x, y) such that x = 2y, (x - 2y)(x + 2y) x x2 - 4y 2 = = x + 2y. If x = 2y, then y = , and x - 2y x - 2y 2 x x + 2y = x + 2 = 2x. So we want g(x) = 2x. 2 xy(x2 - y 2 ) if (x, y) = (0, 0) . Sec. 15.3 #32: f (x, y) = x2 + y 2 0 if (x, y) = (0, 0) f (h, y) - f (0, 0) a.) For every y, fx (0, y) = lim h0 h hy(h2 -y 2 ) -0 2 +y 2 y(h2 - y 2 ) y(0 - y 2 ) y3 h = lim = lim = = - 2 = -y. h0 h0 h2 + y 2 h 0 + y2 y Therefore, fx (0, y) = -y for all y. f (x, h) - f (0, 0) b.) For every x, fy (x, 0) = lim h0 h xh(x2 -h2 ) 2 -0 x(y - h2 ) x(x2 - 0) x3 2 2 = lim x +h = lim = = 2 = x. h0 h0 x2 + h2 h x2 + 0 x Therefore, fy (x, 0) = x for x.
1
fy all (h, 0) - fy (0, 0) . But according to part(b), fy (x, 0) = x h for all x. Thus, fy (h, 0) = h and fy (0, 0) = 0. Plugging these results in yield h-0 = lim 1 = 1. So fyx (0, 0) = 1. fyx (0, 0) = lim h0 h0 h fx (0, h) - fy (0, 0) d.) fxy (0, 0) = lim . But according to part(a), fx (0, y) = -y h0 h for all y. Thus, fx (0, h) = -h and fx (0, 0) = 0. Plugging these results in yield -h - 0 fxy (0, 0) = lim = lim -1 = -1. So fxy (0, 0) = -1. h0 h0 h c.) fyx (0, 0) = lim
h0
Sec. 15.4 #6: f (x, y) = sin3 (x2 y). Then fx (x, y) = 3 sin2 (x2 y) cos(x2 y)(2xy) = 6xy sin2 (x2 y) cos(x2 y), and fx (x, y) = 3 sin2 (x2 y) cos(x2 y)(x2 ) = 3x2 sin2 (x2 y) cos(x2 y). Therefore, f (x, y) = 3x sin2 (x2 y) cos(x2 y) 2y, x . Sec. 15.4 #13: f (x, y) = cos x sin y + sin 2y. So, fx (x, y) = - sin x sin y, and fy (x, y) = cos x cos y + 2 cos 2y. Then, fx (-1, 1/2) = - sin(-) sin(/2) ...