中南大学：《大学数学》课程教学资源（微积分案例题解）练习一

练习13 四. 3.L=x = xInx.(y=x”mx孙2-1 =x"Ix:(y2)2= ylnx·y 四4.F(x,)≈ f(s)ds+e dx OF ax f∫(xy)·y OF f(xy)·x-∫(y) ay K心

练习1.3 四. 3. , z y u = x =   y u x x  z y ln y z ( y ) 1 ln − =  y z x x zy z =   y u x x  z y ln z z ( y ) x x y y y z z = ln  ln 四. 4. =  +  1 0 2 F(x, y) f (s)ds e dx xy x y f xy y x F =    ( ) f (xy) x f ( y) y F =  −  

五.2.=m/2」p2+22x+y2+x2) au 1 2x ax r teti x tz CFs ah 02u 2 +z-x·2xy2+z-x 2 (x2+y+z) (x- +y+z 22 x·2 2. andy (x+ 2 y+z)( 2、2 r t tz 02u 2Z 2xz axa (x y+z (x2+y2+x) K心

五. 2. 2 2 2 u = ln x + y + z ln( ) 2 1 2 2 2 = x + y + z 2 2 2 2 2 2 2 2 1 x y z x x y z x x u + + = + + =   2 2 2 2 2 2 2 2 2 ( ) 2 x y z x y z x x x u + + + + −  =   2 2 2 2 2 2 2 (x y z ) y z x + + + − = 2 2 2 2 2 ( ) 2 x y z x y x y u + + −  =    2 2 2 2 ( ) 2 x y z xy + + − = 2 2 2 2 2 ( ) 2 x y z x z x z u + + −  =    2 2 2 2 ( ) 2 x y z xz + + − =

练习14 2.∵山=P(x,y)dx+Q(x,y)dy, a f P(x,y)=a, o(x, y) af 故 aP 82f 8g a2f ay axay ax ayax OP 00 ay ax K心

练习1.4 一. 2. df = P(x, y)dx + Q(x, y)dy, ( , ) , xf P x y   = ( , ) , yf Q x y  = , 2x yf yP    =  故 , 2y xf xQ    =  . xQ yP  =  

练习15 1.取I={-1,0,0}, af af cos a+cos B+cosy al ax oz ax 七∫x(xo,n)c+fxn,yn)sia1=1 ∫x(x0,J0)c0s的2+∫(x0,Jo)sin62=0 ∫x(x0,y0)+ 2 2y(x0,)=1 ∫x(xo,y0)=3 ,(xn) 2 ∫x(x0,y)+。J (x0,y0)=0 2 grad(x0,y)=√3-j,∴ gradf(xo,y)=2即为所求 K心

练习1.5 一. 1. l = {−1,0,0},  取 cos cos  cos z f y f x f l f   +   +   =   而 . x f   = − 七. , ( , )cos ( , )sin 0 ( , )cos ( , )sin 1 0 0 2 0 0 2 0 0 1 0 0 1    + = + =     f x y f x y f x y f x y x y x y , ( , ) 0 2 3 ( , ) 2 1 ( , ) 1 2 1 ( , ) 2 3 0 0 0 0 0 0 0 0        + = + = f x y f x y f x y f x y x y x y , ( , ) 1 ( , ) 3 0 0 0 0    = − = f x y f x y y x ( , ) 3 , 0 0 gradf x y i j    = − ( , ) 2 .  gradf x0 y0 = 即为所求

练习16 一.2.z=∫( e sin y,"), az e sin y (f,2) f1· e sin y-f2 J a 2 一.5.z=-∫(xy)+yq(x+y), z ar +2ff(xy)+f(xy) y+ yo(x+y), 02z axa 2f() x+f(xy) xy+f(xy) +o(x+y)+yo(+y)=o+yo+f")

练习1.6 一. 2. ( sin , ), x y z f e y x =           − =   2 1 2 sin ( , ) x y e y f f x z x sin . 1 2 2 x y f e y f x =  −  一. 5. ( ) ( ), 1 f xy y x y x z = +  + ( ) ( ), 1 ( ) 1 2 f xy y y x y x f xy x x z = − +   +  +    ( ) 1 ( ) 1 ( ) 1 2 2 f xy x f xy xy x f xy x x y x z = −   +   +     +(x + y) + y(x + y) =  + y( + f )