求解 微積分

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圖片參考：https://s.yimg.com/rk/HA05107138/o/1652520232.png

7.y(x)=x^2*ln(x^2)=2x^2*ln(x)(7.1) 證明limit(x->0){x*ln(x)}=0Set y=ln(x) => e^y=xlimit(x->0){x*ln(x)}=limit(y->∞){1/(y*e^y)}=e^(-y)/y=[e^(-y)]'/y'......羅必達法則=-e^(-y)=-1/e^y=-/∞=0
(7.2) y'(0)=?y'(x)=2x^2/x+4x*ln(x)=2x+4x*ln(x)y'(0)=2*0+4*limit(x->0){x*ln(x)}=0+0=0
(7.3) y"(0)=?y"(x)=2+4*ln(x)+4x/x=6+4*ln(x)y"(0)=6-4*∞=-∞.....Note
Note: y=ln(x) => e^y=x=0 => y=-∞
8.F(x,y)=x^2+y; 1<=x^2+y^2=r^2<=2(8a) x,y範圍cosQ<=x=√2*cosQ, -√2*cosQ<=x=-cosQsinQ<=y=√2*sinQ, -√2*sinQ<=y=-sinQ
(8b) 極大值方程式Fm(Q)=2*cos^2Q+√2*sinQ Fm'(Q)=-4*cosQ*sinQ+√2*cosQ=cosQ*(√2-4sinQ)=0=> sinQ=√2/4, cosQ=√14/4, Q=20.7(deg)Fm"(Q)=-4cos(2Q)-√2sinQ < 0 => max存在max=Fm(Q)=2*14/16+√2*√2/4=7/4+1/2=9/4......ans
(8c) 極小值方程式Fn(Q)=cos^2Q+√2*sinQFn'(Q)=-2cosQ*sinQ+√2*cosQ=cosQ*(-2sinQ+√2)=0=> cosQ=0 => Q=-90(deg)Fn"(Q)=-2cos(2Q)-√2*sinQFn"(-90)=2+√2 > 0 => min存在min=Fn(-90)=[cos(-90)]^2-√2*sin90=0-√2=-√2.......ans (3d) w=∫∫{18-2(x^2+y^2)}dA; 0<=x^2+y^2=r^2<4 =∫∫(18-2*r^2)rdr*dQ=∫∫(18r-2r^3)dr*dQ=∫(9r^2-2r^4/4)dQ=2πr^2*(9-r^2/2)......r=0~2=2*4π(9-2)=56π=> 0<= w <56π


2014-07-05 11:23:57 補充：
=-/∞ 改為 =-1/∞

2014-07-05 11:24:46 補充：
=-/∞ 改為 =-1/∞

2014-07-05 12:15:30 補充：
修改: 取消極小值方程式

直接由Fm'(Q)=-4*cosQ*sinQ+√2*cosQ

=cosQ*(√2-4sinQ)

=0

cosQ=0 => Q=-90(deg)

Fm"(-90)=-4cos(2Q)-√2sinQ

=-4cos(-180)-√2*sin(-90)

=4+√2>0 => min存在


min=Fm(-90)

=2*cos^2Q+√2*sinQ

=2*[cos(-90)]^2+√2*sin(-90)

=0-√2

=-√2......ans