Question

求速度做！！！！高等数学

Answer 1

1. ∂z/∂x=f1· 2x+f2· y=2xf1+yf2

   ∂2z∂x2=2[f1+x(f11· 2x+f12· y)+y(f21· 2x+f22· y)=4x^2f11+4xyf12+y^2f22+2f1

   ∂2z∂x∂y=2x(f11· 2y+f12· x)+f2+y(f21· 2y+f22· x)=4xyf11+2(x^2+y^2)f12+xyf22+f2

2. 原式=∫(0,2π)dθ∫(π,2π)sinrrdr=2π(-rcosr+sinr)|(π,2π)=2π(-2πcos2π+πcosπ)=-6π^2

3. x'=4   y'=-2t    z'=(2/3)t

4. 4×1+(-2t)×2+(2/3)t×1=0      t=-6/5

5. 故切线方程为：[x+(24/5)]/4=[y+(36/25)]/(12/5)=[z-(12/25)]/(-4/5)