微積分積分問題

Question

第一題
The function F(x,y)=4-x^2-y^2-xy-x has one critical point. Determine its location and type.
第二題
Let x+y+z -sin(xyz)=3. Use implicit differentiation to find dz/dy when (x,y,z)=(1,0,2).
第三題
若函數 f(x,y)=ln(3x-y),則 f (0.95,2.03)的近似值為何?
第四題
曲面 Z=ln(2x+y)在點(-1,3,0)的平面方程式

Answer 1

1. F(x,y)=4-x²-y²-xy-xFx(x,y)=-2x-y-1Fy(x,y)=-2y-x-2x-y-1=0-2y-x=0解得critical point為 (-2/3,1/3)又Fxx(x,y)=-2Fyy(x,y)=-2Fxy(x,y)=-1Fyx(x,y)=-1Fxx(-2/3,1/3)Fyy(-2/3,1/3)-Fxy(-2/3,1/3)Fyx(-2/3,1/3)=4-1=3>0且Fxx(-2/3,1/3)=-2<0故知在(-2/3,1/3)有相對極大值2. x+y+z-sin(xyz)=31+dz/dy-(cos(xyz))(xy(dz/dy)+xz)=01-xzcos(xyz)+(1-xycos(xyz))(dz/dy)=0dz/dy∣(x,y,z)=(1,0,2)= (-1+2cos0)/(1-0)= 13. f(x,y)=ln(3x-y) fx(x,y)=3/(3x-y)fy(x,y)=-1/(3x-y)f(0.95 , 2.03)= f(1-0.05 , 2+0.03)≈f(1,2)+(-0.05)fx(1,2)+(0.03)fy(1,2)= ln(3-2)-(0.05)(3)+(0.03)(-1)= 0-0.15-0.03= - 0.184. z=ln(2x+y)∂z/∂x=2/(2x+y)∂z/∂y=1/(2x+y)∂z/∂x∣x= -1,y=3 = 2∂z/∂y∣x= -1,y=3 = 1∴切平面方程式為 (2)(x+1)+(1)(y-3)-(z-0)=0即 2x+y-z-1=0