Question #136781

Find the linearizationL(x,y,z) of the function f(x,y,z) =sin(xy)/z at the point (pi/2,1,1).

Expert's answer

Given, f:R3→Rf:\mathbb{R}^3\rightarrow\mathbb{R} to be a real valued map (x,y,z)↦sin⁡(xy)z(x,y,z)\mapsto\frac{\sin(xy)}{z} , now we have to find Linearization of f(x,y,z)f(x,y,z) at (a,b,c)=(π/2,1,1)(a,b,c)=(\pi/2,1,1)


Since, from Taylor expansion we get


L(x,y,z)=f(a,b,c)+∑x,yz(x−a)fx(a,b,c)L(x,y,z)=f(a,b,c)+\sum_{x,yz}(x-a)f_x(a,b,c)

Now,

fx(x,yz)=yzcos⁡(xy)  ⟹  fx(π/2,1,1)=0fy(x,y,z)=xzcos⁡(xy)  ⟹  fy(π/2,1,1)=0fz(x,y,z)=−1z2sin⁡(xy)  ⟹  fz(π/2,1,1)=−1f_x(x,yz)=\frac{y}{z}\cos(xy)\implies f_x(\pi/2,1,1)=0\\ f_y(x,y,z)=\frac{x}{z}\cos(xy)\implies f_y(\pi/2,1,1)=0\\ f_z(x,y,z)=-\frac{1}{z^2}\sin(xy)\implies f_z(\pi/2,1,1)=-1

Thus,

L(x,y,z)=1+(x−π/2)⋅0+(y−1)⋅0+(z−1)⋅(−1)  ⟹  L(x,y,z)=−z+2L(x,y,z)=1+(x-\pi/2)\cdot 0+(y-1)\cdot 0+(z-1)\cdot(-1)\\ \implies L(x,y,z)=-z+2


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