Question #144509

. If \\(θ(x,y,z)=3x^2 y-y^3 z^2\\),find \\( ˆ‡Î¸\\) (grad θ) at the point (1, -2, -1).

Expert's answer

f(x,y,z)=3x2y−y3z2,M(1,−2,−1)f(x, y, z)=3x^2y -y^3z^2, M(1, -2, -1)


∂f∂x=6xy,\dfrac{\partial f}{\partial x}=6xy,


∂f∂y=3x2−3y2z2,\dfrac{\partial f}{\partial y}=3x^2-3y^2z^2,

∂f∂z=−2y3z\dfrac{\partial f}{\partial z}=-2y^3z

∂f(M)∂x=6(1)(−2)=−12,\dfrac{\partial f(M)}{\partial x}=6(1)(-2)=-12,

∂f(M)∂y=3(1)2−3(−2)2(−1)2=−9,\dfrac{\partial f(M)}{\partial y}=3(1)^2-3(-2)^2(-1)^2=-9,

∂f(M)∂z=−2(−2)3(−1)=−16\dfrac{\partial f(M)}{\partial z}=-2(-2)^3(-1)=-16


gradf(M)=−12i⃗−9j⃗−16k⃗\text{grad}f(M)=-12\vec{i}-9\vec{j}-16\vec{k}


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