scalar field ϕ(x,y,z)=3x2y−y3z2
point M(1;−2;−1)
direction i+j+k with normalized vector n(31;31;31)
To find directional derivative we need to get scalar product grad(ϕ)⋅n
grad(ϕ)=(∂x∂ϕ;∂y∂ϕ;∂z∂ϕ) at point M
∂x∂ϕ=6xy=−12
∂y∂ϕ=3x2−3y2z2=−9
∂z∂ϕ=−2y3z=−16
Then grad(ϕ)=(−12;−9;16)
Directional derivative equal grad(ϕ)⋅n=−12⋅31−9⋅31−16⋅31=−337
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