1) If a scalar field φ(x,y,z)=3x2y-y3z2, calculate the gradient Ñφ at point M(1,-2,1).
If we know the scalar field is "\u03c6(x,y,z)=3x^2y-y^3z^2", we only have to find the first partial derivatives to have the gradient of the function that describes the scalar field:
"\\bigtriangledown \u03c6(x,y,z)= \\begin{pmatrix}\n \\frac{\\partial \u03c6(x,y,z)}{\\partial x}, & \\frac{\\partial \u03c6(x,y,z)}{\\partial y}, & \\frac{\\partial \u03c6(x,y,z)}{\\partial z}\n\\end{pmatrix} = \\begin{pmatrix}\n 6xy, & 3(x^2-y^2z^2) , & -2y^3z\n\\end{pmatrix}"
The gradient can be calculated by substituting (x, y, z) = (1, -2, 1) and we find
"\\bigtriangledown \u03c6(1,-2,1)= \\begin{pmatrix}\n 6(1)(-2), & 3((1)^2-(-2)^2(1)^2) , & -2(-2)^3(1)\n\\end{pmatrix}"
"\\implies \\bigtriangledown \u03c6(1,-2,1)= \\begin{pmatrix}\n -12, & -9 , & 16\n\\end{pmatrix}" .
In conclusion, at the point (1, -2, 1) we find the gradient▽φ(1, -2, 1) to be (-12, -9, 16).
Reference:
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