Find the gradient vector and its modulus of scalar function (x,y,z) = x6y4z0-z3yx3-89 at (3,-2,1)
Answer
Function
V=x6y4z0−z3yx3−89x^6y^4z^0-z^3yx^3-89x6y4z0−z3yx3−89
So
Gradient vector
∇V=(idVdx+jdVdy+kdVdz)\nabla V=(i\frac{dV}{dx}+j\frac{dV}{dy}+k\frac{dV}{dz})∇V=(idxdV+jdydV+kdzdV)
Putting value of function
∇V=i(6x5y4−3x2yz3)+j(4x6y3−z3x3)+k(−3z2yx3)\nabla V= i(6x^5y^4-3x^2yz^3) +j(4x^6y^3-z^3x^3) +k(-3z^2yx^3)∇V=i(6x5y4−3x2yz3)+j(4x6y3−z3x3)+k(−3z2yx3)
Now at point (3,-2,1)
∇V=23382i+23355j+162k\nabla V=23382i+23355j+162k∇V=23382i+23355j+162k
Now it's magnitude
∣∇V∣=(23382)2+(23355)2+(162)2)=33048.5|\nabla V|=\sqrt{(23382) ^2+(23355) ^2+(162) ^2) }\\=33048.5∣∇V∣=(23382)2+(23355)2+(162)2)=33048.5
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