VC3.45

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Note that the coordinates of the point whose position vector is 2i+3j-k\, are (2,3,-1).

Given that \phi(x,y,z)=x^2y^3z^4\,

Therefore,\nabla\phi=[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}](x^2y^3z^4)=2xy^3z^3i+3x^2y^2z^4j+4x^2y^3z^3k\,

=108i+108j-432k\, at the point (2,3,-1).

=108(i+j-4k)\,

The rate of change of phi is maximum in the direction of gradient of phi along the vector 108(i+j-4k)\,

Therefore the magnitude of the maximim rate of change of phi at the point 2i+3j-k\, =|\nabla\phi|\, at (2,3,-1).

=|108(i+j-4k)|=108\sqrt{1+1+4}=324\sqrt{2}\,

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