Let $V:R^{3}\rightarrow R$ be a differential function. Let $$A= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix}. $$
Now, restrict $V$ to the ellipsoid $\{x\in R^{3}: x^{T}Ax=1\}$. Compute the gradient of $V$ in the metric induced by $A$. What is the relation between the partial derivative $(\partial{V}/\partial{x},\partial{V}/\partial{y},\partial{V}/\partial{z})$ and the gradient?
Thank you.
The ellipsoid $${x^2} + 2{y^2} + 3{z^2} = 1$$ may be given a parametrized form like so: $$\psi (u,v) = \left( {\begin{array}{*{20}{c}} {x(u,v)} \\ {y(u,v)} \\ {z(u,v)} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 1 }}\cos (u)\cos (v)} \\ {\frac{1}{{\sqrt 2 }}\sin (u)\cos (v)} \\ {\frac{1}{{\sqrt 3 }}\sin (v)} \end{array}} \right)$$ The partials for this are given by $$\frac{{\partial \psi }}{{\partial u}} = \left( {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 1 }}\sin (u)\cos (v)} \\ {\frac{1}{{\sqrt 2 }}\cos (u)\cos (v)} \\ 0 \end{array}} \right),\frac{{\partial \psi }}{{\partial v}} = \left( {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 1 }}\cos (u)\sin (v)} \\ {\frac{{ - 1}}{{\sqrt 2 }}\sin (u)\sin (v)} \\ {\frac{1}{{\sqrt 3 }}\cos (v)} \end{array}} \right)$$ These two vectors are tangential to our ellipsoid. Now, the function $V$ restricted to the ellipsoid means: $$V(\psi (u,v)) = V(x(u,v),y(u,v),z(u,v))$$ Calculating the gradient for this restriction, we get: $$\begin{gathered} dV = \nabla V \cdot (dx,dy,dz) \hfill \\ = \frac{{\partial V}}{{\partial x}} \cdot (\frac{{\partial x}}{{\partial u}}du + \frac{{\partial x}}{{\partial v}}dv) + \frac{{\partial V}}{{\partial y}} \cdot (\frac{{\partial y}}{{\partial u}}du + \frac{{\partial y}}{{\partial v}}dv) + \frac{{\partial V}}{{\partial z}} \cdot (\frac{{\partial z}}{{\partial u}}du + \frac{{\partial z}}{{\partial v}}dv) \hfill \\ = (\frac{{\partial V}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial V}}{{\partial y}}\frac{{\partial y}}{{\partial u}} + \frac{{\partial V}}{{\partial z}}\frac{{\partial z}}{{\partial u}})du + (\frac{{\partial V}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial V}}{{\partial y}}\frac{{\partial y}}{{\partial v}} + \frac{{\partial V}}{{\partial z}}\frac{{\partial z}}{{\partial v}})dv \hfill \\ = \nabla V \cdot \frac{{\partial \psi }}{{\partial u}}du + \nabla V \cdot \frac{{\partial \psi }}{{\partial v}}dv = (\nabla V \cdot \frac{{\partial \psi }}{{\partial u}},\nabla V \cdot \frac{{\partial \psi }}{{\partial v}}) \cdot (du,dv) \hfill \\ \end{gathered}$$ That is:
$$\nabla (V \circ \psi )(u,v) = (\nabla V(\psi (u,v)) \cdot \frac{{\partial \psi }}{{\partial u}}(u,v),\nabla V(\psi (u,v)) \cdot \frac{{\partial \psi }}{{\partial v}}(u,v))$$
and therefore:
$$\nabla (V \circ \psi ) = \left( {\left( {\begin{array}{*{20}{c}} {\frac{{\partial V}}{{\partial x}}} \\ {\frac{{\partial V}}{{\partial y}}} \\ {\frac{{\partial V}}{{\partial z}}} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 1 }}\sin (u)\cos (v)} \\ {\frac{1}{{\sqrt 2 }}\cos (u)\cos (v)} \\ 0 \end{array}} \right),\left( {\begin{array}{*{20}{c}} {\frac{{\partial V}}{{\partial x}}} \\ {\frac{{\partial V}}{{\partial y}}} \\ {\frac{{\partial V}}{{\partial z}}} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 1 }}\cos (u)\sin (v)} \\ {\frac{{ - 1}}{{\sqrt 2 }}\sin (u)\sin (v)} \\ {\frac{1}{{\sqrt 3 }}\cos (v)} \end{array}} \right)} \right)$$