kaeisky9u

2022-08-13

Let $g:{\mathbb{R}}^{\mathbb{2}}\to \mathbb{R}$ be totally differentiable in the point $a\in {\mathbb{R}}^{\mathbb{2}}$

Let ${v}_{1}:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${v}_{2}:=\frac{1}{5}\left(\begin{array}{c}4\\ -3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$

It is $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{1}}(a)=5\sqrt{2}$ and $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{2}}(a)=1$

With the info given above, how can one find out $\mathrm{\nabla}g(a)$?

Let ${v}_{1}:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${v}_{2}:=\frac{1}{5}\left(\begin{array}{c}4\\ -3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$

It is $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{1}}(a)=5\sqrt{2}$ and $\frac{\mathrm{\partial}g}{\mathrm{\partial}{v}_{2}}(a)=1$

With the info given above, how can one find out $\mathrm{\nabla}g(a)$?

Gauge Howard

Beginner2022-08-14Added 19 answers

Given that ${v}_{1}$ and ${v}_{2}$ are unit vectors, the the direction derivative gives

${\mathrm{\nabla}}_{{v}_{2}}g=\mathrm{\nabla}g\cdot {v}_{2}$

and

${\mathrm{\nabla}}_{{v}_{2}}g=\mathrm{\nabla}g\cdot {v}_{2}$

Use these to build a linear system of 2 equations in the 2 components of $\mathrm{\nabla}g$, and solve. So, if

$\mathrm{\nabla}g=[x,y{]}^{T}$, then

$\frac{1}{\sqrt{2}}}x+{\displaystyle \frac{1}{\sqrt{2}}}y=5\sqrt{2$

and

$\frac{4}{5}}x-{\displaystyle \frac{3}{5}}y=1$

After multiplying the first equation through by $\sqrt{2}$ and second equation by 5, they becomes

$x+y=10$

$4x-3y=5$

It is trivial to solve to get x=5 , y=5

Therefore, $\mathrm{\nabla}g=\left[\begin{array}{c}5\\ 5\end{array}\right]$

${\mathrm{\nabla}}_{{v}_{2}}g=\mathrm{\nabla}g\cdot {v}_{2}$

and

${\mathrm{\nabla}}_{{v}_{2}}g=\mathrm{\nabla}g\cdot {v}_{2}$

Use these to build a linear system of 2 equations in the 2 components of $\mathrm{\nabla}g$, and solve. So, if

$\mathrm{\nabla}g=[x,y{]}^{T}$, then

$\frac{1}{\sqrt{2}}}x+{\displaystyle \frac{1}{\sqrt{2}}}y=5\sqrt{2$

and

$\frac{4}{5}}x-{\displaystyle \frac{3}{5}}y=1$

After multiplying the first equation through by $\sqrt{2}$ and second equation by 5, they becomes

$x+y=10$

$4x-3y=5$

It is trivial to solve to get x=5 , y=5

Therefore, $\mathrm{\nabla}g=\left[\begin{array}{c}5\\ 5\end{array}\right]$

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