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Problem: If $f(x,y)=4-x^{2}-y^{2}$, then the vector $(2,2)$ is orthogonal to the level curve 2 of $f(x,y)$ in the point $(1,1)$.

Solution:

$z=4-x^{2}-y^{2}$

$=\left \langle z=2 \right \rangle$

$2=4-x^{2}-y^{2}$

$=\left \langle \text{Subtract}\ -4\right \rangle$

$-2=-x^{2}-y^{2}$

$=\left \langle \text{Multiply by}\ -1\ \right \rangle$

$2=x^{2}+y^{2}$

$=\left \langle \text{Divide by}\ 2\right \rangle$

$1=\frac{x^{2}}{2}+\frac{y^{2}}{2}$

So $a=b=\sqrt{2}\approx1.4$

Then I plotted the level curve as follows:

1=x^2/2+y^{2}/2

Graphically we see that the vector $(2,2)$ is orthogonal to the level curve of 2 at point $(1,1)$.

But... how do I make this analytically?

InfZero
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    There is a theorem that says that the gradient (and thus also any of its multiples) of a function at a point is perpendicular to the level set (curve in the 2D case). Try to use that theorem. – Hans Engler Sep 14 '16 at 19:32

1 Answers1

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The gradient of your function is orthogonal to level sets $$ \nabla F(x,y)=(-2x,-2y) $$ Evaluating the gradient at the given point $(2,2)$ yields $$ \nabla F(x,y)=(-2,-2) $$ Which is parallel to the given vector $(2,2)$, therefore your vector is perpendicular.

edit: You can see quickly that the gradient orthogonal to level curves using the differential operator/gradient operator. Let $f(x,y)=c$ for some regular value $c$, then $$ df(x,y)\vert_{\vec{p}}=dc=0\Rightarrow (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})\cdot (p_1,p_2)=0 $$ for points along $\vec{p}=(p_1,p_2)$ along the level curve and since $c$ is constant.

operatorerror
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  • I also have found that $(2,2)=-\nabla F(1,1)$ and the point $(1,1)$ belongs to the level curve $x^{2}+y^{2}=2$. Thanks. – InfZero Sep 14 '16 at 19:41
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    sure thing. Is your issue with it you are not convinced that the gradient is perpindicular to the level sets? I would be happy to provide a proof – operatorerror Sep 14 '16 at 19:44