0

The picture attached shows two contour lines of a function $F(x,y)$. Now, the normal vector to the contour at point $A$ is the steepest ascent or decent ($\vec{ AC}$) for a fixed value of length $l$. Now to prove the $\vec{AC} = c \times \nabla F(x,y)$ at point $A$, where $c$ is any real constant. I need to prove $$\vec{AB} = c \times \frac{\partial F}{\partial {x}} (i) \\ \text{and} \\ \vec{BC} = c \times \frac{\partial F}{\partial {y}}\text (j) .$$

So how do I prove that? Please help me prove it using this method.

AMISH GUPTA
  • 1
  • 4
  • You might want to have a look at this question: https://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent?rq=1 – Epiousios Sep 12 '20 at 06:21
  • @Epiousios I have seen that already, thanks, but I needed help with figuring out this particular proof. – – AMISH GUPTA Sep 12 '20 at 23:07

0 Answers0