f-x-y-ax-2-by-2-How-do-you-find-where-the-gradient-is-zero-for-multivariable-funtions-

Question Number 7040 by FilupSmith last updated on 07/Aug/16

f (x, y) = {a x}^{2} + {b y}^{2}

How do you find where the gradient

is zero for multivariable funtions ?

Commented by Yozzii last updated on 07/Aug/16

▽ f = g r a d f = (\begin{matrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{matrix})

A t s t a t i o n a r y p o i n t s, ▽ f = \underline{0}

∴ \frac{\partial f}{\partial x} = 0 a n d \frac{\partial f}{\partial y} = 0 s i m u l t a n e o u s l y a t s o m e (x, y)

F o r f (x, y) = {a x}^{2} + {b y}^{2}, ▽ f = \underline{0} a t x = y = 0 .

T h e r e i s a l s o t h e d i r e c t i o n a l d e r i v a t i v e o f f

\hat{u} ∙ (▽ f) a t a p o i n t (x, y) i n t h e d i r e c t i o n

o f t h e u n i t v e c t o r \hat{u} i n t h e x - y p l a n e .

\hat{u} ∙ (\begin{matrix} \partial f / \partial x \\ \partial f / \partial y \end{matrix}) g i v e s t h e g r a d i e n t i o f

t h e s u r f a c e o f f i n t h e d i r e c t i o n o f \hat{u} .

Commented by FilupSmith last updated on 07/Aug/16

This is extremely interesting!

I^{'} ll have to learn this area of mathematics!