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Hi

I've searched Gradient is like derivative but just in high dimensions now referring to dimensions of gradient,it says gradient is a vector that is perpendicular to function curve. consider function $f$ in 2-dimension if $f(x)= x^2$ then $f'(x) = 2x$ and it means the slope of tangent line in $x$ so now if we want to interpret it by gradient what can we say ? or for finding local minimum or maximum we use $f'(x) = 0 $ but in higher dimensions we use $\nabla f = 0$ and in 2-dimension it's very easy to see $f'(x)=0$ get you local min/max but i can't interpret $\nabla f = 0$ Geometrically as a perpendicular vector for getting the local min/max of $f$ in n-dimension or even in 2 dimention.

I would appreciate for getting simple intuition about it.

Captain
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  • The geometric interpretation of the gradient of a function depends on whether you’re looking at a surface defined explicitly by something like $z=f(x,y)$ or implicitly by something like $F(x,y,z)=0$. See this question for more. – amd Nov 02 '17 at 23:41

2 Answers2

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I like to phrase it in a different way. I like to say that

the gradient is the vector that points in the direction of bigger growth of the function

In one dimension, the gradient (which is just the usual derivative) is either positive or negative. If it is positive, it means that if you are at $x $, then going right will make the value of the function increase. If the derivative is negative and you go left of $x $, the value of the function increases.

Say $f $ is a function of $x,y: f(x,y) $. If the derivative is $(0,1) $ in some point, it means that if you want to increase the value of the function as much as you can, you should go along the $y $-axis and up.

One verifies that this idea is the same as saying that the gradient is perpendicular to the level curves of a function. This doesn't make much sense if $f $ only depends on a single variable. But if you take $f(x,y) = x^2+y^2$ then the level curves are circumferences centered at $(0,0) $ and the gradient is always perpendicular to them: radial direction and pointing outwards.

Of course these interpretations are only true for a neighbourhood of the point in question, but I find them rather explanatory.

RGS
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The gradient at a given point is perpendicular to the level surface of the function. A level surface, or isosurface, is the set of all points a function has a given value. For example for given function $f:R^2 \rightarrow R$ where $f(x,y) = x^2 -y$ the level surface corresponding to zero includes all points such as $(0,0), (1,1), (2,4) , \dots$. This level set of function $f$ is match with the curve of the function $g: R \rightarrow R$ where $g(x)=x^2$. At the given point (1,1), the line $$y= g'(1)(x-1) + 1 = 2x -1 $$ is the tangenet line to this curve. Note that $\nabla f = (2x,-1)$. You can obeser easily that $\nabla f\vert_{(1,1)} = (2,-1)$ is perpendicualr to the tangenet line $y = 2x-1$.

Ehsan
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  • thank you for reply :) but if we want find min of $f(x,y) = 0$ then we find $\nabla f =0$ for $x$-axis therefore level curves are just on $x$-axis such as on $x=1$ the gradient on $x$-axis is $\nabla f =2$ and how this vector($\nabla f(1) =2$) is perpendicular to level curves on the $x$-axis? – Captain Nov 03 '17 at 08:36