Question about grad, nabla (vector operator)

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    Grad Nabla Operator
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SUMMARY

The discussion clarifies the distinction between the gradient (grad F) of a function and its relationship to contours and paths. Specifically, grad F along a path yields a vector tangent to that path, while grad F on a contour equation provides a vector normal to the contour. This is exemplified using the function y=x², where grad f(x) = 2x indicates the direction of increase along the path, and grad F(x,y) = (2x, -1) shows the direction of increase perpendicular to the contour defined by F(x,y) = x² - y = 0.

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aero&astro
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In my notes it says that grad F will give you a vector normal to the contour. Howver I thought grad F would give you a vector tangent because the path is aligned with the vector field. Is it different when talking about contours and paths?

If you find grad F of a function F does that give you a vector tangent or normal to the contour/path?
 
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Hi aero&astro! :smile:

Yep.
grad F on a path will give you a vector tangent to the path.
grad F on an equation will give you a normal to a contour satisfying the equation.

Actually, the gradient gives you the vector in which F increases.
A path "increases" in the direction of the path.
The equation-value of a contour increases perpendicular to the contour.

Consider for instance y=x2.
Its path is given by f(x) = x2 and its contour-equation is F(x,y) = x2 - y = 0.

grad f(x) =2x, which gives you the direction f(x) increases with increasing x.
grad F(x,y)= (2x, -1), which gives you the direction F(x,y) increases with changing (x,y).

In the first case you "walk the path".
In the second case you move to a higher level contour given by F(x,y)=c.
 
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