Why does the gradient vector point straight outward from a graph?

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SUMMARY

The gradient vector in a scalar field points outward from a graph or surface in three-dimensional space, forming a 90-degree angle with the surface at any given point. It measures the direction and rate of change of the scalar field, with its direction indicating the steepest ascent. The directional derivative, represented as \nabla f \cdot \vec{v}, confirms that when the function is constant on the surface (f(x,y,z) = 0), the gradient vector is perpendicular to the surface, as the dot product with any tangent vector results in zero.

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A gradient vector points out of a graph (or a surface in 3D case). Locally, it makes an angle of 90 degrees with the graph at a particular point. Why is that so?

Thanks.
 
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Hi.

The gradient vector measures the change and direction of a scalar field. The direction of the gradient is expressed in terms of unit vectors (in 3-dimensions, say) and points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change.
 
Another way of looking at it is that the "directional derivative", the rate of change of function f(x,y,z) as you move in the direction of unit vector [itex]\vec{v}[/itex], is given by [itex]\nabla f\cdot\vec{v}[/itex]. If the function is given implicitely by f(x,y,z)= 0 (or any constant, then on the surface f is a constant and so it derivative is 0 in any direction tangent to surface: the dot product of [itex]\nabla f\cdot \vec{v}[/itex], with [itex]\vec{v}[/itex] tangent to the surface, is 0 so [itex]\nabla f[/itex] is perpendicular to the surface.
 

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