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Why is the direction of gradient taken as the the direction of steepest ascent ,is there any easy explanation for this,I can't wrap my head around this idea inspite of referring many sources.

Harry Case
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  • Also see this – FShrike Jul 19 '21 at 14:19
  • @Don Thousand ,but does the unit vector always have to be along the gradient vector? – Harry Case Jul 19 '21 at 16:50
  • The unit vector in the other thread is arbitrary. It can point in any direction. But the directional derivative in that direction is maximal when it parallel to the gradient. – Paul Sinclair Jul 20 '21 at 02:25
  • @Paul Sinclair,but doesn't maximum here refer to what percentage of ( here 100%) the gradient vector is along the unit vector,(ie)how much of the gradient vector is along the unit vector ,if the dot product is max ,then it means that the gradient is along the arbitrary vector,it doesn't mean that value of gradient is maximum. – Harry Case Jul 20 '21 at 09:18
  • I don't know how to even interpret that question. The directional derivative of $f$ with respect to a unit vector $\vec u$ at some point $\mathbf p$ is defined by $$f_{\vec u}(\mathbf p) = \left.\frac{d}{dt}\right|{t=0} f(\mathbf p +t\vec u)$$ It is the rate of change of $f$ as you move through the point $\mathbf p$ in the direction of $\vec u$. Some playing with the definition will give you $$f{\vec u}(\mathbf p) = \nabla f(\mathbf p) \cdot \vec u$$ The largest this value gets is when the direction $\vec u$ in which you are travelling is parallel to $ \nabla f(\mathbf p)$. – Paul Sinclair Jul 20 '21 at 20:36

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