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This is going to take a bit of explanation. I am not sure how to ask this question. The math lecture said the gradient points to the direction of greatest increase and if you want to know the direction of greatest decrease then you take the negative.

My intuition tells me that the gradient is either the direction of greatest increase or decrease depending on the function at the particular point you are evaluating. It could be pointing down or up since after all it is nothing more that the derivative of a function.

On the other hand what ever that turns out to be then if you apply the negative it will simply go in the opposite direction.

So I am at odds with the lecture here. By default my intuition tells me the gradient could turn out to be greatest increase or greatest decrease and which ever it is then the negative will reverse the direction.

Thank you

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Remember that the gradient exists in the domain. It does point in the direction of the greatest increase and its magnitude is the slope (rate of change) of the tangent line in that direction. To find the slope, aka directional derivative, in any direction, dot the gradient with a unit directional vector. If the directional vector is in the opposite direction of the gradient, then the dot product will equal the negative of the magnitude of the gradient.

Note that if (and only if) the direction is orthogonal to the (non-zero) gradient, the directional derivative is zero. This is why level curves/surfaces are always orthogonal to the gradient.

Hope this helps.

Ced

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    $\begingroup$ (+1) but maybe sketch why this is true - e.g., suppose $\gamma$ is a curve running through the domain, and $f$ the real-valued function the gradient of which we are considering. Then, by the chain rule, the derivative of $f\circ \gamma$ is given by the dot product of ... $\endgroup$ – peter a g Apr 28 '18 at 14:28
  • $\begingroup$ @peterag, The direction of a point traveling along $ \gamma $ will be given by the velocity vector. The rate of rise or fall of the point on $ f $ will be proportional to the speed along $ \gamma $. So if $ \gamma = \gamma(t) $: $$ \frac{ d( f\circ \gamma) }{dt} = \vec{\nabla} f \cdot \frac{ d \gamma }{dt} $$ Conceptually it can be expressed as: $$ \frac{ d(f\circ \gamma) }{dt} = \frac{ df }{\vec{dr}} \cdot \frac{ \vec{dr} }{dt} $$ Where $ \vec r $ is the position of the point. $\endgroup$ – Cedron Dawg Apr 28 '18 at 14:54
  • $\begingroup$ Please allow me to Use this example please ! .... f(x,y) = - x^2 - y^2. Graph it. A bowl . upside down. Now take the gradient . ...a vector < -2x, -2y> , this will generate a vector field . Now use the point (1,1) from the original function domain to insert in the gradient field , we get <-2 , -2> . This vector points DOWN not up . that is the direction of the greatest decrease in slope not increase. Can you show me where my logic is failing ? $\endgroup$ – Sedumjoy Apr 28 '18 at 17:14
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    $\begingroup$ @Sedumjoy, Sure, from point (1,1) the vector (-2,-2) is pointing towards the origin, which is the peak of your upside down bowl. By radial symmetry, the gradient is pointing in the most uphill direction. Up or down is determined by the change in the value of the function, not the direction of the gradient within the domain. $\endgroup$ – Cedron Dawg Apr 28 '18 at 17:22
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    $\begingroup$ @Sedumjoy, You got it. If I am heading south it doesn't mean I am going downhill, though sometimes it means that. ;-) The radial symmetry refers to the function you gave. It is an upside down parabola spun around the z-axis. It's the same on every radius. $\endgroup$ – Cedron Dawg Apr 28 '18 at 19:01

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