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I seem to have some confusion concerning evaluating the gradient for a given point. The question is subtle so you have to follow the history. Lets assume we know the gradient definition and are given some point. Just as an example. Lets say f(x,y) = ln(xy^2. And the point we are interested in is (1,1,0). So now a standard question would be to determine the direction of the steepest ascent from this point and the maximum rate of increase. Easy enough. Calculate partial derivatives and evaluate at given point to get direction and then take the norm of the gradient to get the maximum rate of increase which is a scalar. So far so good. Initially When you calculate the gradient it does not give you a specific value but a vector with x and y components which form a gradient field. Correct? Then AFTER you evaluate it at the point ( 1,1 ) sure enough you get the direction of the maximum increase in steepness and the maximum value is the norm. First note: I don't see how the evaluation can be anything else because only ONE possibility exists at the point. The direction at that point can only be one vector and the norm can only be one value. Similarly if you define the directional derivative as the gradient doted ( that is the dot product ) with the unit vector u in direction u which equals the norm of the gradient times the cosine of the angle between the gradient and the direction of the unit vector u it is at maximum steepness when cos of the angle is 0. So the maximum steepness for the directional derivative is when u is in the same direction as the gradient. Again my confusion is that it can't be anything else for the gradient at the point. No other choice is available so why bother to calculate the directional derivative at all? The vector field has all the information you need. Subtle question but driving me crazy. The directional derivative appears to me to be useless. Thank you.

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  • $\begingroup$ I'm not sure I'm understanding the question. The directional derivative in a given direction at a point gives the instantaneous rate of change of the function at that point in that direction, and it's true that one of those directions will be the gradient itself. It's true that all the information you need is to calculate the directional derivative is contained in the gradient, but sometimes (in real-world applications, e.g.) you want to extract that information so that it can be put to use. Taking the dot product with the unit direction vector is the way of extracting that information. $\endgroup$ – Kaj Hansen Jul 31 '16 at 3:00
  • $\begingroup$ i accept your comment. Thank you $\endgroup$ – Sedumjoy Jul 31 '16 at 15:56
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To answer your question: Yes, the vector field contains the information but only in an implicit form.

The calculation of the directional derivative is the means by which the information is made explicit. Thus the directional derivative is very important.

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  • $\begingroup$ I accept the answer but the dot product is a cludge, ad hoc. Once you calculate the gradient which is done with partial derivatives there are NO degrees of freedom for the maximum and minimum slopes , they are what they are at the given point therefore knowing the cosine angle for the direction derivative is completely useless as I pointed out in my original question. It is what it is at that given point. $\endgroup$ – Sedumjoy Jul 31 '16 at 15:56

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