So I have always read that taking the gradient of a scalar function gives the direction of maximum rate of increase in that scalar field (for example Wikipedia). For the problem at hand I have a field defined numerically and I need to find the direction in which the rate of decrease of the field is maximum. I thought the gradient could just be the position vector connecting one point to another along that direction,

$$ (x-x_0)\hat{i}+(y-y_0)\hat{j} = \frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j} $$

but later realized that it is obviously wrong because the dimensions are incorrect. So, what exactly is the nature of the gradient vector? It is not simply just the direction. That definition is more suitable for the unit vector of the gradient, $$ \hat{n} = \frac{\vec{\nabla}f}{|\vec{\nabla} f|} $$

So the relationship between $\vec{\nabla}f$ and $\vec{r}$ must be that of a multiple i.e. $\vec{\nabla}f = c \vec{r}$. But, what is c?

So in the problem, starting at a point $(x_0, y_0)$, I am trying to determine the next point $(x,y)$ to move to in the direction of maximum decrease.

  • $\begingroup$ What is $\vec r?$ It appears out of nowhere. $\endgroup$ Jun 19, 2017 at 14:50
  • $\begingroup$ I am using it as standard notation for a position vector. $\endgroup$
    – Shaz
    Jun 19, 2017 at 15:01
  • $\begingroup$ It is not true that $\vec{\nabla}f=c\vec r$ then. I think you meant to say $\vec{\nabla}f=c\hat n$, and then $c$ is the magnitude of the gradient, which is the slope of $f$ along the gradient. $\endgroup$ Jun 19, 2017 at 15:04
  • $\begingroup$ My idea behind writing $\vec{\nabla}f=c\vec{r}$ was to say that the gradient at a certain point in the field is proportional to the length of the position vector between two points in that direction and the proportionality constant is then something related to the amount of change. So amount of change multiplied by the rate of change in a direction would give the position vector. But now I see that wont work because the function can change by the same amount in different directions. $\endgroup$
    – Shaz
    Jun 22, 2017 at 15:29

1 Answer 1


The magnitude of the gradient represents how fast the function changes along the gradient. The gradient vector is the first term in a Taylor expansion of $f$, so if we work around a point $\vec {x_0}$ and consider a nearby point $\vec x$ we have $f(\vec x) \approx f(\vec{x_0}) + (\vec x - \vec {x_0})\cdot \vec{\nabla} f(\vec{x_0})$. The direction of maximum decrease is opposite to the gradient because it makes the cosine of the angle in the dot product $-1$. There is no information here about how far you should go in that direction. That would come from a second derivative.

  • $\begingroup$ Yes I see that now. The change in the function in any direction needs to be specified as well. Thanks a lot. $\endgroup$
    – Shaz
    Jun 22, 2017 at 15:30

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