# Gradient and How to Calculate the Directional Derivative

Level 3 (with higher mathematics)
Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.
Updated by Alexander Fufaev on

The gradient of a scalar function $$f$$ is a multidimensional derivative of the function $$f$$. In a three-dimensional case, you get the gradient by applying the Nabla operator $$\nabla$$ to the scalar function $$f$$. You will understand what is meant by this after the lesson.

## Necessary ingredient: scalar function

A three-dimensional scalar function $$f$$ takes three arguments $$x$$, $$y$$, and $$z$$ and spits out a number (a scalar).

The three variables $$x,y,z$$ are usually spatial coordinates in physics. In general, they do not have to be space coordinates, they can also represent other quantities, such as radial distance and two angles: $$r,\theta,\varphi$$. We assume here for the explanation of the gradient that $$x,y,z$$ represent spatial coordinates.

The function $$f(x,y,z)$$ assigns a number to each point $$(x,y,z)$$ in space. For example, $$f(x,y,z)$$ could be a temperature function $$T(x,y,z)$$ that assigns a temperature $$T(x,y,z)$$ to each position in space.

To better illustrate a scalar function $$f$$, we assume that it depends on two spatial coordinates $$(x,y)$$, that is, it is two-dimensional. Then you can think of $$f(x,y)$$ as a curved surface with mountains and valleys (sort of a landscape).

You are at some location $$(x,y)$$ on this landscape (function) and want to know what the slope would be if you were walking in, say, the $$x$$ direction. The slope in $$x$$-direction is the derivative of the function with respect to $$x$$:

Slope of the function in x-direction
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How do we find the slope of the landscape $$f(x,y)$$ if we go in the $$y$$ direction? We differentiate $$f(x,y)$$ with respect to $$y$$ and get the slope in $$y$$-direction:

Slope of the function in y-direction
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Depending on the location $$(x,y)$$ we are at, the slopes 2 and 3 are of course different:

• On the mountain, the $$x$$ and $$y$$ directions tend to be downhill - that is, both slopes are negative and rather large.

• In a valley it goes uphill - that is, both slopes are positive and rather large.

• And in a plateau, the landscape is flat - that is, the slopes are zero in both directions.

What if we want to consider not only the slope in $$x$$ direction, but also in $$y$$ direction? Then we have to consider two derivatives of the function $$f(x,y)$$:

Derivatives of a scalar function with respect to x and y
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We can consider the derivatives separately or we can combine them to create a new mathematical object that is indispensable in physics and mathematics, namely the gradient in more than one dimension. In our case here: In two dimensions.

To do this, we write down the two derivatives 9 in a column vector:

Derivatives of the function in a column vector
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To keep 10 a bit more compact, we omit $$(x,y)$$ dependence, but keep in mind that $$f$$ may still depend on $$x$$ and $$y$$.

By writing the derivative in a column, we thereby obtained a vector function (also called vector field). This vector has a magnitude and a direction. But before we look at which direction the two-dimensional gradient 10 points, let's rewrite it a bit. To do this, we pull the function $$f$$ out of the vector:

Vector function with pulled out function
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The "vector" in Eq. 11 with derivatives is a so-called operator. Stand-alone derivatives, of course, make not much sense. An operator takes effect only when it is applied to a function, as in this case to the scalar function $$f$$. This operator in Eq. 11 is called the Nabla operator $$\nabla$$:

Nabla operator in two dimensions
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Thus, we can also write the two-dimensional gradient 11 as follows:

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Since gradient 13 is a vector (more precisely: a vector field), it is also called gradient vector or gradient field.

In physics, which describes our three-dimensional world, the gradient is usually also three-dimensional and looks like this:

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Here we have simply added the $$z$$ dependence to the $$f$$ function and added the derivative of $$f$$ to $$z$$ as the third component in the gradient.

## Gradient points to the steepest ascent

To understand why the gradient vector $$\nabla f$$ points in the direction of the steepest ascent, we exploit the so-called directional derivative. For this we take a unit vector $$\boldsymbol{v}$$ pointing in any direction. The only important thing is that it is a unit vector, that is normalized.

Remember that the magnitude $$|\nabla f|$$ is a constant that we cannot change. Why not? Because $$f$$ is fixed. As I said before, you can think of the $$f$$ function as a landscape, with rigid hills and valleys. On this surface there is the unit vector $$\boldsymbol{v}$$ whose direction we can change and thus scan the landscape $$f$$.

The slope of the shadow of $$\boldsymbol{v}$$ on the plane, we get with the help of the following scalar product:

Directional derivative
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The scalar product 19 is the directional derivative of the function $$f$$ in the direction of $$\boldsymbol{ v }$$.

The result of this scalar product is a pure number, namely the slope in the direction of $$\boldsymbol{ v }$$. For example, if you choose $$\boldsymbol{ v }$$ as the unit vector in the $$x$$ direction: $$\boldsymbol{ v } = \boldsymbol{\hat e}_{\text x}$$, then the scalar product 19 gives the slope in $$x$$ direction.

The question is, why does 19 indicate the steepest ascent? We can try around and use different unit vectors $$\boldsymbol{ v }$$ in the scalar product 19. The steepest ascent has that unit vector $$\boldsymbol{ v }$$ which yields the largest scalar product 19.

Instead of laboriously inserting all conceivable unit vectors into 19, there is a more sophisticated way to find out the maximum slope in the direction of $$\boldsymbol{ v }$$. To do this, we need to rewrite the scalar product 19. To do this, we use the geometric definition of the scalar product:

Scalar product using an angle
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Thus we can write the directional derivative 19 as follows:

Direction derivation by using an angle
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Here the non-bold $$v$$ is the magnitude of the vector $$\boldsymbol{v}$$. Now it is much easier to determine the maximum slope in the direction of $$\boldsymbol{ v }$$. We have assumed that the vector $$\boldsymbol{v}$$ is normalized. If it is normalized, then it has the magnitude: $$v = 1$$. Thus Eq. 21 becomes:

Directional derivative with unit vector
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So far so good. The only way we can vary the slope is by the angle $$\theta$$ enclosed by the vectors $$\nabla f$$ and $$\boldsymbol{v}$$. The cosine in Eq. 22 has its largest value at $$\theta=0$$: $$\cos(0) ~=~ 1$$. So we set $$\theta=0$$ to get the largest slope:

Steepest ascent for directional discharge
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By setting $$\theta=0$$, we have aligned the vectors $$\nabla \, f$$ and $$\boldsymbol{v}$$ parallel to each other (see Illustration 5). That means: Now vector $$\boldsymbol{v}$$ points in the same direction as vector $$\nabla \, f$$. Also, we made the scalar product 21, that is, the slope, as maximum as possible. And as you see in 21: The maximum slope is the magnitude $$|\nabla f|$$ of the gradient vector. Consequently, the gradient vector $$\nabla f$$ points in the direction of largest ascent!

## Calculate directional derivative in 4 steps

If you don't want to calculate the direction of the steepest ascent 23, but the ascent in any direction $$\boldsymbol{v}$$, then you have to do the following:

Directional derivative in any direction
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Here we no longer assume that the vector $$\boldsymbol{v}$$ is normalized. Therefore, we must always normalize it by dividing by the magnitude $$|\boldsymbol{v}| =: v$$ of the vector. Follow these 4 steps to calculate the directional derivative in any direction:

1. Calculate the gradient $$\nabla f$$ of a given scalar function $$f$$.

2. Normalize the vector $$\boldsymbol{ v }$$ by dividing it by its magnitude: $$\boldsymbol{ v } / v$$.

3. Form the scalar product between the gradient $$\nabla f$$ and the normalized vector $$\boldsymbol{v}/v$$.

4. Insert specific values for the variables $$x, y, z$$. By doing this you set the specific position $$(x, y, z)$$ where you want to calculate the slope in the direction of $$\boldsymbol{v}$$.