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## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 9: Divergence

# Divergence notation

Learn how divergence is expressed using the same upsidedown triangle symbols that the gradient uses. Created by Grant Sanderson.

## Want to join the conversation?

• Isn't it strange that the formula for divergence is the same as the formula for the directional derivative?
I can get the reason behind the formula of each one independently, but I can't make the connection why the two different ideas (divergence-directional derivative) have exactly the same formulas! •   They have different formulas:
The divergence formula is ∇⋅v (where v is any vector).

The directional derivative is a different thing. For directional derivative problems, you want to find the derivative of a function F(x,y) in the direction of a vector u at a particular point (x,y). It can be any number of dimensions but I'm keeping it x,y for simplicity. The directional derivative is the -dot product- of the GRADIENT of F with the UNIT VECTOR of u: ∇F(x,y) ⋅ u(unit).

I always believe that working through actual problems is the best way to understand something, so for the sake of your understanding and my own, let's do an example:

Suppose F(x,y) = x^2 * y. What is the derivative of F(x,y) in the direction of (1,2) at the point (3,2)?
Firstly we have to calculate the gradient of F(x,y). The gradient of F(x,y) is basically the vector field of all of F(x,y)'s partial derivatives: (∂F/∂x, ∂F/∂y). ∂F/∂x = 2xy. ∂F/∂y = x^2.
Since we want to evaluate the directional derivative at point (3,2): ∂F/∂x = 2(3)(2) = 12 and ∂F/∂y = (3)^2 = 9. Thus ∇F(x,y) = (12,9)
Now, we have to evaluate the derivative at the direction of (1,2), so let's assign vector u the direction (1,2). The unit vector of u is:
u/|u| = (1,2)/√ (1^2 + 2^2) = (1,2)/√ 5 = (1/√ 5 , 2/√ 5 ).
The directional derivative is :
∇F(x,y) ⋅ u(unit) = (12,9) ⋅ (1/√ 5 , 2/√ 5 ) = 12/√ 5 + 18/√ 5 = 30/√ 5

Calculating divergence is much simpler:
If we want to calculate the Divergence for F(x,y) = (x^2 * y, xy) at (5,4), all we need to do is take the dot product of F(x,y) with the (∂/∂x, ∂/∂y) operator:
Div (F(x,y)) = ∂/∂x (x^2 * y) + ∂/∂y (xy) = 2xy + x = 2(5)(4) + (5) = 40 + 5 = 45. No unit vectors vectors or directional vectors needed.

I hope this oversized explanation helped a little bit
• Imagining the divergence as a dot product implies that the divergence can't be defined when the number of components in the input and output are different.
How can we think of this intuitively ? • When we have a function which has the same number of input and output components, we can think about it as it takes a point in space and associates a vector with this point which tells a direction of movement and speed of a particle placed in this point. That is what vector fields are all about. Therefore the output vector and the input point should be in the same space otherwise associating a vector with a point will make no sense. Hope it helps!
• I have a really weird question. According to the definition for dot product, a. b = |a||b|cost, where t is the angle b/w the vectors a and b.

Now, my question is "What happens if I sub nabla for a and some vector field v for b?" It seems so meaningless and absurd to ask what angle a partial differential operator makes with respect to a vector field. • Video says you take partial derivatives of a function and then add them together. But..it looks like you are taking only the partial derivative of one variable, say x, of only one function, P in the video. It's like the partial derivative with respect to the other functions Q and R (for variable x) is just..skipped. Question: so does the divergence formula require partial derivative of a variable, x or y or z, with respect to only 1 function, either Q or R or P? I mean...do we just pick and choose? How about situations where you have two variables, x and y, but three functions, P, Q, and R? • A function in (x,y) which has three outputs (P(x,y), Q(x,y), R(x,y)) is an example of a parametric surface. We don't need to define divergence for a parametric surface, because divergence is a concept solely modeled for vector fields. It is valid however to take partial derivatives of parametric surfaces with respect to its two parameters, which has been discussed in an earlier series.
• If the input and output have different dimensions, could you just think of the term at the mismatched dimension(s) to be zero in the respective set for the dot product and just have that dimension not weigh in on the divergence? • does the divergence/curl value of a vector field change on changing the coordinate system? suppose there is a field whose divergence is zero in Cartesian, then will the divergence remain the same in cylindrical/spherical coordinate systems?
(1 vote) • Does the nabla vector mean anything on its own? What do partials with respect to x, y, z (the different components of nabla) mean if there is no function to operate on?
(1 vote) • The nabla-operator, like all operators, doesn't mean anything on its own. Operators are just a symbol that represents a certain operation, so they only make sense when they're accompanied by something to operate on. You can compare the nabla-operator to a factorial operator (!): the operator on its own has no meaning, but when you use the operator on a number (like '5!') it does.
(1 vote)
• I get tripped up by notation a lot. For instance, I sometimes see a nabla symbol squared (for the Laplacian operator), and I have to remember if that means that I'm squaring the result of a single operation, or if I'm applying the operator twice. Same with 2nd derivatives and 2nd partial derivatives!

Is it preferred to use "div" or "nabla dot" when writing this operation?
(1 vote) • I usually prefer writing $DivF$ when solving problems. $\nabla \cdot F$ is only a way to write the formula for divergence shorthand (and a shorthand way to remember it too. It helps in deriving it if you forget the formula). However, if you can remember how to calculate divergence given a vector field, you don't need to use $\nabla \cdot F$

As for second derivative and second partials, see that second partials are just second derivatives with d replaced by $\partial$. So, I think that shouldn't be too confusing
(1 vote)