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## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 9: Divergence

# Divergence formula, part 1

How does the x-component of a vector field relate to the divergence? Created by Grant Sanderson.

## Want to join the conversation?

• When P is increasing, the partial derivative of P with respect to x may not be positive.
Indeed, we could imagine a function P like P(x,y) = 2y. In this case, we really don't care about the x-variable, but P can increase !

So why does he only talk about the partial derivative of P with respect to x ? What about the one with respect to y ?

(I think the name of the variables 'x' and 'y' have no link with the x and y directions in a graph. In our case, the right-left direction is given by P, and the up-down direction by Q)
• You are correct that P could increase if P(x,y) = 2y. However, it would not increase with a change in the x-input. Thus, the divergence in the x-direction would be equal to zero if P(x,y) = 2y. In this example, we are only trying to find out what the divergence is in the x-direction so it is not helpful to know what partial P with respect to y would be.

Where the divergence looks like this:
<-- <------- <------------ O ------------ > -------> -->

It's clearly divergent from 'O', but it doesn't look like dP/dx > 0.
For x>0, dP/dx < 0; decreasingly negative
for x<0, seems like dP/dx < 0 but increasingly negative.
• Another little thing. For x<0, dP/dx>0, because as x increases, the value of P also does that (at least in the shown portion)
• Can be P have a positive divergence at a point by having P as zero at the point, less positive to its right and more positive to its left?
• What is P? I dont know if that comes from an earlier video, but i missed it.
(1 vote)
• P is just a point in the vector field with a certain value in horizontal and (not in this case) vertical component.