Last post we discussed vector fields and their gradient. This time, we will be discussing whether or not vector fields (more specifically their functions) are conservative. Our first goal will be to determine potential functions (reverse of taking the gradient), the curl (tendency to rotate around a point), and divergence of vector fields). I will note that these are much more difficult than the last post, so again, bear with me as I try to explain the concepts. If there is something you think I should explain better (without having to teach you basic calculus I or II) drop a comment. The better I’m able to teach this, the better I’ll understand it, the better I’ll probably do on my exam.
First, we must discuss conservative vector fields. A vector field is considered conservative if it can act as a gradient to another function. This is slightly harder than taking just the gradient. The easiest way to test if a vector is conservative is through the second partial derivatives test. If the vector field we are looking at is conservative, then it would be the first derivative of the potential function (f), or in other words, it would be
∴ P(x,y)=fx & Q(x,y)=fy
Remember that F is not just one equation, it’s a pair of equations (P and Q) made up of x’s and y’s. The second partial derivative test requires that fxy=fyx. So the easy way to test is to take the necessary derivatives of F(x,y). In this case, we want to take the derivative of P in terms of y and Q in terms of x. In equation form, it would look like this.
∂/∂y*P(x,y)=fxy & ∂/∂x*Q(x,y)=fyx
Here we see that the Py should equal Qx because fxy would equal fyx if f does exist. So, if Py and Qx don’t equal, we know that there is not a function f. If the two are equal, we know we must then try to form a potential equation.
To find a potential equation, we must integrate both sides of the vector F. going back to our first equation, we will integrate P(x,y) in terms of x, as it is the x component of the function F and Q(x,y) by y as it is the y component. For this part of the lesson, I will use a simple example to illustrate finding a potential function. Let’s use F(x,y)=<y,x>. We’ll walk step by step through bullet points.
- Using this equation, we have P(x,y)=y and Q(x,y)=x.
- Taking the integrals ∫P(x,y) dx and ∫Q(x,y) dy.
- ∫P(x,y) dx = ∫y dx = xy + g(y) & ∫Q(x,y) dy=∫x dy=xy + h(x)
Before I continue, let me explain the g(y) and h(x). When you take the indefinite integral of a function in calculus II, you always had to add a constant to the function. In this case, not only is a constant possible, but also functions. For example, if you were to take the derivative of xy+g(y) in terms of x, you would end up with x+0. This being a possibility, we must include it in our integral. As it turns out, it’s not important in this integral, but I’ll do another example in a second where it will matter.
So, having this: ∫P(x,y) dx = ∫y dx = xy + g(y) & ∫Q(x,y) dy=∫x dy=xy + h(x), we need to match like terms. Obviously the “xy” term is in both equations, therefore we know it is part of our potential function. The terms g(y) and h(x) do not match up with any other terms, so therefore we can assume they equal zero. It is important to note that if there are other terms that exist, they must either match up with the g(y) or h(x) terms, or you need to recheck if the vector is conservative.
As a second example, we’ll use something slightly more complicated. In this case, let’s use the function F(x,y)=<y+2x,x+2y>. Let’s use the same bullets as above for consistency.
- Using this equation, we have P(x,y)=y+2x and Q(x,y)=x+2y.
- Taking the integrals ∫P(x,y) dx and ∫Q(x,y) dy.
- ∫P(x,y) dx = ∫(y+2x) dx = xy + x2 + g(y) & ∫Q(x,y) dy=∫(x+2y) dy = xy +y2 + h(x)
Now we need to match parts again. As you can see, the part “xy” appears in both equations, therefore we know it must be in our potential function. Now look at the h(x) and x2 parts. Those two can be equal because x2 is a function of x. The same goes for h(y) and y2. Because of this, we can put together a potential function that looks like this:
f(x,y) = xy + x2 + y2 + c
Because we are just looking for one possible equation, the value of the constant “c” can be zero. The reason we can use any value of “c” is because the gradient will be the same at any value of “c” because the derivative of “c” is zero for all constants. It doesn’t affect the gradient.
Now we have a potential function for our gradient field. The ability to find these potential functions can become useful in advanced physics. Fields, such as gravitational fields, are conservative fields. It will also come in handy when we start doing line integrals in future installments of this series.
Curl of a vector field is the tendency of a vector field to rotate in a counterclockwise order. To explain this, let’s use the same example as our last post – the ocean current vector field.
If you look on the right side (Atlantic Ocean), you can see two different areas where the water swirls in a circle. Above the quick current, around the latitude of -37 and longitude 321, you can see one of these circles. Notice that the arrows rotate in a counterclockwise fashion. This rotation has a positive curl. The curl isn’t very strong as the current in the area is inside a level curve that touches a dark blue arrow (very weak current speed).
If you look at the second circular motion (around latitude 55, longitude 326), you can see the arrows go in a clockwise direction. This means the curl of the vector field in this area is negative. Again, you can see the current is not very strong in this area, meaning the curl is only slightly negative.
Curl can be calculated using the cross product operator. For those of you not familiar with cross products, it is one way of multiplying vectors (I won’t be showing how to do it here as it was something covered weeks ago). In this case, the two vectors we shall cross are the gradient operator (∇) and our vector field function. Let’s use a function F(x,y,z) = < P(x,y,z), Q(x,y,z), R(x,y,z)>. Therefore,
curl F = ∇ X F(x,y,z) = < Ry-Qz, -(Rx – Pz), Qx-Py >
There is one special property to the curl function. If the curl function equals the zero vector (<0,0,0>), the vector field is conservative. If you notice, Qx-Py, setting this equal to zero, you get Qx =Py. This is the same equation from our second partial derivatives test performed in the potential equations section. The curl function is just the 3D version of the same test.
Divergence is much like curl, but instead of taking the cross product of the gradient operator and function, you take the dot product. The dot product will become more useful in the days to come, so right now, I’ll just leave it as a simple equation using the same function F as above:
div F = ∇ • F = Px + Qy + Rz
You’ll notice that this isn’t a vector for an answer as dot products produce scalars. I’ll go into the practical uses of these when I find them. They have more to do with volumetric flow, but I’m just not that far into integrals yet.
The next installment will discuss different type of line integrals. I’ll look at arc lengths, line integrals of parametric curves, line integrals of differential form, and maybe even Green’s Theorem. Again, leave comments if something isn’t quite clear, as I definitely have a feeling there is more of that in this than the last post.