Calculating the work through a path with a vector field

Today I want to tell you a simple nonsense to understand and that can be worth to us to give us a few laughs.

Some time ago I told you the difference between a vector field and a scalar field. But I did not tell you a practical example of use.

This is the reason why today I want to talk about the calculation of the work that has to do a particle to move through a path in a vector field.

For this, the best way is an example. So let’s say we have a vector field of the following form (which is also very important and curious) \overrightarrow {F} = -y \widehat {i} + x \widehat {i} .

This field, if we draw it in an x, y coordinate axis, we will see that it is circles whose vctor is greater the further away we are from the center. Exactly as it is represented in the image that title this post.

Now imagine that within this vector field we have a particle that follows a certain path at a time. That is, the particle follows a parametric curve such that y = 0, x = t. That is, it moves along the X axis.

What work will you do? Let’s first define the work. The work itself is the energy we must give to a particle to modify its energy, in this case, its kinetic energy within the field. Then:

T=\overrightarrow{F}\cdot \bigtriangleup \overrightarrow{r}

That is, the classic formula tells us that it is the Force that acts on the particle (or the sum of them) by the distance that it travels, except in this case we have put it in vector mode, nothing serious.

Now, if the particle follows a curve in the field we will have the work now:

T=\int_{c}^{ } \overrightarrow{F}\cdot \bigtriangleup \overrightarrow{r} = \lim_{\triangle \overrightarrow{r_{i}} \to 0}\sum_{i}^{ } \overrightarrow{F}\cdot \triangle \overrightarrow{r_{i}}

Where, what we have done is to divide the curve into small pieces and calculate the sum of the “mini works” (through the integral that is the limit), nothing that any of you could not do.

Yes we now take and make that:

\triangle\overrightarrow{r}=\frac{\triangle \overrightarrow{r}}{\triangle t}\triangle t=\frac{d\overrightarrow{r}}{dt}dt

The only thing we have done is to divide it and add it by the increments of time it takes the particle to traverse the curve (and, since they are infinitesimal increments, they are practically derived), in the end, we have to:

T=\int_{t_{0}}^{t_{1}}\overrightarrow{F}\cdot \frac{d\overrightarrow{r}}{dt}dt=\int_{t_{0}}^{t_{1}}\overrightarrow{F}\cdot \overrightarrow{v}dt

Where, we must remember that it is a vector product.

Returning to the example we have since we have done the most complicated work, let’s calculate:

Where, obviously, we have first replaced the vector field by its values ​​and then these by the curve, and we have also made the derivatives of the curve in time.

The case is that the result obtained is something that we know if we draw the path of the particle in an axis of coordinates with the field. We will observe that the field forces are always perpendicular to the curve (to the trajectory) and, as we know, the work thus done is 0.

By varying the curve we can calculate other types of work (if need be, of course) and we can even limit them to a specific time point in order to calculate, with precision, the work we will perform (or that the field will perform for us, To the particle Remember that if the work comes out positive, we have to do it, while if it comes out negative, it is the field that does it.

This is an example of calculating work a little more complicated than teaching the children in institutes but it is easy enough to understand for these and see how you can calculate such work in a vector field that does not have to be uniform and With a movement of a particle that does not have to be linear.

I recommend that you play and make examples of, for example, if the curve, the path is x = 0, y = t (ie follows the field line completely), x = t, y = t (this is very curious, and looking at the graph, by simple observation you will know why (trick: polar coordinates) or any other that happens to you.