Differential Equations - Direction Fields - Pauls Online Math Notes

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Section 1.2 : Direction Fields

This topic is given its own section for a couple of reasons. First, understanding direction fields and what they tell us about a differential equation and its solution is important and can be introduced without any knowledge of how to solve a differential equation and so can be done here before we get into solving them. So, having some information about the solution to a differential equation without actually having the solution is a nice idea that needs some investigation.

Next, since we need a differential equation to work with, this is a good section to show you that differential equations occur naturally in many cases and how we get them. Almost every physical situation that occurs in nature can be described with an appropriate differential equation. The differential equation may be easy or difficult to arrive at depending on the situation and the assumptions that are made about the situation and we may not ever be able to solve it, however it will exist.

The process of describing a physical situation with a differential equation is called modeling. We will be looking at modeling several times throughout this class.

One of the simplest physical situations to think of is a falling object. So, let’s consider a falling object with mass \(m\) and derive a differential equation that, when solved, will give us the velocity of the object at any time, \(t\). We will assume that only gravity and air resistance will act upon the object as it falls. Below is a figure showing the forces that will act upon the object.

Before defining all the terms in this problem we need to set some conventions. We will assume that forces acting in the downward direction are positive forces while forces that act in the upward direction are negative. Likewise, we will assume that an object moving downward (i.e. a falling object) will have a positive velocity.

Now, let’s take a look at the forces shown in the diagram above. \({F_G}\) is the force due to gravity and is given by \({F_G} = mg\) where \(g\) is the acceleration due to gravity. In this class we use \(g\) = 9.8 m/s2 or \(g\) = 32 ft/s2 depending on whether we will use the metric or Imperial system. \({F_A}\) is the force due to air resistance and for this example we will assume that it is proportional to the velocity, \(v\), of the mass. Therefore, the force due to air resistance is then given by \({F_A} = - \gamma v\), where \(\gamma > 0\). Note that the “–” is required to get the correct sign on the force. Both \(\gamma\) and \(v\) are positive and the force is acting upward and hence must be negative. The “–” will give us the correct sign and hence direction for this force.

Recall from the previous section that Newton’s Second Law of motion can be written as

\[m\frac{{dv}}{{dt}} = F\left( {t,v} \right)\]

where \(F\left( {t,v} \right)\) is the sum of forces that act on the object and may be a function of the time \(t\) and the velocity of the object, \(v\). For our situation we will have two forces acting on the object gravity, \({F_G} = mg\). acting in the downward direction and hence will be positive, and air resistance, \({F_A} = - \gamma v\), acting in the upward direction and hence will be negative. Putting all of this together into Newton’s Second Law gives the following.

\[m\frac{{dv}}{{dt}} = mg - \gamma v\]

To simplify the differential equation let’s divide out the mass, \(m\).

\[\begin{equation}\frac{{dv}}{{dt}} = g - \frac{{\gamma v}}{m} \label{eq:eq1}\end{equation}\]

This then is a first order linear differential equation that, when solved, will give the velocity, \(v\) (in m/s), of a falling object of mass \(m\) that has both gravity and air resistance acting upon it.

In order to look at direction fields (that is after all the topic of this section....) it would be helpful to have some numbers for the various quantities in the differential equation. So, let’s assume that we have a mass of 2 kg and that \(\gamma= 0.392\). Plugging this into \(\eqref{eq:eq1}\) gives the following differential equation.

\[\begin{equation}\frac{{dv}}{{dt}} = 9.8 - 0.196v \label{eq:eq2} \end{equation}\]

Let's take a geometric view of this differential equation. Let's suppose that for some time, \(t\), the velocity just happens to be \(v = 30\) m/s. Note that we’re not saying that the velocity ever will be 30 m/s. All that we’re saying is that let’s suppose that by some chance the velocity does happen to be 30 m/s at some time \(t\). So, if the velocity does happen to be 30 m/s at some time \(t\) we can plug \(v = 30\) into \(\eqref{eq:eq2}\) to get.

\[\frac{{dv}}{{dt}} = 3.92\]

Recall from your Calculus I course that a positive derivative means that the function in question, the velocity in this case, is increasing, so if the velocity of this object is ever 30m/s for any time \(t\) the velocity must be increasing at that time.

Also, recall that the value of the derivative at a particular value of \(t\) gives the slope of the tangent line to the graph of the function at that time, \(t\). So, if for some time \(t\) the velocity happens to be 30 m/s the slope of the tangent line to the graph of the velocity is 3.92.

We could continue in this fashion and pick different values of \(v\) and compute the slope of the tangent line for those values of the velocity. However, let's take a slightly more organized approach to this. Let's first identify the values of the velocity that will have zero slope or horizontal tangent lines. These are easy enough to find. All we need to do is set the derivative equal to zero and solve for \(v\).

In the case of our example we will have only one value of the velocity which will have horizontal tangent lines, \(v = 50\) m/s. What this means is that IF (again, there’s that word if), for some time \(t\), the velocity happens to be 50 m/s then the tangent line at that point will be horizontal. What the slope of the tangent line is at times before and after this point is not known yet and has no bearing on the slope at this particular time, \(t\).

So, if we have \(v = 50\), we know that the tangent lines will be horizontal. We denote this on an axis system with horizontal arrows pointing in the direction of increasing \(t\) at the level of \(v = 50\) as shown in the following figure.

Now, let's get some tangent lines and hence arrows for our graph for some other values of \(v\). At this point the only exact slope that is useful to us is where the slope horizontal. So instead of going after exact slopes for the rest of the graph we are only going to go after general trends in the slope. Is the slope increasing or decreasing? How fast is the slope increasing or decreasing? For this example those types of trends are very easy to get.

First, notice that the right hand side of \(\eqref{eq:eq2}\) is a polynomial and hence continuous. This means that it can only change sign if it first goes through zero. So, if the derivative will change signs (no guarantees that it will) it will do so at \(v\) = 50 and the only place that it may change sign is \(v = 50\). This means that for \(v>50\) the slope of the tangent lines to the velocity will have the same sign. Likewise, for \(v50\) we will have negative slopes for the tangent lines. As with \(v