How To Draw The Line From A Slope Field

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Section 1-2 : Direction Fields

This topic is given its own department for a couple of reasons. Outset, understanding direction fields and what they tell united states 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 exist done here before we become into solving them. So, having some information about the solution to a differential equation without actually having the solution is a squeamish idea that needs some investigation.

Next, since nosotros need a differential equation to work with, this is a good section to prove yous that differential equations occur naturally in many cases and how we get them. About every concrete situation that occurs in nature can exist described with an appropriate differential equation. The differential equation may be easy or difficult to go far at depending on the situation and the assumptions that are fabricated virtually the state of affairs and we may not ever be able to solve it, however it volition exist.

The process of describing a physical situation with a differential equation is chosen modeling. We volition be looking at modeling several times throughout this grade.

One of the simplest physical situations to call back of is a falling object. So, allow'due south consider a falling object with mass \(m\) and derive a differential equation that, when solved, will give u.s. the velocity of the object at any time, \(t\). We volition assume that just gravity and air resistance will human activity upon the object as information technology falls. Below is a figure showing the forces that will deed upon the object.

Earlier defining all the terms in this problem we need to set some conventions. We will presume that forces interim in the downward direction are positive forces while forces that deed in the upward management are negative. Likewise, nosotros volition presume 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 higher up. \({F_G}\) is the strength due to gravity and is given past \({F_G} = mg\) where \(g\) is the acceleration due to gravity. In this grade nosotros use \(k\) = 9.8 one thousand/s² or \(chiliad\) = 32 ft/due south² depending on whether we volition apply the metric or Imperial system. \({F_A}\) is the forcefulness 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 so given by \({F_A} = - \gamma v\), where \(\gamma > 0\). Note that the "–" is required to get the right sign on the force. Both \(\gamma\) and \(v\) are positive and the forcefulness is acting upward and hence must be negative. The "–" will requite united states the correct sign and hence management for this force.

Recall from the previous department that Newton'due south Second Law of motion can be written every bit

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

where \(F\left( {t,5} \right)\) is the sum of forces that act on the object and may exist a function of the fourth dimension \(t\) and the velocity of the object, \(v\). For our situation nosotros will take 2 forces interim 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 Police force gives the following.

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

To simplify the differential equation let's dissever out the mass, \(m\).

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

This then is a get-go 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 social club to look at direction fields (that is afterward all the topic of this section....) it would be helpful to accept some numbers for the various quantities in the differential equation. Then, 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 simply happens to be \(v = thirty\) m/south. Note that nosotros're not proverb that the velocity ever will exist 30 k/southward. All that we're saying is that permit's suppose that by some chance the velocity does happen to be 30 g/s at some time \(t\). So, if the velocity does happen to exist 30 m/due south at some time \(t\) we can plug \(v = 30\) into \(\eqref{eq:eq2}\) to get.

\[\frac{{dv}}{{dt}} = iii.92\]

Recall from your Calculus I grade that a positive derivative means that the role in question, the velocity in this case, is increasing, and then if the velocity of this object is always 30m/s for any time \(t\) the velocity must exist increasing at that time.

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

We could go on in this fashion and option different values of \(five\) and compute the slope of the tangent line for those values of the velocity. Withal, let's take a slightly more than organized approach to this. Allow'due south first place the values of the velocity that volition have zero slope or horizontal tangent lines. These are like shooting fish in a barrel enough to find. All we need to do is set up the derivative equal to nada and solve for \(five\).

In the case of our instance we will have only i value of the velocity which will have horizontal tangent lines, \(5 = 50\) m/due south. What this means is that IF (once again, there's that word if), for some time \(t\), the velocity happens to exist l k/s then the tangent line at that point will exist horizontal. What the gradient of the tangent line is at times earlier and later this point is not known still and has no bearing on the slope at this detail time, \(t\).

And so, if nosotros accept \(five = 50\), we know that the tangent lines will be horizontal. We announce this on an axis system with horizontal arrows pointing in the direction of increasing \(t\) at the level of \(v = l\) as shown in the following figure.

Now, let'southward get some tangent lines and hence arrows for our graph for another values of \(v\). At this bespeak the but exact slope that is useful to the states is where the gradient horizontal. Then instead of going after exact slopes for the remainder of the graph nosotros are just 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 instance those types of trends are very easy to get.

Offset, detect that the correct hand side of \(\eqref{eq:eq2}\) is a polynomial and hence continuous. This ways that information technology can only change sign if it get-go goes through nix. So, if the derivative volition alter signs (no guarantees that information technology volition) it volition do so at \(v\) = 50 and the but 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 aforementioned sign. Also, for \(v<l\) the slopes volition besides have the same sign. The slopes in these ranges may have (and probably will) have different values, only we exercise know what their signs must be.

Let'southward first by looking at \(v<fifty\). Nosotros saw before that if \(v = xxx\) the slope of the tangent line will exist 3.92, or positive. Therefore, for all values of \(v<50\) we volition have positive slopes for the tangent lines. Also, past looking at \(\eqref{eq:eq2}\) we can run across that as \(v\) approaches fifty, always staying less than 50, the slopes of the tangent lines volition approach goose egg and hence flatten out. If we move \(v\) away from 50, staying less than 50, the slopes of the tangent lines will get steeper. If you want to get an idea of only how steep the tangent lines become you can always pick specific values of \(5\) and compute values of the derivative. For example, we know that at \(five\) = 30 the derivative is 3.92 so arrows at this bespeak should have a slope of around iv. Using this information, we can now add in some arrows for the region beneath \(v\) = fifty as shown in the graph below.

At present, allow's look at \(v>50\). The first thing to practise is to find out if the slopes are positive or negative. Nosotros will do this the same way that we did in the last bit, i.e. pick a value of \(v\), plug this into \(\eqref{eq:eq2}\) and meet if the derivative is positive or negative. Note, that yous should NEVER assume that the derivative volition modify signs where the derivative is zero. It is easy plenty to check and so you should e'er do so.

We demand to check the derivative and then permit's use \(5\) = 60. Plugging this into \(\eqref{eq:eq2}\) gives the slope of the tangent line as -ane.96, or negative. Therefore, for all values of \(v>50\) we will have negative slopes for the tangent lines. Every bit with \(5<50\), by looking at \(\eqref{eq:eq2}\) we tin run across that every bit \(five\) approaches l, always staying greater than 50, the slopes of the tangent lines will approach nil and flatten out. While moving \(v\) away from fifty again, staying greater than 50, the slopes of the tangent lines volition become steeper. We tin now add in some arrows for the region above \(v\) = l as shown in the graph below.

This graph above is called the management field for the differential equation.

So, only why exercise we care most direction fields? There are ii nice pieces of information that tin be readily institute from the management field for a differential equation.

1. Sketch of solutions. Since the arrows in the management fields are in fact tangents to the bodily solutions to the differential equations we tin apply these as guides to sketch the graphs of solutions to the differential equation.
2. Long Term Behavior. In many cases we are less interested in the bodily solutions to the differential equations as we are in how the solutions behave as \(t\) increases. Direction fields, if we can get our hands on them, can be used to find information about this long term behavior of the solution.

So, back to the direction field for our differential equation. Suppose that we want to know what the solution that has the value \(five\left( 0 \right) = thirty\) looks like. We can become to our direction field and get-go at 30 on the vertical axis. At this betoken we know that the solution is increasing and that every bit it increases the solution should flatten out because the velocity will be approaching the value of \(v\) = l. And so we start cartoon an increasing solution and when we hitting an arrow we only make sure that nosotros stay parallel to that arrow. This gives us the figure below.

To get a amend thought of how all the solutions are behaving, allow's put a few more solutions in. Adding some more solutions gives the figure beneath. The set up of solutions that we've graphed below is often called the family of solution curves or the set of integral curves. The number of solutions that is plotted when plotting the integral curves varies. You should graph plenty solution curves to illustrate how solutions in all portions of the direction field are behaving.

Now, from either the direction field, or the direction field with the solution curves sketched in we tin see the behavior of the solution as \(t\) increases. For our falling object, it looks like all of the solutions will approach \(5 = 50\) every bit \(t\) increases.

Nosotros will oft want to know if the behavior of the solution will depend on the value of \(five\)(0). In this case the behavior of the solution will not depend on the value of \(5\)(0), simply that is probably more of the exception than the rule so don't expect that.

Let's take a look at a more complicated case.

Instance ane Sketch the management field for the following differential equation. Sketch the set of integral curves for this differential equation. Determine how the solutions acquit every bit \(t \to \infty \) and if this behavior depends on the value of \(y\)(0) depict this dependency. \[y' = \left( {{y^2} - y - 2} \right){\left( {one - y} \right)^ii}\]

Show Solution

First, do not worry about where this differential equation came from. To exist honest, nosotros just made it up. Information technology may, or may not describe an actual physical situation.

This differential equation looks somewhat more complicated than the falling object instance from above. Even so, with the exception of a little more work, it is not much more complicated. The first step is to determine where the derivative is zero.

\[\begin{align*}0 & = \left( {{y^ii} - y - 2} \right){\left( {i - y} \correct)^2}\\ 0 & = \left( {y - 2} \right)\left( {y + 1} \right){\left( {ane - y} \right)^ii}\end{align*}\]

Nosotros can now meet that we take three values of \(y\) in which the derivative, and hence the gradient of tangent lines, volition be zero. The derivative volition be null at \(y\) = -1, ane, and ii. And then, allow's first our direction field with cartoon horizontal tangents for these values. This is shown in the effigy below.

At present, we need to add together arrows to the four regions that the graph is now divided into. For each of these regions I will pick a value of \(y\) in that region and plug it into the correct hand side of the differential equation to run across if the derivative is positive or negative in that region. Again, to get an accurate management fields you should pick a few more values over the whole range to see how the arrows are behaving over the whole range.

\(y < - 1\)

In this region we tin use \(y\) = -two as the test indicate. At this betoken we have \(y' = 36\). So, tangent lines in this region will have very steep and positive slopes. Also as \(y \to - 1\) the slopes volition flatten out while staying positive. The figure below shows the direction fields with arrows in this region.

\( - i < y < ane\)

In this region we can use \(y\) = 0 as the exam point. At this point we take \(y' = - 2\). Therefore, tangent lines in this region volition accept negative slopes and plain not be very steep. So what practice the arrows look like in this region? As \(y \to 1\) staying less than 1 of course, the slopes should be negative and approach zero. As we move away from 1 and towards -1 the slopes volition start to get steeper (and stay negative), but somewhen flatten dorsum out, once more staying negative, as \(y \to - 1\) since the derivative must approach zero at that point. The figure beneath shows the direction fields with arrows added to this region.

\(1 < y < ii\)

In this region nosotros volition use \(y\) = 1.5 as the test betoken. At this point we have \(y' = - 0.3125\). Tangent lines in this region will besides take negative slopes and plain not be as steep as the previous region. Arrows in this region will conduct essentially the same as those in the previous region. Near \(y\) = i and \(y\) = 2 the slopes volition flatten out and equally we move from one to the other the slopes will go somewhat steeper earlier flattening dorsum out. The figure below shows the direction fields with arrows added to this region.

\(y > 2\)

In this last region we volition use \(y\) = three every bit the test betoken. At this point we take \(y' = 16\). And so, as nosotros saw in the first region tangent lines will starting time out adequately flat almost \(y\) = 2 and then as we move mode from \(y\) = two they will get fairly steep.

The complete direction field for this differential equation is shown below.

Here is the set of integral curves for this differential equation.

Finally, let's take a look at long term behavior of all solutions. Different the showtime example, the long term behavior in this instance will depend on the value of \(y\) at t = 0. Past examining either of the previous 2 figures we tin arrive at the following beliefs of solutions as \(t \to \infty \).

Value of \(y\)(0)	Beliefs as \(t \to \infty \)
\(y\left( 0 \right) < one\)	\(y \to - 1\)
\(1 \le y\left( 0 \correct) < 2\)	\(y \to 1\)
\(y\left( 0 \right) = 2\)	\(y \to two\)
\(y\left( 0 \right) > 2\)	\(y \to \infty \)

Do not forget to acknowledge what the horizontal solutions are doing. This is often the most missed portion of this kind of problem.

In both of the examples that we've worked to this point the right manus side of the derivative has just contained the function and Non the independent variable. When the right hand side of the differential equation contains both the office and the contained variable the beliefs can be much more complicated and sketching the management fields by hand tin be very difficult. Computer software is very handy in these cases.

In some cases they aren't too hard to do by hand however. Let'southward take a await at the post-obit example.

Instance 2 Sketch the management field for the post-obit differential equation. Sketch the set of integral curves for this differential equation. \[y' = y - ten\]

Show Solution

To sketch direction fields for this kind of differential equation we commencement identify places where the derivative will exist constant. To practise this we set the derivative in the differential equation equal to a constant, say \(c\). This gives us a family of equations, called isoclines, that we tin plot and on each of these curves the derivative volition be a constant value of \(c\).

Discover that in the previous examples we looked at the isocline for \(c = 0\) to get the direction field started. For our case the family unit of isoclines is.

\[c = y - 10\]

The graph of these curves for several values of \(c\) is shown below.

Now, on each of these lines, or isoclines, the derivative will be abiding and will accept a value of \(c\). On the \(c = 0\) isocline the derivative will ever take a value of zero and hence the tangents will all be horizontal. On the \(c = 1\) isocline the tangents volition always accept a slope of one, on the \(c = -two\) isocline the tangents volition ever have a slope of -2, etc. Below are a few tangents put in for each of these isoclines.

To add together more arrows for those areas between the isoclines outset at say, \(c = 0\) and motility up to \(c = one\) and as we do that we increment the slope of the arrows (tangents) from 0 to 1. This is shown in the figure below.

Nosotros can then add in integral curves as we did in the previous examples. This is shown in the figure below.

Source: https://tutorial.math.lamar.edu/classes/de/directionfields.aspx

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