## 17Calculus Differential Equations - Slope Fields

1st Order

2nd/Higher Order

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Calculus 1 Practice

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Alternate Name For Slope Fields - Direction Fields Slope Fields are two dimensional representations of the slope of a differential equation.

Here is an example of what a slope field looks like. The little lines all over the graph are what we call the slope field lines and they indicate the slope at those points. The three graphs are three possible solutions to the differential equation that was used to produce the slope field.

As you can probably tell from the graph, slope fields are best drawn by a computer. There are many online graphing sites out there. You can also use winplot to plot slope fields.

Here are couple of videos explaining slope fields and how to graph them. The first video is much better and more accurate than the second. So if you have time for only one, the first one is best to watch.

This first video is an indepth introduction to slope fields and a few other introductory topics. He does not actually solve any differential equations using regular techniques in this video. He spends time explaining the concept of differential equations, which is important to understand. Here is what he covers in this video.

Topics Covered in this Video

Geometric View of $$y'=f(x,y)$$

Direction Fields (Slope Fields)

Integral Curves

Use of Isoclines to Hand-Draw a Direction Field

Existence And Uniqueness Theorem

Examples
1. $$y' = -x/y$$
An interesting observation he makes from this example is that you can't tell what the domain of the solution is from the differential equation. You need to either solve it or have a computer graph it for you and try to extrapolate domain information from the graph.

2. $$y' = 1 + x - y$$
This example brings up a few interesting points.
First, an isocline can also be an integral curve (c=1).
Second, two integral curves can't cross at an angle. He does a good job of explaining this important concept when he discusses what's going on between c=0 and c=2.
Third, two integral curves can't be tangent or touch. He discusses the Existence and Uniqueness Theorem to explain this third idea.

3. $$xy' = y-1$$
This is an interesting example that initially seems to violate the existence and uniqueness theorem but he shows that it doesn't.

Bottom Line: This is a great video to help you get a feel for first order differential equations and direction fields. The lecturer does a good job of explaining the concepts and this lecture is full of good solid content.

### MIT OCW - The Geometrical View of y'=f(x,y): Direction Fields, Integral Curves [48mins-55secs]

video by MIT OCW

Here is another video that shows how to plot a slope field by hand. It is not the best video in the world but it has some good techniques. Make sure you read the comments we've written about the video before accepting everything that he does in this video. That said, this video is included to give you another perspective. Here is a rundown of what is in the video.

He starts off this video with the example $$\displaystyle{ \frac{dy}{dx} = \frac{-4x}{y} }$$
He draws the slope field and then uses the separation of variables technique to find the solution. One comment about the graph; he really needs to put a least one number on each axis to indicate the scales. Good teachers will require this on your graphs.

The way he draws the slope field in this video is very commonly taught. When he draws the integral curves, he draws them right through the x-axis where he said there were no slopes. That's wrong! There are slopes there but the slopes are infinite. And what kind of graph do you know has infinite slopes? That's right, vertical lines. This holds for all points on the x-axis except the origin.

Then when he proceeds to separate variables, he says that he uses algebra to move the dx to the other side. AAAHHH! NO! That is not what is going on. The dy/dx is not dy divided by dx, d/dx is an operator on y. For more details read the section on the derivative page about notation. Sorry to be so picky but you need to understand the concepts to do well in calculus.

Finally he integrates both sides of the equation to get $$\displaystyle{ \frac{y^2}{2} = \frac{-4x^2}{2} + c }$$. Then he multiplies both sides by 2. Now he replaces the 2c with just c which, again, is bad notation and can be confusing. He should replace 2c with another constant, like k since $$2c \neq c$$.

Once I got through this video, I debated whether or not to post it here. Some of the video contains misleading information and techniques. However, I thought it was good to see how he drew the slope field. So I left it here. Just a warning: Do not take everything he says at face value. This is true with any teacher, lecturer, professor or video. Check what they say with what you know. Remember instructors are human too.

Bottom Line: If you want to watch this video for an alternate way to draw slope fields, it is okay to watch. However, there are many errors and some bad concepts. The previous video is much better and has a better (and more precise) way to draw slope fields.

### MidnightTutor - Slope fields and differential equations [17mins-5secs]

video by MidnightTutor

Stability of Differential Equations Solutions

At this time, we do not cover stability of ordinary differential equations. However, you can read this page for some information.

Practice

You CAN Ace Differential Equations

Here is a great wikibooks page explaining slope fields in more detail: WikiBooks - Slope Fields

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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