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### Course: AP®︎ Calculus AB (2017 edition) > Unit 10

Lesson 4: Slope fields- Slope fields introduction
- Worked example: equation from slope field
- Worked example: slope field from equation
- Worked example: forming a slope field
- Slope fields & equations
- Approximating solution curves in slope fields
- Worked example: range of solution curve from slope field
- Reasoning using slope fields

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# Slope fields introduction

Slope fields allow us to analyze differential equations graphically. Learn how to draw them and use them to find particular solutions.

## Want to join the conversation?

- Are there no practice exercises, or leveling up or any such thing for the differential equations mission? Is it all only videos? How can I make sure that I retain what the video's teach me?(19 votes)
- If you want good practice with differential equations, I would recommend going to MIT OpenCourseWare online. Navigate to their differential equations classes and you will find course notes, exams, and practice questions with solutions. While it's not as fun as Khan Academy, it will definitely give you a chance to test your abilities.(61 votes)

- I have two questions:

1) Couldn't Sal have just multiplied the right hand side by dx and the left hand side by y, took the integral and solved it that way or was a slope field really necessary?

2) Is there an abbreviation for undefined?(15 votes)- i think it was just to get the idea of slope fields across. some functions are hard to figure out by hand, but if you draw a slope field and then if the existence uniqueness theorem applies, you can draw out an aprox solution by hand. RK4 method is what programs like maple uses to aprox a solution to differential equations. Sorry i couldnt really explain existence uniqueness theorem, but i think google would be your friend(9 votes)

- A small confusion , while estimating the solution using slope field you give arbitrary values to x and y then we find dy/dx at that point .So my doubt is how can y take different values for a particular value of x ( no more a function ), also, that (x,y) might not satisfy our function so how can it still give a correct result(10 votes)
- The thing is, differential equations don't only have one function as their solution. They can have an infinite number of solutions. In this case, the solution in general may be written as:

y^2 + x^2 = c;

where c is any constant. So by adjusting c, we can make an indefinite number of functions that satisfy the solution. In fact, square root of c represents the radius of the circle. (More about that in conic sections)

Hope that made sense :)(25 votes)

- so he is plugging in random numbers to the derivative?(13 votes)
- What if the equation were dy/dx = 2x? And there isn't a y?(7 votes)
- Then the slope field will be independent of y. It will look like a lot of "columns" of lines all with the same slope. So on the x-axis the lines will be horizontal, for x=1/2 they'll be diagonal lines, etc.

We can solve dy/dx = 2x directly (by integration), giving y = x² + C.

The result is a family of parabolas - a different one for each value of C.(7 votes)

- Hi, i'm just wondering, other than visualizing solutions, are there any other uses for slope fields?(6 votes)
- i dont think so. i think slope fields are just a way to visualize the antiderivative.(5 votes)

- how do i make a slope feild(3 votes)
- First, draw your axes. Then pick a bunch of points and draw lines with the slope at each point.(6 votes)

- I always face a difficulty when drawing solution curves through slope field.How do I know which points my curve goes through?The slope field gives a visual idea about what my solution may look like but i can not figure out how to draw my exact solution curves by making the curves tangential to the slope field always.:) would be grateful if i get a reply.(3 votes)
- at4:51, why did Sal not draw the solution passing through the points that he drew?(2 votes)
- See that dy/dx = -x/y can be solved to get x^2 + y^2 = c, where c is the constant you get when you integrate, and here, it is also the radius of the circle (as x^2 + y^2 = r^2 defines a circle of radius r, and r^2 here is just c).

So, this slope field essentially gives you circles for every radius c. The one Sal drew was specifically for c = 1 (Hence, it passes through points (1,0), (0,1), (-1,0) and (0,-1)). If he increases the radius, it'll pass through other points (Imagine blowing up the circle he drew. You'll see that it'll pass through those points at some time)

If it helps, see this(https://www.desmos.com/calculator/gdnnkokiqk). It's the slope field given in the video along with a circle. As you vary the value of c with the slider, you see that it passes through different points, and the lines at those points will always correspond to the slope of the circle.(3 votes)

- Is the solution to a differential equation always a function.But i see that in this video Sal has got a solution to the differential equation which is essentially a relation and not a function.Please clarify.Thankyou!(2 votes)

## Video transcript

- [Voiceover] Let's say that we have the differential equation
dy dx or the derivative of y with respect to x is
equal to negative x over y. Let's say we don't know how
to find the solutions to this, but we at least want to get a sense of what the solutions might look like. And to do that what we
could do is we could look at a coordinate plane, so
let me draw some axes here. So let me draw relatively
straight line alright. So that's my y-axis and this is my x-axis. Let me mark this as one, that's
two, that's negative one, negative two, one, two,
negative one, and negative two. And what I could do is since
this differential equation is just in terms of xs and
ys and first derivatives of y with respect to x, I
could sample points on the coordinate plane, I could look
at the x and y coordinates substitute them in here, figure out what the slope is going to be. And then I could visualize
the slope if a solution goes through that point what
the slope needs to be there and I can visualize that
with a line segment, a little small line segment that has the same slope as the slope in question. So let's actually do that. So let me setup a little table here. I'm going to do a little table here to do a bunch of x and y values. Once again I'm just sampling some points on the coordinate plane
to be able to visualize. So x, y and this is dy dx. So let's say when x is zero and y is one, what is the derivative
of y with respect to x? It's going to be negative zero over one so it's just going to be zero. And so at the point zero one
if a solution goes through this point its slope is going to be zero. And so we can visualize
that by doing a little horizontal line segment right there. So let's keep going. What about when x is one and y is one? Well then dy dx, the derivative
of y with respect to x is negative one over one, so
it's going to be negative one. So at the point one comma
one if a solution goes through that point, it would
have a slope of negative one. And so I draw a little line segment that has a slope of negative one. Let me do this in a new color. What about when x is one and y is zero? Well then it's negative one over zero, so this is actually
undefined, but it's a clue that maybe the slope there
if you had a tangent line at that point, maybe it's vertical. So I'll put that as a
question mark, vertical there. So maybe it's something
like that if you actually did have, I guess it
wouldn't be a function, if you had some kind of
relation that went through it, but let's not draw that just yet. But let's try some other points. Let's try the point
negative one negative one. So now we have negative negative one, which is one, over negative one. Well you'd have a slope
of negative one here. So negative one negative one you'd have a slope of negative one. What about if you had one negative one? Well now it's negative one over negative one, your slope is now one. So one negative one, if a
solution goes through this its slope would look like that. And we could keep going, we
could even do two negative two. That's going to have a
slope of one as well. If you did positive two positive two, that would be negative two over two. You'd have a slope of
negative one right over here. And so we could do a bunch
of points, just keep going. Now I'm just doing them in my head, I'm not going on the table. But you get a sense of
what's going on here. Here you're slope, what if
it was negative one one? It's going to have a slope of one. So at this point your
slope negative one one so negative negative one is one over one, so you're going to have a slope like that. At negative two two same exact idea, it would look like that. And so when you keep
drawing these line segments over these kind of sample
points in the cartesian or in the x-y plane, you
start to get a sense of well what would a solution have to do. And you can start to visualize
that hey maybe a solution would have to do something like this. This would be a solution,
so maybe it would have to do something like this. Or if we're looking only at
functions and not relations I'll make it so it's very clear. So maybe it would have to
do something like this. Or if the function started
here based on what we've seen so far maybe it would
have to do something like this. Or if this were a point
on the function over here, it would have to do something like this. And once again I'm doing this based on what the slope field is telling me. So this field that I'm
creating where I'm sampling a bunch of points and I'm visualizing the slope with a line segement. Once again this is called a slope field. So hopefully that gives you kind of the basic idea of what a slope field is. And the next two videos we'll
explore this idea even deeper.