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## AP®︎/College Calculus BC

# Worked example: equation from slope field

AP.CALC:

FUN‑7 (EU)

, FUN‑7.C (LO)

, FUN‑7.C.1 (EK)

Given a slope field and a few differential equations, we can determine which equation corresponds to the slope field by considering specific slopes.

## Want to join the conversation?

- If you had no answer choices, how would you determine this? It seems like you are just guessing and checking?

If you had infinite slope field lines, could you calculate the function?(8 votes)- Observe that you can draw infinitely many possible graphs for a given slope field. A slope field doesn't define a
*single*function, rather it describes a*class*of functions which are all solutions to a particular differential equation. For instance, suppose you had the differential equation:

𝑦' = 𝑥

By integrating this, we would obtain 𝑦 = (1/2)𝑥² + 𝐶. Observe that there are an infinite number of functions 𝑦 that satisfy the differential equation (because we can choose any constant of integration). This is also reflected by the fact that the slope field permits infinitely many curves to be drawn through it.

So if you had*all*of the slope field lines, chances are you could narrow down the what the graphs of all of the solutions to a particular differential equation must look like. However, if you wish to know the explicit form of the solutions, you are most likely out of luck. You run into the same problem you get when integrating many functions. That is, sometimes there simply does not exist an elementary form for the solutions.

For instance, if you wanted to find the antiderivative of 𝑒^(-𝑥²), you'd be hard-pressed to find an elementary closed form solution. Because there is none. In fact, we*define*a function called the error function to be directly related to the antiderivative of 𝑒^(-𝑥²) because there is no elementary function (a function in terms of polynomials, rational functions, logarithms, exponentials, trigonometric functions, etc.), that is an antiderivative of 𝑒^(-𝑥²).

Likewise, when solving a differential equation, often times, the solution to the equation does not have an elementary form. In those cases, the best we can do is to compute things numerically, or just define a whole new function to be the solution to a particular differential equation.

Comment if you have questions!(3 votes)

- how can i know the slopes (( the shape of this lines )) ?? is there a kind of equation or law ??(0 votes)
- It's just the slope of lines, so there are some "rules" that are derived from the equation of lines:

A horizontal line represents a slope of`0`

A line that "grows" (like "/") represents a positive slope, the steeper it is, the greater the slope.

A line that "falls" (like "\") represents a negative slope, the steeper it is, the more negative the slope.

A line that grows that is exactly at 45° represents a slope of`1`

.

A line that falls that is exactly at 45° represents a slope of`-1`

.

With this guidelines you can approximate the value of the slope in the slope field.(14 votes)

- How do you solve the differential equation if we can't separate the variables?

dy/dx = x+y(2 votes)- Well, I don't believe there is a way to "solve it" like you're hoping. This graph will give us the slopes of the function, so even if we can't solve for the original function, we can at least see what it looks like. Hope this helps!(3 votes)

- Aren't you able to rule out the rational options off the bat, since (0,0) has a slope defined? 0/0 wouldn't result in a slope on the slope field right?(3 votes)
- What does Sal mean when he said the solution depends on the points it contains?(2 votes)
- Basically, the points he is talking about are the initial conditions.(2 votes)

- could you solve this more analytically by looking at the line, whose equation is x+y=-1, and then realising that dy/dx=-1 at the points which lie on the line, so straight away, substituting dy/dx for -1, the differential equation x+y=dy/dx pops out?(1 vote)
- No, you cannot. That works for this case, but what if there were another possible solution in which dy/dx = (-x-1)/y? Then, both equations would be true for the line, but only dy/dx = x + y would work for the entire field.(4 votes)

- I just happen to remember what function would create a field like this, but I suppose it won't work every-time, especially in ℝ≥4?(1 vote)
- At4:32, Sal said that the slope at x=-1 and y=-1 appears to be negative, which matches the yellow equation, which is -2. However, the yellow dy/dx = 2,
**not**-2, so I'm wondering where he picked up the negative.(1 vote) - the shape of the graph is depended on the starting point and anything else ? i still cannot get how sal draw that?(0 votes)
- You have to find the starting point, yes, and then you just follow the slope field. Each line/arrow on the field represents about where the graph should be pointing at that point in space, so just start from the point you have and follow the direction of the lines as you get to them.(3 votes)

- Can the slope of a function be defined where the function does not exist? For example the slope of
`f(x) = x^2`

at x = 1 and y = 5 and what does this value mean in relation to the function?(1 vote)

## Video transcript

- [Voiceover] We've already seen scenarios where we start with a
differential equation and then we generate a slope field that describes the solutions
to the differential equation and then we use that to
visualize those solutions. What I want to do in this
video is do an exercise that takes us the other way,
start with a slope field and figure out which differential equation is the slope field
describing the solutions for. And so I encourage you to
look at each of these options and think about which of
these differential equations is being described by this slope field. I encourage you to pause
the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do
it is I'm just going to find some points that seem to be
easy to do arithmetic with, and we'll see if the slope
described by the differential equation at that point is
consistent with the slope depicted in the slope field. And, I don't know, just for simplicity, maybe I'll do x equals one and y equals one for all of these. So, when x equals one
and y is equal to one. So, this first differential
equation right over here, if x is one and y is
one, then dy/dx would be negative one over one or negative one. Dy/dx would be negative one. Now, is that depicted here? When x is equal to one
and y is equal to one, our slope isn't negative one. Our slope here looks positive. So we can rule this one out. Now, let's try the next one. So, if x is equal to one
and y is equal to one, well then dy/dx would be equal
to one minus one or zero. And, once again, I just
picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative
five and negative seven. This just makes the
arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope
here, so we can rule that out. Once again, for this magenta
differential equation, if x and y are both equal to one, then one minus one is once
again going to be equal to zero. And we've already seen this
slope is not zero here, so rule that one out. And now here we have x plus y, so when x is one and y is one, our derivate of y with
respect to x is going to be one plus one, which is equal to two. Now, this looks interesting. It looks like this slope
right over here could be two. This looks like one. This looks like two. I would want to validate
some other points, but this looks like a really,
really good candidate. And you can also see
what is happening here. When dy/dx is equal to x plus y, you would expect that as
x increases for a given y your slope would increase and as y increases for a
given x your slope increases. And we see that. If we were to just hold y constant at one but increase x along this line, we see that the slope is increasing. It is getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And, in general, we see
that the slope increases as we go to the top right. And we see that it decreases
as we go to the bottom left and both x and y become
much, much more negative. So, I'm feeling pretty good about this, especially if we can
knock this one out here, if we can knock that one out. So, dy/dx is equal to x over y. Well, then when x equals
one and y equals one, dy/dx would be equal to one, and this slope looks larger than one. It looks like two, but since we are really just eyeballing it, let's see if we can find something where this more clearly falls apart. So, let's look at the situation when they both equal negative one. So, x equals negative one and
y is equal to negative one. Well, in that case, dy/dx
should still be equal to one because you have negative one over one. Do we see that over here? So, when x is equal to negative one, y is equal to negative one. Our derivative here looks negative. It looks like negative two, which is consistent with this
yellow differential equation. The slope here is definitely
not a positive one, so we could rule this one out as well. And so we should feel pretty confident that this is the differential
equation being described. And now that we've done it,
we can actually think about well, okay, what are the solutions for this differential
equation going to look like. Well, it depends where they start or what points they contain. If you have a solution
that contains that point, it looks like it might
do something like this. If you had a solution
that contained this point, it might do something like that. And, of course, it keeps going. It looks like it would asymptote towards y is equal to negative
x, this downward sloping. This essentially is the line
y is equal to negative x. Actually, no that is not the
line y equals negative x. This is the line y is equal
to negative x minus one, so that's this line right over here. And it looks like if
the solution contained, say, this point right over here, that would actually be a solution to the differential equation y is
equal to negative x minus one and you can verify that. If y is equal to negative x minus one, then the x and negative x cancel out and you are just left with
dy/dx is equal to negative one, which is exactly what is being described by this slope field. Anyway, hopefully you
found that interesting.