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## AP®︎ Calculus AB (2017 edition)

### 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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# Worked example: range of solution curve from slope field

Given the slope field of a differential equation, we can sketch various solutions to the equation. In this example, we analyze the range of a specific solution.

## Want to join the conversation?

- at1:00Sal says it looks like its not going to actually get to 4, but how can one know that, any more than one can look at the initial numbers in a decreasing series and unknown limit and assume it will not get to zero (and therefore diverge)?(25 votes)
- So we know that the solution is never going to cross y=4, because near y=4 the first derivative goes very close to zero, which means the function will just keep going right roughly horizontally, to meet even more horizontal slopes, etc. So there is never a negative slope near that area to push the function over the line y=4.

How do we know that the solution never reaches 4? I'm not completely sure about this, but this is how I think of it.

As the function gets closer and closer to y=4, it's slope goes more and more horizontal, meaning it approaches y=4 slower and slower. As the function gets infinitely closer to y=4, it's slope gets infinitely closer to zero, and so it approaches y=4 infinitely slow. So, as the domain goes to positive infinity, the function gets closer and closer to y=4 but also approaches slower and slower, and so it never really reaches y=4 (until infinity?).(17 votes)

- OK, quite interesting, but why? And what is initial condition and what is range of solution? What are we doing here? What meaning does it have? What application could it be?(6 votes)
- Let me explain this to you. To start with, initial condition isn't a new concept. It is talking about a "condition" (you can think of it as a location, simply, in this particular exaple) where function is "starting" from (for example, function f(x) starts from (0,6)).

In this chapter, We are looking at some diffrential equation (equation about functions, not variables) and are using slope field to solve it in situation like this: we are not given numerous values of the equation, (or the numerous value from`f'(x)`

) so that we need to solve it by, basically, putting all the possible values of the function's slope in certain bounds. and using that field to find approximate value of the asked function.

hope this helped you some bit! If not, please leave a comment below!(7 votes)

- What if I am ask to write the equation for the integral curve that passes through that point (0,6)(6 votes)
- The equation for this slope field is given in the previous video as

dy/dx = y(4-y)/6

This equation is indeed separable, but after you separate you will have to deal with a rather awkward integral if you want to find the integral curve, or the exact solution, which isn't the point of this video. Of course that doesn't stop us from solving it anyway

∫ 1 / (4y-y^2) dy = ∫ 1 / 6 dx ⇒ (ln(y)-ln(y-4)) / 4 = x / 6 + C

We can plot in y(0)=6 to find C and our integral curve already, but if we want to solve for y(x) it will be a little cleaner to get rid of some fractions and logarithms first

e^(ln(y)-ln(y-4)) = e^(4x/6 + C)

y / (y-4) = Ce^(2x/3)

Keep in mind that e^(a-b) ⇔ e^a / e^b. Now we can easily substitute (0,6) to find C

6/2 = Ce^(2*0/3) = C*1

thus C = 3

To find y(x) we'll treat the right hand side as a function u(x) for the sake of simplicity

y / (y-4) = u (x)

y = u(y-4) = uy - 4u

Move uy, extract the y and divide the rest to get the solution

y(1-u) = - 4u

y = 4u / (u -1)

y(x) = C4e^(2x/3) / (Ce^(2x/3) - 1)

Control that we have the correct answer by substituting x=0 and C=3 to get y=6

y(0) = 12e^0 / (3e^0 -1) = 6

Interestingly, we can see how y → 4 as x → ∞, because 4∞ / (∞ -1) = 4. And similarly, as x → -∞, y → 0. Also, notice that x ≠ ln(1/√27) as it is the vertical asymptote for this specific solution.(5 votes)

- When the problem gives the initial condition, can there be no points to the left of that point? For example, if we took off the domain (x >= 0) for this problem, would the range still be (4, 6] or would it be (4, ∞)? Thanks!(5 votes)
- If you took off the domain, the range would be (4, +∞); as you can see, at x < 0 and y > 6, the slope approaches infinity. Therefore, the graph would go up forever with no limit. Hope this helps!(3 votes)

- At9:49pm,Why y can't be equal to 3 or 2 or 1 I mean less then 4?(3 votes)
- Because your initial condition started at 6. The range is based off of the initial condition.(3 votes)

- But how about the lines below y=4 and x>0 ? why we don't care about the lines below?(3 votes)
- The initial condition is that (0, 6) is on the line. Therefore, the lines below y = 4 don't matter because the line from (0, 6) never goes below the line y = 4. The reason we don't care about x < 0 is that the problem says "The range for x >= 0". Hope this helps!(2 votes)

- Is the fact that we must approach a slope of zero but that we can't reach it a consequence of the paradox that 3Blue1Brown describes in his Essence of Calculus playlist? The one referring to the fact that a rate of change requires two points but that a derivative is the rate of change at a single point? Does this show an inherent weakness in a calculus description of the world?(1 vote)
- You seem to be referring to the numerical approximation of the derivative.

When we are measuring something continuous there is always going be error. I wouldn't call this a weakness as this issue is always going to be there by nature You can try increasing numerical precision of variable to improve the approximation and consequently set the value of h appropriately to deal with trade off between truncation error and round off error.

It is also worth considering using central difference as it generally performs better than forward difference.(2 votes)

- How do I figure out if a differential equation [ dy/dx = ((x^2 * y) + (y^2 *x)) / 3x +7 ] will have horizontal segments and when they will occur(1 vote)
- dy/dx =((x^2 * y) + (y^2 *x)) / (3x +7) will have horizontal segments when dy/dx = 0

Which means when (yx^2+xy^2)/(3x+7)=0 we'll have horizontal segments. 1/3x+7 doesn't affect us, if 3x+7 = 0 (x=-7/3) the whole thing is undefined.

So basically we're looking for solutions for both x and y when (yx^2+xy^2)=0

if x=0 => y*0^2+0*y^2 = 0, that's our first solution

if y=0 => 0*x^2 + x*0^2 = 0, our second solution

and then (yx^2+xy^2)=0

means that if yx^2 = -xy^2 then dy/dx =0

simplifying:

divide both sides by x:

yx = -y^2

divide both sides by y:

x=-y, that's our third (and last) solution!

To summarize, dy/dx = (yx^2+xy^2)/(3x+7) will have horizontal segments if:

x = 0, y = 0 or x = -y.

Here's a graph showing the slope field for your equation, and the solutions for when it has horizontal segments.

https://www.desmos.com/calculator/ql5onaikzl

Basically the key realisation is to figure out that all you need is to find the conditions that has to be met for dy/dx to be equal to 0.

Hope that helped.(2 votes)

- Shouldn't have sal wrote the interval as [6,4)? why did he write it backwards? the question is asking for the value of x bigger than 0.(1 vote)
- Intervals are always written with the smaller number first. So, even though Sal showed that the numbers go from 6 and approach 4, the interval would still be written as (4,6] and not [6,4)(2 votes)

- I am puzzled. The slope field is y(x) or not? as at (0,6) y(x) should be around -2 at it should be increasing to 0 gradient at x>=0(1 vote)
- I'm not sure what you are trying to ask. What do you mean by "increasing to 0 gradient"?(1 vote)

## Video transcript

- [Instructor] If the initial
condition is zero comma six, what is the range of the solution
curve y is equal to f of x for x is greater than or equal to zero? So we have a slope field here,
for a differential equation. And we're saying, okay,
if we have a solution where the initial condition
is zero comma six, so zero comma six is
part of that solution. So let's see, zero comma six, so this is part of the solution. And we want to know the
range of the solution curve. So solution curve, you
can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be
greater than or equal to zero, so it's going to include
this point right over here. And as x increases, you
can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing
at a slower and slower rate. And it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger, larger, it's gonna get infinitely
close to y is equal to four, but it's not quite gonna get there. So the range, the y-values
that this is going to take on, y is going to be greater than four. It's never gonna be equal to four. So I'll do, it's going
to be greater than four. That's gonna be the
bottom end of my range. And at the top end of my
range, I will be equal to six. Six is the largest value
that I am going to take on. Another way I could have written
this is four is less than y is less than or equal to six. Either way, this is a way
of describing the range. The y-values that the
solution will take on for x being greater than or equal to zero. If they said for all x's, well, then you might have
been able to go back this way and keep going. But they're saying the
range of the solution curve for x is greater than or equal to zero. So we won't consider those
values of x less than zero. So there you go, the curve
would look something like that. And you can see, the highest
value it takes on is six, and it actually does take on that value 'cause we're including x equaling zero. And then it keeps going
down, approaching four, getting very, very close to four, but never quite equaling four.