Stuart S. answered • 04/19/23
Patient, Knowledgeable Math Tutor; Aims to Make Math Fun and Intuitive
It's difficult to re-create the formula of a slope field just by looking at it, and very difficult if you don't have a tool to try some formulas and see what they look like. So before we begin, I would highly recommend checking out this link, which is a Demos graph that lets you try play around with different formulae and see their slope fields:
https://www.desmos.com/calculator/jkajcrvdhm
The problem is definitely much easier to tackle when you can actually try some answers!
In any case, let's get into it. When you're trying to reverse-engineer a formula for a graph, whether it's an explicit function, implicit function, or slope field as we have here, you want to look first for special features the graph displays, as these contain critical information about the formula. Special features include:
- Discontinuities
- Zeroes
- Asymptotes
- Maxima/Minima
There are other features to look for depending on the context, but this is a great list to start with, and we will actually only need to look for discontinuities and zeroes here.
So the first thing I notice is that along the x-axis, the slope lines aren't drawn, which means they're completely vertical. This means at y=0, the slope is always infinity. This is what a discontinuity in a slope field looks like, and it tells us that there must be a "y" in the denominator of our formula.
There don't seem to be any other discontinuities, which tells us that the denominator probably only contains y's. If there were things like (x+1) or (y-3) in the denominator, we would see more discontinuities where those terms equal 0.
So, moving down the list, are there any zeroes? Remember that a zero in a slope field is anywhere the lines are horizontal, and we can find them by looking for where the slope changes from positive to negative.
It's not very easy to see from the picture, but there are some zeroes here. We can see the lines at x= -2 have positive slope, but at x= -1 they have negative slope, and then at x=1 they have negative slope but at x=2 they have positive slope. So somewhere around x=-1.5 and x=1.5, the slope field is equal to 0. (We're not going to get exact values with a rough picture like this, so ±1.5 is a good guess.)
That's enough information to start putting guesses together. Again, we know that there must be a y in the denominator, and we know that the numerator is 0 at x= -1 and x=1. So, maybe the function is dy/dx = (x+1.5)(x-1.5)/y? Let's use Desmos like I mentioned before and try it!
https://www.desmos.com/calculator/tcellckz91
Hmm, that didn't work so well. The values around x=0 seem pretty good, but as we move farther horizontally and x gets larger, the slopes get way too big. All of the "x" terms are in the numerator, so either the numerator is too big, or the denominator is too small. We need both of the zeroes that we have in the numerator so it would be easier to make the denominator bigger. We could change it to y², since that has the same discontinuity at y=0 that we saw earlier. Let's try that:
https://www.desmos.com/calculator/ztkocxw61d
And hey, that looks pretty much right! Again the graph provided is not exact, so we can't sweat the details too much. I'd say dy/dx = (x+1.5)(x-1.5)/y² is a great answer to the question.
Hopefully you learned some tricks for identifying slope fields; like I said at the start, it is not an easy thing to do in general and there's way more that I didn't touch on in this answer. But you can see that just looking for discontinuities (places where the slope is vertical) and zeros (places where the slope is horizontal) was enough to solve the entire question in this case! It's a very powerful trick.
