Topic 7 3 7 4 Slope Fields

Added: 2026-05-12

[00:00:05]okay welcome back today we are going to talk about uh the next two sections seven three seven four which have to do with uh slope fields um so we are going to estimate solutions to a differential equation where before we were checking to see uh they were given to us and we were asked to check to see if they were solutions now we're going to try to estimate solutions later we're going to learn how to solve differential equations but sometimes that can be quite difficult sometimes it's impossible so we're going to take a graphical approach first and this is called creating slope fields which shows the general shape of the solutions to the differential equation so at first we were given them and we were saying okay if check to see if these are solutions now they're saying find a solution and this is the graphical approach so this is our differential equation here we've got our first derivative in terms of x and y okay we're going to sketch a slope field of this differential equation now again this first derivative represents the slope of the tangent line to the curve at any point um so what we're going to do is we're just going to draw in many tangents we're going to plug in some coordinates into this first derivative and it's going to give us the slope of the tangent line so we're going to create little mini tangents so it's easiest to start where the first derivative is equal to zero that's the easiest place to start okay so where is our first derivative gonna be equal to zero our first derivative is going to be equal to zero when x is equal to zero so what i'm gonna do is i'm just gonna plug in put some mini tangent lines at x equals zero so at x equals zero if i plug in zero two if i plug in zero one zero zero if i plug in any of these coordinates into here i'll get 0 divided by something so the slope at x equals 0 the slope at that point i'm going to just draw a little mini tangent line at that point okay now where is the where is this going to be undefined where is our first derivative undefined well that's any place where y is equal to zero okay so our first derivative is undefined when y is equal to zero so this right here is where y is equal to zero so the first derivative is not always undefined on all of the solutions but in this one it is because we have a y in the denominator okay or it could be here as well okay little mini tangents all right and then you can just pick you know any x value um you know on the a b tests they may have you draw six um mini tangent lines on the bc test i have not seen them actually have you draw these um on a free response so but we're going to go through and do this anyway because you do have to know how to come up with solutions on the bc test so let's just go to this point right here this point right here is 1 1. so if i plug in 1 1 i get a slope of negative one-half the first derivative is equal to negative one-half so if the graph were to go through this point what does a negative one-half slope look like well negative one slope would be at a forty-five so again any we're just guesstimating what a negative one-half slope would look like if the graph were to go through that point okay um we can do this point here if we plug in one negative one that gives us a negative one-half slope if you plug in this point right here that would be negative one negative one you're going to get a negative one-half slope so you can go through and you can plug in every single one of these points in for x and y and figure out what the slope of the tangent line would be okay um you know that might take you a little while so i'll just go ahead and give you a couple of a couple more um this right here is going to be the tangent line would be here because if we plugged in 1 2 that's going to be negative 1 4. so it's it's going to be a little bit flatter than a negative one-half or a positive one-half okay so a little bit flatter and again these are just guesstimations okay so and we can even do some more if you want but i think that's plenty okay it says sketch a slope field of the differential equation on the coordinate plane plug in x and y into the differential equations and then draw a short line segment by estimating the appropriate slopes which is what we just did okay now most of the time they're drawn in for you if it was a free response question sometimes like i said on the a b test they'll ask you to put in 6 or 12 or something like that okay so those that's that's part a part b says sketch one possible solution curve to the differential equation that passes through the point 0 1 so it's giving us a point on the actual graph and these are possible tangent lines to that graph okay so i'm going to go ahead and i'm going to sketch in um i'm going to go through 0 1 so where is 0 1 0 1 is right here so i know that my solution is going to go through that point well if my solution is going to go through that point okay it's going to be tangent to these so i know that the solution curve is going to start something like this because i know it goes through this point and i know that these are little tangents to the graph okay so i'm just going to go ahead and complete see what i can do to complete this if i go with this curve now again here's some more possible tangents here so i know that the curve this is a possible solution to the differential equation okay okay that's kind of looks like an ellipse right so the equation of this graph would be a solution to this so that is sketching one possible solution it goes through this point and satisfies this differential equation okay so if i were notice here this is where you're undefined here okay so that is a possible curve all right so let's take a look at some multiple choice so this is the differential equation it says shown to the right is a slope field so here's your slope fields these are all little mini tangents that are drawn for you which of the following differential equations satisfies this so if i were to take any of these notice this right here this is a this looks like y that looks like y equals x plus one that's what it looks like but i'd like to start with these horizontal lines right here see these little mini tangent lines that means that the first derivative i'd like to start here that means that your first derivative is going to be equal to zero at all of those coordinates so what would that coordinate be right there okay that coordinate right there looks like it's 1 1 that looks like it's 2 2 that looks like it's 3 3. so basically what it's saying is that x if if our x is equal to 1 and our y is equal to 1 our first derivative is equal to zero so which one of these would satisfy that one plus one no one minus one this looks like it might satisfy that but we have to make sure they're all satisfied so let's keep going how about this the opposite of one plus one that satisfies that um or two two and then this one um one one satisfies it but two two does not so we can cross off this one we can cross off this one okay so now we have two we're down to two choices so let's go to something else that we recognize okay what else do we recognize um let's look at another point let's look at say 0 1.

[00:10:55]at 0 1 when we plug 0 1 into the derivative it should give us this slope right here that looks like about a slope of 1. so let's just go ahead and let's plug in 0 1 okay i plug in 0 1 into this one if i plug in 0 1 i get a slope of first derivative i get a slope of negative 1.

[00:11:24]does this look like a slope of negative 1.

[00:11:27]nope how would i plug in zero one here i get a slope of one does this look like a slope of one yes and you can pick another point if you want to you can go oh um i don't know this one right here looks like one two so when i plug in x equals one y equals two which one of these gives me that tangent line slope of one so see it looks like c is the answer okay so let me kind of review what i just did okay so let's go back up to here what did i do to sketch a possible curve okay what we did was we sketched a graph through the given point okay and we're also so the given point and using the short line segments well what are those short line segments the short line segments are your mini tangents possible tangents to the curve okay we use those to guide us okay so in the first one i actually had to draw in the mini tangents on this one all the many tangents are drawn for us we just have to find out which one of these differential equations will give us these mini tangents and again there's so many many tangents where do you start so start with the easy slope i'd like to start with zero hopefully there's a slope of zero somewhere okay so on this graph the slope it looks like the points where the tangent line is equal to 0 are 1 1 2 2 3 3 etc so i can eliminate these again i need to plug in an x and a y and i'm going to get a first derivative equal to zero so that gives me that narrows it down from four choices down to two choices and then i can pick another slope you could pick anything you want on here but i'm looking at these mini tangents and again these mini tangents look like the slope is approximately equal to one so this is going to give me when i plug in each one of these points i'm going to get a first derivative equal to 1 because the slope of the tangent line is equal to the first derivative okay so right now this narrows it down to b or c but now when i go to get another here if i go to this point or this point or this point again you're just estimating this point right here let's just say that is zero one looks like it's zero one so looking at the point zero one i need my d y d x to equal one so of the two equations that are left which one gives you a slope of one and the answer is c because the opposite of zero plus one is equal to one okay so you have to understand what these little mini tangents are all right let's take a look at the next one the bunch of mini tangents here which looks kind of crazy doesn't it so what are we going to do here it says the graph to the right shows a slope field of a function f of x and again we're trying to find what f of x is okay that's that's our goal is to try to find the solution curve something that satisfies our derivative let's see what else it says it says if the initial condition is 0 6 so that means 0 6 is actually on our solution curve so where is that at 0 6 is right here and you can see that little mini tangent right here that's that point right there is on the curve there's our tangent line right there okay it says what is the range of the solution curve for the function for x is greater than or equal to zero okay well if we know that it goes through we know that it starts at that point that's the only thing that we know and these are little mini tangents and look at this right here this looks like a horizontal asymptote at y equals four so based on everything that's given the only thing that we know is that this point is on the graph and it's going to be tangents to some of these one of these in here and it's going to approach this so that's what the a possible solution curve would look like so it says what is the range of the solution curve well what is the range of this if you remember range is from bottom to top so our range of our solution curve it's approaching four doesn't get to four and the only thing we really know is that it goes up to six anything other than that we would be making an assumption okay let's just keep looking at some examples what is a general solution to the differential equation that generates the slope field here so now we're giving a slope field here okay they don't really give us a coordinate um but let's look at these equations over here let's look at the the answers to kind of guide us it looks like we have y equals sine or cosine or negative sine or cosine graphs so let's look at our mini tangents here you see we've got a mini tangent to the curve right here there's a little mini tangent there okay there's a little little mini tangent here and mini tangent here okay so if you kind of look what graph is that you kind of sketch something in that looks like i can draw a curve tangent what breath is that that's the sine graph well they didn't give us a coordinate that the actual graph went through so we don't know if we should be sketching here or we should be sketching up here or up here or down here we don't really know so here's the sine graph and it could be plus any constant so we can translate this up and down and still you're not going to be tangent to all of these you're just going to be tangent to some of them so it looks like e would be the answer for that one okay so we found a solution by using slope fields and if you recall back in section seven one actually seven two um they gave us the solution to verify okay we're going to eventually get to the the trying to find the solution ourself this is the l this is the um the graphical way of trying to find the solution okay all right let's look at example number five last example here's our differential equation describe all points in the x y plane for which the slopes are positive which the slopes are positive so in order for our slopes to be positive what must be true well if x squared times y minus 1 needs to be positive what must be true well x squared is always positive so that means that y minus 1 has to be positive because if this is always positive and you want your all your slopes to be positive then this also has to be positive which means what that y has to be greater than one okay so what else must have to be true well neither one of these can be zero right because we want our slopes to be positive and zero is neither positive or negative so x squared oops x squared is not oh is not only always positive but it cannot equal zero because at any time x squared is equal to zero zero times anything is going to give you 0. and again we want all of our slopes to be positive so what must be true okay so that's this is the first thing that must be true y must be greater than one not equal to because that would cause it to be zero and the second thing that must be true is x squared cannot equal zero so if those conditions were satisfied you would be giving me a slope field where all the slopes are positive okay so again what is what are these differential equations just think of this what is d y d x well that's a derivative right d y d x that is called a is a differential equation that's a first first order differential equation because that's the first derivative so d y d x is a first order differential equation okay so think of it as a formula to produce slopes or tangent lines slopes little mini tangents at any point okay so that's how the slope fields are created you're plugging in points and you're trying to find what the possible the slope of the tangent line would be at those points if the graph were to go through it so that's seven three and seven four dealing with slope fields