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Sketch three approximate solution curves to the given Slope Field generated by computer software:...

Question

Sketch three approximate solution curves to the given Slope Field generated by computer software:

Sketch three approximate solution curves to the given Slope Field generated by computer software:



Answers

Sketch three solution curves for each of the slope fields in Figure 11.18.

In the problem, the three solution cards for the slope field in figure so on and figure that is 11.16. So, so this is the question figure and the three solution carbs is soon here, so this is the answer to the problem as Figure 11.16. Now we have the other solution. So here the three solution Cobbs Food the yeah, slope fields In figure 11.17 edge soon here. So this is The answer for the figure 11.17. Hence this is the overall answer.

So in this problem, we're given a bunch of to slope feuds and were asked to draw three solution curves. Now, to be a solution, curve to a slope field, the slope field has to be providing tangent lines to that curve. So you can think of this as basically a game of connect. The dots were also given some sort of inclination about the direction that the graph is going to have to take. So, for example, this would be a perfectly valid solution curve, because all of those blue vectors are tangent to that red line at each of those points. Another valid curve would be something of a similar shape. But maybe a little further down and a little wave year like that, um, and and the And if we needed 1/3 1 we could once again grab something from the middle. But something I'm gonna do is one of these on the outside, where we have something that goes kind of down like that Sorry, down like that. But then also comes back up here. I kind of like an ass in tow. On this other side, we play basically the same game. If we're on the top, we can have something that looks sort of like a sine wave if we're on the bottom as we get closer to the bottom, we have something a lot like that, but that gets flat for a bid. And then if we're in the center, we would actually be able to stay tangent to most of those lines with just a straight line.

Actual equation. We were asked to use this occasion. Grab the slope feel solutions. The creation were given one before the draft. But a slip of you so particular take Why to be some constant c Every young numbers X Any remember half that well of wine going to be 1/2. Why one? In other words, say it's easy. Why zero at every point along in particular value. See slow any Why it c zero going to you zero So we have wars on a one. Hey you! Why is it 0.5? Have a all right by 11 Why is it 2.5? Have a good one is very Why is it warm? What? Why is it one from five quarters? Have a look off one. Do you see this? Increases is broken extending from the way already today. Wrong It is anything away from the organ? Likewise. Why is negative? No one had a stroke of May 25. Get this is very rough. Both feel bad around here. I think the general idea Foreign three solutions. It's different initial values. For example, let's take initial value. Why is your people zero? They don't have a point goalless and like, are you? We'll see if this is simply the bars on one. It's just the horizontal one, Mother. Initial condition. Why you one I'll draw this red. You're one. And you worked too. Graphic. Simply follow you See that? This is a That's access, baby, isn't there? Looks a little bit right now. One more initial value will take a negative one. Now we've got one for below. And on the and apparatus Marine, we have a 0.0 native one. You said his expert. His name. It's a purchase. The absolute Y zero. Mr Perfect is that purchase. This is another kind of experiential ocean support and reflected across the X axis.

So for this problem were asked to verify our solutions for, um, problem number three to be able to determine if the slope up fields are going to be correct. So the purpose of having to do this is solving them using differential equations and a lot of the integral properties we've already learned. So that way we don't have to do so Fields forever single problem to determine what the's values they're gonna look like given initial conditions. So the going gets started into this. The first thing I want to do is I want to rewrite my white prime to be D Y over de X and then equals one minus y. This makes it a little bit easier, because now what I want to do is I want to go ahead and separate out all of my y variable, um, items over with on the same side. And then, of course, my, um de X items. So what I want to do is I want to leave one minus y together. I don't want to split these up, so I'm going to go ahead and, um, multiply d X to the other side. So now I'm left with D Y equals one minus Y d x. And then I want to divide one minus y to the other side. So now I have my why variable of one side, my ex variable on the other. So what I want to do is I want to go ahead and integrate both sides so my left side is gonna be based off of in terms of why, and the left side's gonna be in terms of X. So one thing I'm gonna have to do at this point, as I'm gonna have to do some use substitution. So what I want to do is I want to do you equals one minus y. So now if I take the derivative of this, I'm gonna have the derivative of negative Why? Which is going to be negative one. Do you? I'm sorry. Do you? Why? So that if I want to solve this in terms of d y, I can rewrite this of as negative D'You so negative d y equals d. You can also be d y equals negative, dear. So that negative can switch. So now I'm ready to go ahead and go through and switch these out so I'm doing the integral of D U. And this is gonna be negative Over you equals the integral of d X. I'm not gonna rewrite DX because that one's a pretty simple integral. So now what I'm ready to dio is I know that the integral of I can rewrite this first off because I don't need that negative inside the integral. But I know that the integral of one over you is going to be held in of you and the integral of D X is just going to be X plus a constant. Now, technically, I could go ahead and do Ellen of you Plus C. However, I'm gonna go ahead and not do that because whenever I go to solve this with the initial values, this one constant is going to take care of all of that combined. So I don't really need to separate all of those out. So now what I want to do is I want to divide by wine, so I'm gonna have negative X minus C. And then what I want to do is I want to raise all of this to the E power. So what's gonna happen is this e and Eleanor going to cancel some of you left with you, and I'm gonna have e to the negative negative x minus c. So what I want to do instead is I'm gonna move this constant outside of that because again, like I said, when I start filling all this and it's going to take care of all of that on its own, So now I'm gonna hav e to the negative X now, Earlier, I substituted you as one minus y. So what I want to do now is I want to rewrite that back in there. So when a un substitute this and I want to sell for, why equals some in a subtract one on either side and then divide by negative one. So in other words, I'm just gonna change the sign of every term in the equation. And so this will be my the equation I'm going to use. However, I still have my see value, my constant. That is gonna very based upon the initial, um, values that I'm given. And so for problem three. There were three different initial conditions given, and I'm gonna use this equation toe look at each one of thes so scrolling down to the 1st 1 I'm given that why of zero equals negative one and from problem three, this was the slope field that I had created. So what I want to do is I want to go ahead and get put in my y equals C E to the negative X plus one. So I am given that X equals zero and why equals negative one, some going to substitute those in. So I know that Edith zero is simply just one. So now I have negative one equals C plus one. So I'm going to subtract one toe either side, and I'm given that the value of C is negative two. So my equation for this one would be one minus two e to the negative X or negative to eat of the AK negative X plus wine. So this would be my equation. So then the question also asked us to compare this to what we came up with. So I went ahead and graft the equation, and this is what it looks like. So if we could go back up to the slope field I had started at one, uh y equals one and I ended up at the equation. Why over here were y equals one. So if I go here to the equation, we cross where? Why is negative one And we go up to where? Why is one until we meet it. So this would verify that this equation does, in fact, match our slope field. So going on to part two or part B, if I used the equation, why equals one plus c eat of the negative X and for this initial condition and given x zero and why is one so similarly to what I did in the next? In the last one, I know that e to the zero is one. So again I'm left with one equals one plus c. I'm going to subtract one on either side. So that's going to give me that. C is I zero. So this tells me that my equation is going to be why equals so since this is zero, why is gonna equal one? So if I look at my slope fields, my entire equation is on were Y equals one. And if I look at the one that I've already graft, I can see that why equals one? So this, in fact, does confirm that the slope of this that this slope field graph is correct. And then finally, for my last example, I have Weiss of zero equals two. So if I have, why equals one plus c eat of the negative X and for my initial condition, my X equals zero. Why equals two? So I'm going to substitute those end to find out what see is and just like the previous to eat of the zero is wine. So I'm left with two equals one plus c. I can subtract one on either side and that gives me that C equals one. So my equation would be one plus e to the negative x. So, given that if I look at my slope field that I had gotten for problem number three, I noticed that I crossed the y axis at two and I curve down into where I get where y equals one. So if I go down to the graph for my y equals one plus eat of the X, this is what would happened if we noticed across the Y axis at why equals two and curved down until I hit were Y equals one. So this matches my slope fields graph into this confirms that these are in fact, the same.


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