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Exercise 8.2.52: For the systems below:Find all critical points (equilibria).Determine the linearized systemt for each Of the points in (i.). Classify each of the ...

Question

Exercise 8.2.52: For the systems below:Find all critical points (equilibria).Determine the linearized systemt for each Of the points in (i.). Classify each of the critical points (equilibria) . If the critical points are not . spirals or centers, find all eigenvectors: Sketch the global phase portrait. Include the eigenvectors for the linearized system at each critical point and draw arrows n the solution curves to indicate the direction of flow: x' =x-xty,y = 212_8.2 STABILITY AND CLASSIF

Exercise 8.2.52: For the systems below: Find all critical points (equilibria). Determine the linearized systemt for each Of the points in (i.). Classify each of the critical points (equilibria) . If the critical points are not . spirals or centers, find all eigenvectors: Sketch the global phase portrait. Include the eigenvectors for the linearized system at each critical point and draw arrows n the solution curves to indicate the direction of flow: x' =x-xty,y = 212_ 8.2 STABILITY AND CLASSIFICATION OF ISOLATED CRITICAL POINTS 277 b) x' = -3x+y,y' =-y+x2 c) x' =-x+y?,y' =x+2y d) x' = 2x+ y,y' = ~y+x2



Answers

Identify the coordinates of any local & absolute extreme points and inflection points and graph the function. y= x-2sin(x) for 0 less than or equal to x less than or equal
to
2pi.

Okay. This question gives us this nonlinear system and wants us to find a way to take this system and approximate a solution. So really if you're doing this with a calculator, all you do is graph each of these and just find the intersection that way. But in our case let's try and estimate it out to break lee. So what we should do here first is let's just eliminate one of these variables. So if we write these equations again we can write this as why minus two Synnex is equal to zero or okay, Y plus three X is equal to two for the other equation. And see we have y right here on both of these. So let's just subtract the two equations to eliminate it. So we'd get negative two. Synnex minus three X. Is equal to minus two. So multiplying by negative one. We want to find where to sin X plus three X is equal to two. And now looking at this, we have to figure out what we want to do with that sign term and a really bad approximation, but still one that will use if we're kind of small is we say that sine of X behaves approximately like X when we're close to zero. So if I just replaced this with an axe mm we can get a rough solution. So we'd get five X. It is roughly equal to two or X is roughly equal to five or 2/5. And that's kind of small. So our approximation won't be absolutely terrible. And then we just have to back substitute for why? So we said that Y plus three X is equal to two. So then why plus 6/5 would be equal to two? So why would be equal to 15th minus 6/5 which would be 4/5. So this is our approximate solution. And if we wanted to we could actually compare this probably just by graphing and as most to see what the actual solution is and doing that I pulled up a day's most graph are actual solution is pretty close. Its approximate actual solution is more or less 0.404 Yeah. And then 0.787 and noticed that that's very close. 20.4 comma 0.8, which is are approximated answer. So this is actually pretty good.

And and we are told to find the critical points. And if it's if those points are maximum so find critical points and if they are maximum ins so first thing I'm gonna do is I'm going to rewrite this just to make it easier for me because I like to work with fractional exponents that this route notation because then it's just applications of the the the power rule and they were gonna get to the quotient rule but here we go. So we want to find the derivative of this, that'll tell us our tangent points. So we'll do take the derivative of it. DDX So we're going to get Past this function through that little derivative bit. So you get 1/3 times X cubed. This is three x squared. We subtract the one so we get minus two thirds times the derivative of the inner which is three X squared minus six X. So that's our function. But we can and this is F prime dex. So we can rewrite this as let's see what we rewrite this as we can rewrite this as X squared. Well before I do there's gonna be three X squared minus six X. All over three times X cubed minus three X squared to the two thirds because that negative exponents will be the reciprocal. Um But I'm just going to cancel out the three with this three and that six will become a tooth to make working with this little easier. So the critical points will be where F prime is equal to zero. So we want to find F prime of zero. And so we don't really care so much right now with the denominator is going to be, We just want to see what is this thing called? zero. So You said equals zero. And that gives us X squared minus two X zero. And we can we can look at this and see what the values are. I can see already X zero. It's kind of one that stands out. But you can also To the old at two x too. Bye bye. X If X is not zero, That is and you get X is two. There we go. So we get our values of X0 and X is two. Those are critical X values. But we want to find the points. So we substitute these in for original f function zero. Well this is actually zero because 0 cubed 0 -3 times zero, And F two is let's see what we get, we get the cube root of eight right minus three times two squared is four. So last 12. Uh this and this is the 1 3rd power. So this is negative for the one third. So our critical points are 00 and two comma negative for To the 1/3. Those are critical points. Now the question of the max of the men's. So for that we're going to I'm gonna clean up the board here that because what we're gonna do is going to take the derivative of this derivative function of our first derivative to get the second derivative and use the second derivative test and see if it's a maximum. So we're going to take the derivative of this thing. I'll do it in this color. I'm going to pass this function through the another derivative. And this is going to be a variation of the we're not a variation but an application of the quotient rule. So it's the denominator times the derivative of the numerator. My words are faster than my writing times the derivative of the numerator which is two X. And this too. Yeah. Mhm minus the derivative of the numerator, X squared minus two X times the derivative of the denominator. So this is going to be kind of beastie. We get two thirds. We're taking the derivative of this. Execute these three X squared. This becomes negative one third times the derivative of this new This inner functions we get three X squared six X. Holy Cow. All over dying. Mhm X cuba is three X squared To the 4/3 because we square. And this is F double prime X. That's how the second road of test works is we substitute in our X. Y is here. And if it's positive then this is a minimum. If it's a negative then it's a maximum. So that's what we're gonna do. We're gonna sub in double price F double prime of zero. So this I'm not gonna you can see the subsidy in 0000. That's zero terms that cake. I'm gonna racist. And second but these are all going to zero out because this zero and if we do this here zero squared minus two times zero is zero. So all this is going to go out. But hold on it doesn't mean it's zero because look at the denominator we get 0/0. So that is indeterminate form. So this means It's probably not differential there. So something going on at that 00. So not a I can't really tell there's nothing it's not differentiable. That's what that means. So let's look at F prime prime a double prime that is of two. So here we're gonna substitute in to appear. No, I'm just gonna kind of go not do every substitution because we don't need to because here's zero. Or excuse me, two squared is four -2 times two is 4 sides zero. And then we don't even need to worry about these because this is all zero times that stuff. So it just cancels out. So it's fun when we get let's see this is we get eight- Let's see that's going to be three times 4, 12. All to the 2/3 and then we get two times 2 is 4 -2 involved. That is over. Let's see this is going to be eight. This here's another 12 To the 4/3 and we really only care if it's positive or negative, it's going to end up being positive and here's why because you get negative for Yeah, two thirds times two all over negative. Four To the 4/3. And what we're gonna get to see some some nice things canceling out. We can get this to can't divide out with this. So it's that four here, this will go to two. So we get to over negative for two thirds. And what's this going to be? This is you might say that's a negative. No, it's not because we square it and then we can we can break this up by squaring the denominator. So we get two over, this is going to be negative four squared. So 16 to the 1/3. It ends up being that sound you heard was my book just fine. Mm is .79 37 which is positive, which means that this is a minimum and this we can't really tell here. So let's look at the graph and sure enough, here's the original function. It's not differentiable here because it's a point. And here is that the minimum value of access to that we found and there you go. So at that zero, it doesn't it being a local maximum, but not a not differentiable at that point. So there you go

The inherent formula for the area edge of a triangle of sites length X. Y. M one satisfies the equation. 16 times age squared equals F. Of X Y. Where f of X Y is equal to negative one plus two X squared minus X to the fourth plus two Y square plus two X square one square minus Y to the fourth. We want to find all the critical points of F. And classified each one of them. So first we get a five the critical points. And for that we need to calculate the derivatives, the partial derivative of F respect to X and Y. So the Persian derivative of F respect X. At any point X. Y. Is equal to for ex minus four X cube plus for ex why square. And that's all. Now respect to why partially rhythm of F. At any point is for why glass or X square Why minus four wife to the queue to the third power. So to calculate the critical points of F. Okay, we have to solve the following system of equations. The system is formed by equating each partial little bit is zero. So we get four eggs minus for X cube plus for x Y squared equals zero. And for why plus four eggs square Y minus four. Y cube equal. Super. Now we can divide by four. Each of the equations to simplify a little bit. And we get x minus X cube plus X. Y squared equals zero. And why plus X square Why minus Y. Q. Equals zero. And we get in the first equation we get a common factor eggs out and we get X times one minus X square plus white square. That is zero. And the second one we get y common factor get one plus X square minus why square equal zero. Okay so we got to solve this system here and we have some possibilities. The first one? Yeah. Eastern ex and why both be equal to zero? That case we we'll have solution because putting eggs zero and 10 here we'll get inequality. That is true. So this generates the point 00 as a critical point. Now we can have this possibility X equals zero. And why different from zero? In that case if Y is different from zero. Following the second equation here is one. We got to have that. The other factor that is one plus x square Minnesota square gotta be zero. So we get from here that X equals zero and one plus X square mints with squared equals zero. And because X equals zero, putting that here we get one minus Y squared equals zero. And so why square is zero? And from there why is more or less one And that give us two critical points. Both with X. A quint zero here and why equal to one? Give us one point equal to negative one. The other point. So okay we get here points 01 and zero anti one. The option is X different from zero and Y equals zero. So you now the first equation here this one because X is different from zero. We must have that. The other factor is zero, that is one minus X squared plus Y square, so one minus X squared plus y square zero. And why he's zero. And because why zero put it that here we get one minus X were equal to zero, that is X square equal to one. And so X is equal to more or less one. And again we have two critical points, both with white gold zero. So we get the point 10 and negative one cereal. And the fourth possibility is X different from zero and why different from zero? That is both different from zero. And is in both equations of the system here we get that. The other factors get to be zero, that is one minus x squared plus one square gonna be zero and one plus X squared minus one square. Going to be sued. So we're at that here, that's the first factor they get to be zero and the other is one plus a square when it was square goods here. Okay. And now from this two equations, we get from the first one, X squared minus one square equal one. Put in those terms to the right and putting this one to the right, we get x squared minus y squared equals negative one. And because we have the same quantity here, this would imply that one equal nearly one, which is false. So this two equation has no solution. So there is no solution. Two the equations, one minus x squared plus y squared equals zero and one plus x square minus one squared equals here. Or put another way is not possible. That one minus X squared plus y square and one plus x squared minus one square b zero simultaneously. So we have found finally 00 as the first critical point. Then these two critical points 010 91. And these two others 10 and 91 0. So we can say that the five critical points of F okay, are 00 10 anti 10 01 and negative zero negative one. Now we want to classify those critical point now for that we use the following result. Mhm. If a B he's a critical point of F. Yeah. And we're supposed that critical point of F of X. Y. And the second order derivatives which we know are three second order respect to X. Second order with respect to Y and second order mixed respect to X and Y. And these those three derivatives of F are continuous. Not necessarily in the whole plane, but in the region containing the critical point, let's say some region containing the critical point A. B. And if we define D of X Y as not X wise, Ary of Abyan meant of the critical point. Okay, that is the second order partial derivative respect X at a B times a second order derivative of F respect to Y. At the critical point of B minus the mixed second order derivative of F at a b square. So we calculate this number which we call the we have the following options. Yeah. Number I if T is positive and the second order for partial derivative, respect to eggs at the critical point is positive, then if has uh relative mm minimum at the critical point, maybe if the is positive and the second order partial derivative is negative at the critical point, then it has a relative maximum at the critical point. Okay. If T is negative, it has a subtle point at a B. Mhm. That is a point where the function has cratered and less value than the value of the point. There is there is a point that is not maximum or minimum. And if T equals zero, there is no conclusion. That is. We could have either a maximum relative maximum or minimum or set a point or whatever classification at the point. The critical point maybe. So more information is needed or more work is needed to classify the griddle point at that case. So these are the options and in this case is important to remark that it doesn't matter if you use the second order partial derivative respect to wise it will be the same thing. And that is because this quantity being positive became proof easily that the second order but derivative of f respect to X and Y have the same sign. So we can use any of them in the items. One and two. Okay, so now we calculate for each point for each critical point we calculate the and use these uh information here. So we start with critical points here. Zero. Mhm. Okay. Okay. Okay. There D equals for that. We need to calculate first critical point. The second orders are preservative. We have only the first order partial derivative here. This one. This one. So now we calculate the second order partial derivatives. So you had that. Okay, We are going to use this notation here for the second overpressure derivative. Remember that? This is the same as this. So we're gonna use this compact notation in this case for the second older pressure, the relatives breeze equal to oh four minus 12 X square plus four was square. Second order pressure derivative respect. Why of F is equal to four plus four X square minus 12 Y square. And the mixed second order brush, a derivative with respect to X and Y is eight X. Y. Remember in this case because they are continuous, all these functions are continuous because they are pretty normals. And then we're going to have that the mixed order process narrative respect to X and Y and X. That is in the inverse order. It's got to be the same. Okay, given that these old conservatives are continued, in fact, they all continuous implies that we can use this classification theory in here without a problem with that because we had in the theory um uh ask for continuity of the second order process derivatives in regions around the critical points. And that will be the case in this example. Now we can made the calculation. So we start with critical 0.0 and here we have the equals. Okay, uh do you serve it to you? 00 is for times deserve a tive 00 is four also minus and the mixed, personally rated at 00 is zero square, so is 16. She's positive. And now we even because it's positive if you see here we get to evaluate now the second older pressure derivative at that point. And we knew because we do it here that this is for which is positive. So these two things implies that if has a relative minimum at Uh huh 00 So that's the first result. Yeah. Now. Uh huh. Now we can talk about mm all other critical points left which are once you're through 91 0 01 and zero negative one because we can see that um Yeah okay at all those points we're going to have the same value for X square on my square which appear in the second order pressure conservative respect to X and respect why? So for example at 10 will have four minus uh 12. That's gonna be negative eight. And the other second order personal narrative will be the same. Mhm. Okay. So we're going to do it separately in order to be sure that it is clear what we're doing here. So with zero for example. Mhm. So in this case the is so you get the uh deserve a tip here is four minus 12. It's near the eight. This one here at 10 is eight. Yeah minus. And the mixed order the mixed second order for security will be zero because it's X. Time Y zero. So again negative 60 for that is is negative. So we have a set of white. Let's see the other critical point. Anti want zero. And as we said the calculations of they made the second order pressure there it it will be the same value because we have X square and white square all around. So it's going to be the same again we have a subtle point. Mhm. As to no 01 Yeah now X zero always one. So you had eight. The first one and the second through this one year that's 21 is negative eight. Okay. And a mixed partially already zero. And as we see here, the result is again negative 64. Okay. So we have a subtle point. And for the critical point, yeah, 09 81 it is clear that he is going to is going to be again eight times negative eight minus serious square. That is negative 64. Which is negative seven point. So it has saddle points at 10 21 0 sierra one and 0 81. And sorry. Mhm. Okay. The end. Okay. Better. So we have these results for these four points and the other critical points here, zero. It has their relative minimum. So we have classified the five critical points of the function.

Yeah, that's probably want to find any local and absolute extreme points for why is april two x minus two sinners. Yeah. Just on the interval 0 to 2 pi. I remember looking for local and absolute extreme. We use what is called the extreme value theorem and the extreme value theorem says that the absolute maximum and the absolute minimum value of a function on a closed interval and 0 to 2 pi. Look you're either at the zero at the two pi read any critical points that lie in between. So let's look first for critical points. Some critical points where derivative is zero. And so we have dy the act is able to one minus two co signers, we said that equals zero. And then when we solve this this gives us the pro side effects is equal to one half. And so you look on your unit circle for where the co sign is equal to one half and they would be at pi over three and five pi over three. And so those are your critical points. Those are critical points. So we need to check Y of zero. That was our left in points. 500 minus two times a sign of zero. That's just zero. We need to check why of pi over three which is pi over three minus two. Sine of pi over three. Uh huh. Which is pi over three minus route three. Okay, you need to check our 553 So this is five pi over three which is five pi over three minus to sign five pi over three. So this is five pi over three plus route three. And then we need to check to pie. Yeah this will be two pi minus to sign of two pi. Sign into pirates zero. And this is just this this is just equal just equal to buy. And so these are all of your options for absolute men in absolute maximum values. So the smallest of these will be your minimum value. And so looking for the smallest of these. Uh huh. That would be a pi over three minus route three. That your minimum by. And then they just want to be your maximum bet and then be five pi over three plus route three inside for maximum back. So we have the minimum value and we have the maximum value. Now we also want to find any inflection points and inflection points are where the second review of Israel. Now we knew that dy dx is you for the one minus two. Cosign X. And some mines are second derivative. We just take the derivative of this. It would just be too synapse. And setting that equal to zero, tells us that X zero high or two pi zero to buy our boundary points. That means we have an inflection that actually equals pop inflection at X. Equals pine. And so in order to draft this again, we're just graphing in on 0 to 2 pi. Was graphing up on 0 to 2 pac. When we plug in zero we got zero. When we identify over three, that's when we had our minimum value. We put in power three. We had our minimum by sort of a pi over three. That was our minimum. I had to be less than zero because this one was a small one. So it's down here we poured in five pi over three. They gave us our largest value according to pilot to give us two pi back and then a pie. We had an inflection point. So pirate needs a change from concave up. The sun came down like so and so here a pie. It changes common cavity for us. That's our graph should look something like this


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