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Find the following limnits: (a) Lituj 2) (3,3)-(2-3)E-+ (b) Jiljl (5,0)-(4,3) T-w-12+0 2. By cousidering dilferett paths ol approach, show that the fuuetion f (~,0)...

Question

Find the following limnits: (a) Lituj 2) (3,3)-(2-3)E-+ (b) Jiljl (5,0)-(4,3) T-w-12+0 2. By cousidering dilferett paths ol approach, show that the fuuetion f (~,0) = bns ED linnie a8 (,9) (0,0)

Find the following limnits: (a) Lituj 2) (3,3)-(2-3) E-+ (b) Jiljl (5,0)-(4,3) T-w-1 2+0 2. By cousidering dilferett paths ol approach, show that the fuuetion f (~,0) = bns ED linnie a8 (,9) (0,0)



Answers

Find the number of paths from $a$ to $e$ in the directed graph in Exercise 2 of length
$\begin{array}{llll}{\text { a) } 2 .} & {\text { b) } 3 .} & {\text { c) } 4 .} & {\text { d) } 5 .} & {\text { e) } 6 .} & {\text { f) } 7}\end{array}$

Okay, so for this problem were asked a couple of questions. First, we need to find the Grady. INTs of F as well as the derivative of the path function are so we'll start with the finding the Grady int of f, which is going to be composed of the partial derivatives with respect to X and why? So if we take the partial derivative of F with respect it X, we're going to get ah, why squared? And then if we take the partial derivative with respect to why we're just using the power rule in treating X is a constant just as a reminder. So we'll get to X y. And that is the Grady int of F um, Now, to get the derivative of the path function are we're just gonna take the component wise derivative as usual s. So this is just gonna be some power rules here for one half a t squared, uh, bringing the two down and subtracting one from the power we're just gonna get t and then for T Cube. We're gonna get three t squared. So that is the derivative of the path function are on DNA. Now. They want us to take the derivative of the path function are composed with F. So to do this, we can use the formula. We can take the Grady int of f and just multiply it, uh, by the derivative of the path function are of tea. So we already have the excuse me, and we're gonna be evaluating. Ah, the Grady int of f at are of tea. So what will be composing the function first and then taking the Grady int of it? So I'll go ahead and show that right now. So we already have the Grady int and the Grady INTs of f is why Squared and two x y, and we're going to plug in our X and Y values from our of tea. I remember that are of tea is equal to one half two squared and cubed. So we're gonna plug that into the greedy int. So we have t cubed squared and then to X is one half t squared. And why is t cute? And we're just gonna be doing some algebra here and clean that up a little bit? Um, we're multiplying. Ah, taking the powers and multiplying them because we haven't exponents raced to an exponents. Eso we're going to end up with this will be t to the sixth. And then, uh, this two and one half will go toe one, and then the powers, uh, will add to five. So we get to the fifth. Okay? And now we're ready to go ahead and take this Grady int evaluated at our of tea and take the dot product with the derivative of our E. And this will be t to the seventh plus three t to the seventh, which was just before tea to the seventh. And now, finally, we want to evaluate this At T equals one and negative one. So we just plug each of those values in for T. So for the first one of four, well, that's just gonna be equal to four when we plug in positive one. And then for negative one negative number Racer nod. Power is negative. So to be negative for and that's the final part of the problem.

F as a function of X and Y are vector in T. They want us to find LF and our prime of T and then find the derivative with respect to T of F f r t. All right. So dlf that's a vector is the derivative with respect to X comma derivative with respect to why? Okay. Our prime of T just find the derivative of each piece. So T and three t squared. Okay, So then Dell f dot r prime equals t y squared plus six T squared X Y. So at uh maybe a chef found the values first. Let's see. Um So this is T times Y squared. And why is t cubed plus 60 square times X? Which is one half T squared times Y, which is t cubed. So you get t to the seventh plus three T square T to the seventh 42. The 7th. So Dlf dot our prime At T equals one is 4 and l f dot our prime at T equals minus one is minus folk.

Right there in this problem, we're given a function f and this path see, in part, they were simply has to find the ingredient of F and C problem of tea. So the great end of it was Follow the definition, Which is that partially deaf by ex marshal of battle I So this will be nice and quick. The partial of F with respect to X, you can see it's just y squared partial of f with respect to why gonna treat ex like a constant and we will get to X y. So that's the Grady int and for C prime of tea in derivatives of these vector value functions are very straightforward. We just take the derivative and each slot. So the derivative of 1/2 t squared is T and the derivative of T cubed, of course, is three t squared. And we're done with party no art be. We're asked to evaluate. She's a derivative respected T, uh, half of CFT at two different points. T equals one and T Eagles with minus one. Okay, so before we even plug in points, uh, the chain rule for paths company tells us that we look at the greedy int of f. Then we'll dot that results to the dock product with the derivative The path. See, in other words, just thinking about what this means. Looking at her answer from Part eight Ingredient of F uh, waas, easily of why squared looks why? And if we dot that with her answer for her seat prime was t three t squared, we find Will this kit I squared Times T plus C two Ex Wives and three T Square and I'll give a six x y t squared. Okay, now it's it's a matter of looking in our two points so on T equals one. That means we'll plug in tea for one for all the teas. And moreover, when tea is one, we know exactly what X and y are also won t is one we can see that X is gonna be 1/2 times one square half. And why is going to be in one of the third, which is one So T equals one? We can see. Let's just see what we get. So Y squared is one times one is 16 times in half kinds, one times one squared, so one plus half of 61 33. That'll give us four incidents. The T equals one case actor ever comes out before and finally was to the t equals minus one case. We could have, like a negative one for all the tea's handing. And that's just look at our original definition of see when t equals negative. One x will be 1/2 of negative one squared. So once again we have. And why? Negative under the third is negative one. So no negative one sport is one to use. Negative one. So, okay, why is where t will give us minus one plus six times, half times, negative one and then T square will be a positive one. So the negative one minus three or negative before and we're done, hopefully it helped.

In this problem was just going to determine whether or not there is a path from the first for sex to the second protect but given figure. So first we're looking at from A to B and there is a direct path from A to B. So, yes, this is a path. There's a past, then really you're looking from B to a We can go from being that being and back to a here and over to a Sorry. So this one also has a path from B to B. We can used maybe from being Teoh over to a and back to be so there is a path here. Keep in mind that some of these may have more than one possible path. Um parte de from a trying to get from a t e. Um, we can do from a to be t e or we could have also them from a to d Teoh even one work. So there is a path here run be to do. We can go from B E to Dean. Still a path this exist here now from state to D, we can go from C you mean city that shows that there is a path from de do you? Well, we can go over from duty and then Bactine t So, yes, there is a path their own. So Okay. Now from E T. A. That's because there is a direct path from each either Maybe a little longer one. But this one is the shortest. I think so, yes. You have a bath there. And finally, from E to speed we can go from you a being and then to see so the results of our final


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