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Puwet Ling Pulet lrom last week: From last week the functiondist(x,y) = Just pick point; make Rucs_ somewhere between Hickory; Boone and Asheville: (Based upon your...

Question

Puwet Ling Pulet lrom last week: From last week the functiondist(x,y) = Just pick point; make Rucs_ somewhere between Hickory; Boone and Asheville: (Based upon your coordinates, in miles, from last week ) Your guess: P = miles; miles) Now, evaluate your total distance. Use multivariable calculus to: Find the critical points Ok, solving this pair of equations by hand hurd' recommem that you Use Web App to solve the pair of non-linear equations_ GeoGebra will do this. So will Wolfram $ Alpha

Puwet Ling Pulet lrom last week: From last week the function dist(x,y) = Just pick point; make Rucs_ somewhere between Hickory; Boone and Asheville: (Based upon your coordinates, in miles, from last week ) Your guess: P = miles; miles) Now, evaluate your total distance. Use multivariable calculus to: Find the critical points Ok, solving this pair of equations by hand hurd' recommem that you Use Web App to solve the pair of non-linear equations_ GeoGebra will do this. So will Wolfram $ Alpha You might find easier #pp Go look? Use the Discriminant verily which ones are minimums. (Upload photo of your answers and your work ) Compare your guess to Your calculated minimum. How close YouT and your distance? That how many miles of Power Line were YOu able save by using calculus? Puwer Line Puzdle d} from last week: Make guess for two points and follow the guidelines and answer questions similar to #2 above: Hints: See the hint above in 2 for using the web to solve the system of equations. You can skip the portion about verifying (the discriminant)



Answers

In Exercises $69-74,$ you will explore functions to identify their local extrema. Use a CAS to perform the following steps:
$$\begin{array}{l}{\text { a. Plot the function over the given rectangle. }} \\ {\text { b. Plot some level curves in the rectangle. }}\end{array}$$
$$\begin{array}{l}{\text { calculate the function's first partial derivatives and use the CAS }} \\ {\text { equation solver to find the critical points. How do the critical }} \\ {\text { points relate to the level curves plotted in part (b)? Which critical }} \\ {\text { points, if any, appear to give a saddle point? Give reasons for }} \\ {\text { your answer. }}\end{array}$$
$$\begin{array}{l}{\text { d. Calculate the function's second partial derivatives and find the }} \\ {\text { discriminant } f_{x x} f_{y y}-f_{x y}^{2} \text { . }} \\ {\text { e. Using the max-min tests, classify the critical points found in part (c). }} \\ {\text { Are your findings consistent with your discussion in part (c)? }}\end{array}$$
$$\begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3, \quad-3 / 2 \leq x \leq 3 / 2} \\ {-3 / 2 \leq y \leq 3 / 2}\end{array}$$

In this question, we are supposed to use the C s to find a critical points The first and second derivatives you should get ah showing as such. So now we want to find a discrimination that this community is given by f x x times y y minus f x y square. So a second C in this case of discrimination has Bull X and Y variables the critical point you should have gotten from PCs iss at the origin 00 The discrimination of the origin is also equal to zero because it does not have any terms independent off both x and why so since the discrimination ist equals zero, we cannot make any conclusions on a critical point. Zero Cyril.

And this problem we are asked um what we're given the problem where we want to find the minimum distance between this function here or G one equals zero and G two equals zero. So I plotted those here and G1 equals zero is just a line here. And this is kind of like a rotated problem but rotated um in the plane. So that again, Z here is corresponding to um corresponding to why for this curve and w corresponding to X. No other way around that makes it here. What did I say that? Right? Um Yes, he is correspondent to Why W correspondent to X. Yeah, that makes sense. And so you know, we won't use Lagrange multipliers to form this new function age and then the rest of it I did in Mathematica because that's just what I'm familiar with. Obviously pretty much any computer algebra system should be able to do this. So, you know the notation maybe a little unfamiliar, but hopefully you can, I'll explain what's going on here. So if you have to find F as a function of X, Y, W and Z. So what we're doing is f is actually the distance between two points, any two points on these two on two paths. And so what we're doing is trying to find a minimum distance. And so I define G one MG too, that I defined H. So H is a function of X. Y. W. Z. Lambda one lambda too. And then I took the gradient. So in Mathematica there's this gradient function and you give it the function you want to take the gradient of and then the parameters um that you want to differentiate with respect to. And so that gives us a list or an array of 1, 2, 3, 4, 5 6 equations that will depend on X Y W. Ramsey number one and number two. So we have six equations and six unknowns that we need to solve. So what I did here is this is just a little little thing why I basically set every one of these components of this factor um to zero. So we have now we have six equations, all this grouped into an array. And then I asked, I asked Mathematica assault mathematical to solve them and just look for real solutions. And so it was able to spit out some nice clean answers. And so we can see that on the one curve, The point is -18 78. So that's this point here. And then on the other curve there parabolic curve we have ah let's see here. This is X, w correspondence to X is one quarter And Z is 1/2. So that's this point here. And so the distance is 9/32 units. So that's um when we plug in, plug in all these this solution into um into this function, the distance function. And so what we see is that this is the line connecting the two points. And interestingly what you can see is that um this line is perpendicular to both of these curves and I think, I can't remember, I think that's always going to be the case when when you have the minimum between any two points because yeah, I can't remember exactly why that is, but I'm pretty sure that's the case that that any time you find the distance between the minimum of the distance between two curves, that the line connecting those points will be perpendicular to both curves. I'm pretty sure that's the case anyway. Um So again, you can see here we just minimized and obviously this looks looks right. Um This is kind of like if you think about how, how the distance is changing as you move away from this point, it certainly seems like the distance between any two points. You know, it's going to be growing, you know, So obviously this is not good or you know, this is not so anyway, um it looks about right and this is your solution.

In this question. We want off way. We want to use D. C as to identify the critical points off this function the first and second derivatives that you should get a shortness such, um, the critical points you should get from the C S u should have three off them. So we want to look at that this criminal for each off the three points. So for the first point, the discrimination, it's more than zero. So to determine whether it is a local maximum, a local minimum, we need to look at this second derivative refreshment two X, and we find that it is less and zero so we can conclude that it is a local maximum for the other critical Prime X is also e to the power of negative 0.2. That this community is more than they're also again, we need to look at the second derivative with respect to X to determine whether it is a local maximum, a local minimum. So in this case, the second derivative ISS more than zero. So we can conclude that this point is a local minimum. And for the final critical point, um, it is at the origin the discriminative ISS undefined because because there's going to be alone. Looker natural logarithms off zero. So what we can do is to look at if we want to deviate exam wise, like me, positive and slightly negative. So I'm looking at the function when ex and why. We know that when X and Y equals zero, it is a critical point. But when we look at when ex and why I slightly positive the function becomes negative because the lawn, the natural logarithms off a number less than one is negative. And then when we when we hear the function to be, when we have X and y to be slightly negative, the function becomes positive. So it is in fact, a point off inflection. But this is not covered in in this part of the taxable. So it is. It is. It is not a settle point, a local maximum, no local minimum. So we simply conclude that negative e to call on the grease jumping to zero is a maximum and e to the hole. They don't want to zero. It's a minimal

In this question. We want to use PCs to find, um, the critical points and the first and second of it, if you should get a showing here, Um, we're also interested to find that this criminal so that this criminal is given by f x x times y y minus f x y square. So in this case, I'm not gonna expand it, verifying the various critical values. I'm simply gonna substitute the values off pecs and why into this expression. So the critical points you should have gotten from the BCS Ashaninka. So they are four critical points. We want to find out this criminal for each off them. So the first for the first critical point that discrimination is negative. So we can conclude that it is a settle point for this second this criminal. So the second discrimination excess sickles and negative two y 030 We find that the discriminates positive, so to determine whether it is a local maximum, a local minimum. We need to look at the second over different respect to X in this case, and second derivative is negative, so we can conclude that it is a local maximum next at the origin of this criminal is zero is equal to zero. So we cannot make any conclusions about this critical point off. Our last critical point exists. Pickles to Weiss equals zero. That this criminal is positive. The second there were different respect. Toe X is positive, so we can conclude that this point is a local minimum.


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