http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 21 Jul 2014 06:48:10 -0500http://calc3.askbot.com/question/137/post-test-analysis/For those of you who want to join me in trying to figure out what happened on the test, did anyone manage to solve number 5? I got a monstrous set of equations after factoring out the $\mathrm{e}^{-(x^2+y^2)}$ that went up to sixth powers of $y$ and couldn't figure out what went wrong. I'm not stressing too much about it but I'm curious how we can solve it without getting those high rank polynomial equations. (I can't quite remember what the function was).Fri, 18 Jul 2014 15:26:28 -0500http://calc3.askbot.com/question/137/post-test-analysis/http://calc3.askbot.com/question/137/post-test-analysis/?answer=139#post-id-139Wahoo, I'm pretty sure thats the gradient I got, so at least I feel better about that. :) But i think when i worked it out farther I was ending up with things to the 7th power and some nonsense like that. I reworked this problem a couple times, and still kept getting crazy answers, so I just moved on and hoped it was right.Fri, 18 Jul 2014 15:39:37 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=139#post-id-139http://calc3.askbot.com/question/137/post-test-analysis/?answer=141#post-id-141To add to Justin's answer here, I think the graph was both helpful and misleading. The way we interpret where the maximum and minimum are on the graph, we tend to put them in the middle of our peaks and valleys,, so if we compare this to what we see on the graph vs our numbers, the visual max/min doesn't quite align correctly with those points $(0,0) (1,0) (-1,0) (0,1) (0,-1)$. I agree with all of the mathematics above though, and that is also the answer that i calculated.Sat, 19 Jul 2014 02:16:35 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=141#post-id-141http://calc3.askbot.com/question/137/post-test-analysis/?answer=140#post-id-140I don't remember the problem exactly, but I believe it went something like this: > Find all maxima of the function $ f(x, y) = (x^2-y^2)e^{-(x^2+y^2)}. $ Even if this is not the exact same problem, the concepts should still apply. The first thing I did was find both of the first derivatives. $$ f_x = (2x)e^{-(x^2+y^2)} + (-2x)e^{-(x^2+y^2)}(x^2-y^2) $$ $$ f_x = (-2y)e^{-(x^2+y^2)} + (-2y)e^{-(x^2+y^2)}(x^2-y^2) $$ I then set them equal to zero: $$ (2x)e^{-(x^2+y^2)} + (-2x)e^{-(x^2+y^2)}(x^2-y^2) = 0 $$ $$ (-2y)e^{-(x^2+y^2)} + (-2y)e^{-(x^2+y^2)}(x^2-y^2) = 0 $$ Factoring both of the equations results in: $$ (2x)e^{-(x^2+y^2)}(1 - (x^2-y^2)) = 0 $$ $$ (-2y)e^{-(x^2+y^2)}(1+ (x^2-y^2)) = 0 $$ Since $e^{-(x^2+y^2)}$ never will equal $0$, I can divide both sides of both equations by it, yielding: $$ (2x)(1 - (x^2-y^2)) = 0 $$ $$ (-2y)(1+ (x^2-y^2)) = 0 $$ With some simplifications, this is: $$ (2x)(1 - x^2 + y^2) = 0 $$ $$ (-2y)(1 + x^2 - y^2) = 0 $$ Okay, so here is where the art of solving systems of equations comes into play. We know that the first equation is equal to zero when $2x = 0$. The solution to this equation is obviously $x = 0$. Plugging in $x = 0$ into the second equation yields: $$ (-2y)(1 - y^2) = 0 $$ This means that either $-2y = 0$ or $1 - y^2 = 0$. First possibility: $$ -2y = 0 $$ $$ y = 0 $$ Second possibility: $$ 1 - y^2 = 0 $$ $$ y^2 = 1 $$ $$ y = \pm 1 $$ I find it useful to think of a "tree" of values to keep track of this. x = 0 | +---- y = 0 | +---- y = -1 | +---- y = 1 We will now repeat this same process with $y$, since we know that the second will equal zero when $-2y = 0$. Again, this has an obvious solution: $y = 0$. Plugging in $y = 0$ into the first equation yields: $$ (2x)(1 - x^2) = 0 $$ This means that either $2x = 0$ or $1 - x^2 = 0$. First possibility: $$ 2x = 0 $$ $$ x = 0 $$ Second possibility: $$ 1 -x^2 = 0 $$ $$ x^2 = 1 $$ $$ x = \pm 1 $$ Let's add it to our wonderful tree of life! +---- x = 0 | | | +---- y = 0 -> (0, 0) | | | +---- y = -1 -> (0, -1) | | | +---- y = 1 -> (0, 1) | +---- y = 0 | +---- x = 0 -> (0, 0) | +---- x = -1 -> (-1, 0) | +---- x = 1 -> (1, 0) Removing any duplicates, we are left with the points: $(0, 0)$, $(0, -1)$, $(0, 1)$, $(-1, 0)$, $(1, 0)$. An inspection of the contour plot (which was given on the exam) shows us that the maxima occur at $(-1, 0)$ and $(1, 0)$. Here is the contour plot for reference (courtesy of WolframAlpha).  For what it's worth, this also tells us that minima occur at $(0, -1)$ and $(0, 1)$. This leaves $(0, 0)$, which the contour plot tells us is a saddle point. ## Update/Comment I completely agree with Kyouko and Christina. I too spent a long time worrying that it didn't line up, but then I realized what the caption to the graph said (the function didn't match up exactly with the problem).Fri, 18 Jul 2014 16:12:06 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=140#post-id-140http://calc3.askbot.com/question/137/post-test-analysis/?answer=142#post-id-142Just wanted to chime in here, too. I also got (-1,0) and (1,0) for the maxima and I too noticed that this did not line up visually with the image on the exam. I got so hung up on this fact that I reworked the problem way too many times! I finally noticed that the function for the image did not exactly match the function given in problem number 5. As I remember it, the problem in 5 was exactly what Justin has in his answer: $$ f(x, y) = (x^2-y^2)e^{-(x^2+y^2)} $$ The image, 2a, I think it was, looked more like this, but I am not 100% sure about the numbers: $$ f(x, y) = (x^2-y^2)e^{-(2x^2+3y^2)}. $$ This gave us an image that looked more like this:  I left my answer as (-1,0), (1,0) because I had already spent way to much time on this problem. I made a note of this and am hoping for the best...I am sure Dr. McClure will be fair... Sat, 19 Jul 2014 06:19:32 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=142#post-id-142http://calc3.askbot.com/question/137/post-test-analysis/?answer=146#post-id-146I got (1,0) (-1,0) as well. I remembered what McClure said in class about using graphs in order to do less work, which led me to figure out that the maxima of this function live on the x-axis.Mon, 21 Jul 2014 06:48:10 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=146#post-id-146http://calc3.askbot.com/question/137/post-test-analysis/?answer=138#post-id-138I was bothered by that one on the test as well. I definitely got it wrong on the test but looking at it afterward I'm pretty sure you were on track if you had some crazy gradient. The function was $(x^{2}-y^{2})e^{(-x^{2}+y^{2})}$. Looking at it after the test I'm pretty sure the gradient should come out to $$\langle2xe^{-x^{2}+y^{2}}(x^{2}-y^{2}-1),2ye^{-x^{2}+y^{2}}(-x^{2}+y^{2}+1) \rangle$$. That answer was also confirmed by wolfram alpha so if you had some crazy algebra you were dealing with I imagine you were on the right track.Fri, 18 Jul 2014 15:32:10 -0500http://calc3.askbot.com/question/137/post-test-analysis/?answer=138#post-id-138