text:
MathSciNet bibliographic data MR515534 (80a:20054) 20G99 McDonald, B. R. Automorphisms of ${\rm GL}\sb{n}(R)$${\rm GL}\sb{n}(R)$. Trans. Amer. Math. Soc. 246 (1978), 155–171. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# 3D intersection algorithm for cylinders

The problem

Input is a list of $$N$$ cylinders in 3D space, and output should be a list of $$M≤N(N-1)/2$$ pairs of cylinders that intersect. ($$M$$ depends on input data, obviously.)

If it matters, the cylinders are very thin (diameter less than 1% of the length for all cylinders), and a solution for “rounded cylinders” would work for me (it probably simplifies the geometry calculations). A “rounded cylinder” is a cylinder with half-spheres at the extremities; formally, for a start point $$S$$, an endpoint $$E$$ and a radius $$r$$, the rounded cylinder $$(S, E, r)$$ is defined as the set $$\{P|∃ Q ∈ [S,E], ||PQ|| ≤ r\}$$.

The obvious solution

It is easy enough to do in $$O(N^2)$$ time and $$O(max(M, N))$$ space: the pseudocode of my current implementation is (for rounded cylinders):

``Ncyl = length(cylinder_list) output = {} for i = 1, 2, ... Ncyl:   for j = i+1, i+2, ... Ncyl:     (S1, E1, r1) = cylinder_list[i]     (S2, E2, r2) = cylinder_list[j]     find P∈[S1, E1], Q∈[S2, E2] such that ||PQ|| is minimal  # this is the costly line, says the profiler     if ||PQ|| < r1 + r2:       add (i, j) to output return output ``

Better performance?

Any algorithm will have a (time and space) worst-case in $$O(N^2)$$ (at least) because the output list can itself be of that length. However, the above algorithm guarantees $$O(N^2)$$ time even on “friendly” data because it tests all possible intersections.

In my use case, the cylinders are fairly spread apart in space (the longest cylinder is less than one tenth of the diameter of the whole set of cylinders). Furthermore, they occupy a small fraction of space and $$M\sim N$$ (for values of $$N$$ up to 2000 or so, above that it times out). This suggests to me that there could be an improvement by some “sweeping plane” algorithm similar to Bentley-Ottmann. However, I did not find a straightforward way to do Bentley-Ottmann in 3D (in 2D after sweeping you end up ordering points on a line which is easy enough, but in 3D there is no obvious ordering for a plane).

format: plain


text:
Making practice work with distance learning

My story with distance learning, so far, looks like this:

• You need a plan, and a conversation with Justin Reich helped me form one. Create assignments that kids can work on more-or-less independently, and then try to check in with as many as you can.
• I needed materials for that plan, so I put together assignments with examples, scaffolding and extension challenges. I wrote a post summarizing what I understood after three days of this plan, which was not much.
• After a week spent thinking about the kind of work kids could do in our distance learning set-up, it became clear that I was at risk of coming in contact with precious little mathematical thinking. And when I was getting kids to hand in work, it was often after they had completed an assignment, i.e. kind of late to help them with it.
• During my second week of distance learning, I focused on the workflow. A big goal was to figure out if I could get kids to upload photos of their work, something I deemed essential.
• After that week, I wrote a second post about how I was making sure kids were able to send their work to me. I mentioned three paths that were working OK — typing in the chatbox of Zoom (with all messages privately sent to me), typing in a google doc, or uploading images into a google doc.

And, now, at the end of Week 3, I need to largely retract that second post.

The main reason was because I was largely dismissive of tech tools in those first two weeks. I mean, everything we’re using is a tech tool. But I meant the apps, the endless stream of tech tools that people have been recommending over the past few weeks. Off the top of my head, those include: CueThink, FlipGrid, OneNote, Microsoft Whiteboard, GoFormative, Equatio, and so many more.

I ignored these tools at first for two reasons:

• Who’s got the time?
• I don’t want kids to have to learn a new tool.

But there are two tools that stand out, and those are Desmos and Classkick. Go on over to Rachel’s blog and read her on-target comparison of the pros and cons of each tool. She also has examples of materials she has adapted for Classkick, and she’s a great designer of Desmos custom activities.

My main purpose in writing this post is to apologize for the earlier mistake. At the end of this third week of distance teaching, I want to summarize what my classes currently look like.

***

Students log on to Google Classroom. Twice a week they have live classes on Zoom, and I post the meeting link there.

The other two days, I have a day-labeled assignment waiting for them.

Whole-group interactive lessons on Zoom are probably the smoothest part of this. I use slides, and I’ve become pretty adept at annotating them using Zoom’s tools. I pepper students with questions that they then respond to in the chat. When I set the chat to private, students get a direct channel for sharing their thinking with me. This is a wonderful picture into who is participating, what people are thinking. Teaching is basically a conversation, and the chat makes sure we’re able to have it.

Then, we move into practice. That’s when I have started to lean extremely heavily on Desmos and Classkick.

These tools are simple for students to use because, as Rachel notes in her post, kids can just click a link and go to the activity. They are simple for me to use, because I can take my existing resources and post them online.

Twice this week, I took activities I wanted my 4th Grade students to work on and brought them into Desmos activities. Nothing fancy. First, a decimals worksheet:

As students were working on these practice problems, I was able to watch what they were doing and find ways to have conversations with them. Desmos lets you test a new tool for typing little feedback comments (though kids frequently don’t see them). Most important is the big-picture view of where kids are, something that roughly stands in for those moments when you’re looking around at your students and just watching and figuring out what is going on, can they do this thing?

I next took one of my favorite puzzle pages from the Beast Academy books and ported it into a different Desmos activity:

The Desmos teacher dashboard is, once again, extremely helpful.

Because I have knee-jerk skepticism about tech tools, I was initially dismissive of Classkick. But once I saw it in action, I realized that, in the distance learning context, it is very similar to Desmos. It isn’t built for math (so no math type) but it is built for letting teachers import worksheets, have kids work on them through the computer, and enable kids to ask and receive help on specific problems.

This time I was going even more basic: I just wanted to post a review worksheet for students to work on independently this afternoon. I took a page out of my new favorite collection of worksheets and quickly turned it into a Classkick assignment. It looked like this:

This afternoon, while students were working, I was able to monitor their thinking as it came in. (I was “on call” for questions in Classkick, where kids can raise their hands and request help through a chat box. The chat is great — it feels like AOL Instant Messenger.)

Here was a sample of my view of things:

This is an individual student’s work:

I can leave comments through that chat function, or I can leave notes on the slide itself. A student raised her hand to ask for help, so I came in and left and note and an unfinished diagram on her slide:

***

And that’s it, basically. It avoids the awkward need for students to take pictures of their work with their web camera. I’m still open to students turning in their work that way, but I’m not currently encouraging it. These tools seem to do the trick better.

So, in sum, that’s where I’m at. I’m currently using these tech tools reluctantly but enthusiastically. We’re living in a world where you necessarily have to use a tech tool for your teaching. All I’ve done is realize that a web camera and Google Docs are often clumsier for math practice than these other tools.

So, in short, my teaching this week used chat to make whole-group lessons interactive. Then, for practice or assignments, I used Desmos or Classkick, both of which make student thinking more visible. Which enables me to then make informed decisions about how to respond.

None of which is nearly working as well as teaching in an actual classroom would. But it’s much better than when kids were working on their own, invisible to me, for the longest time. So this is a step forward, and where my teaching is at right now.

What are the fastest, simplest ways to get student work while using Zoom and Google Classroom?

In case you missed it, here is the current situation:

• NYC, where I live, is at the center of the coronavirus epidemic in the United States, and our governor is saying we are 14-21 days from the apex of the epidemic.
• But, we’re still teaching!
• I teach math to young children (3rd and 4th) as well as to older students (8th and 9th).
• My wife is also a teacher.
• My two young children are what some people would call “loud.”
• And what if one of them slips off the bed while they’re jumping on it? And we have to go to the hospital?

This is my second week doing distance learning. I previously wrote about what I had figured out in the first few days of this wild experiment, and my school now has each class doing two live sessions and two asynchronous sessions every week. The live sessions are on Zoom; I post materials on Google Classroom.

By the end of last week it was clear that my next, big problem was to figure out how to make more of my students’ work visible. To me, I mean.

Before going forward: I am not interested in your favorite app. At least, not yet. There are dozens of apps that aim to help students and teachers communicate over the computer. I’m sure some of them are useful, but I just can’t deal with a new tech tool right now.

Here is what I have figured out.

****

Before Class

It took me a few days to realize that it’s much easier to organize Google Classroom around classwork instead of announcements. Here is how I’ve started to organize my Google Classroom pages:

I’m organizing things this way mostly so that students can clearly see what they are supposed to do on each day. The instructions are clear: go to Classwork, to the opening assignment, start the next if you are ready, etc.

The most important thing, though, is that each learning activity becomes its own “assignment.” During week 1 I was creating large documents that students were working on over multiple days. This was good in one sense, because I had to post only one thing. But it became very difficult to monitor the progress of kids through the assignment at all. And then it became tricky to modify the plan in the middle of the week by adding on other bits of classwork.

This is a compromise position now, and I think it’s working. It’s certainly easier for me to figure out what I’ve assigned. This also seems to be the correct way to share solutions and notes, rather than trying to organize them by topic.

During Class

There are basically three good ways to collect student thinking during class.

1. The chat box in Zoom.
2. Creating a separate doc for every student, and asking them to type in the doc.
3. Taking a picture of their work and posting it.

Imagine these three ways of collecting work plotted on a grid. One axis keeps track of how easy it is to share your idea with the teacher. Chat is easiest, moving to a doc and typing is still pretty easy, but it’s pretty tough to take a picture of your work and post it to your teacher.

The other axis is about how easy it is to express yourself using a tool. Drawing on a piece of paper is the easiest, most natural approach. You can type at length in a doc, and a chat allows you only to type quick responses.

Here is how I’ve been using these three methods for communicating with kids over the course of the lesson.

Opening Problem

I saw a common mistake in my students’ work yesterday, so this morning I wanted to talk about similar triangles couched between a pair of parallels.

This was my “opening” assignment. But how should I collect this work?

Students need to be looking at the image while responding (or else having to hold a lot of stuff in their heads while shifting back and forth between screens). It’s also not hard to type responses to these questions. And I wanted to ask a series of questions, and that might be hard to type in a chat.

I went for “type your response in a google doc” and I think it went OK. Here is what I saw from the students.

My 4th Graders can do this, my students with shaky tech can more or less do this. Monitoring them is a little bit slow, but it’s OK. As I monitor, I start saying things over Zoom about the work. “Hey Anna, got your work. I left you a little comment.” “Jay, good stuff; here’s what you should try for the last one.” You can also leave little private comments on the side that kids can see pretty quickly. Classroom is good at this.

During a Live Lesson

This is where Zoom chat is great. You don’t want kids going back and forth between Zoom and some doc during the live, most interactive part of the lesson. I’ll quote my last post:

There is a way to sort of do whole-group instruction that opens up a tiny window into how students are thinking. It uses the chat box.

You can change the chat settings so that all chat messages are sent privately to the host of the meeting. At first, I did this because I wanted to cut down on random chat chatter. But then I realized it’s a private way that students can respond to questions.

Here is a routine I’m finding useful during whole-group discussions:

1. Ask a question. I state it, and if possible I also write the question in the chat box. That way, everyone receives the question even if they missed it when I said it.

2. Then, everyone types in their answer to the question. As the answers come in, I acknowledge receipt. “Thanks, Emma.” “Got it, Jake.” I’ll comment on wrong answers without calling individual students out. “Careful, if you’re writing (x – 2)(x -3) that’s factored form.” I can also address individual students: “Tommy, what you wrote is fine.”

3. Then, I’ll share the answer. I’ll take questions, rinse and repeat for however much whole-group time I’ve decided on for the lesson. Attention spans for this vary by class and by age, of course. More than 15 minutes is probably pushing it, I think.

Classwork or Asynchronous Assignments

If you can get away with using some online practice tool, that’s great. In general these tools are less useful in this situation than I would have hoped. The problems do not ramp up in difficulty; they’re only useful for practicing something that students are already reasonably proficient at. I use Deltamath, but it doesn’t work for my 4th Graders. I’m looking at IXL for the little kids, but it looks like it’ll be very tricky to coach kids through signing up and getting used to the tool — I’m holding off on that for now. ASSISTments is too buggy for me to use right now, and I find it difficult to navigate their reports on what the students did.

If you can get away with using a doc and asking kids to respond in it, by all means. But this is incredibly limiting. To see why, let me take you back to my Sunday night. My 4th Graders were supposed to work independently on an assignment on Monday afternoon. I decided to do review; I found a useful set of fractions worksheets from the Math in Focus books. I really wanted to use them:

But! How are kids going to fill this out in a doc? Do I need to recreate this from scratch? That would take so much work.

Same with an assignment for my 8th Graders — I wanted them to try sketching a parabola. Same with my geometry students — I wanted to see their work setting up proportions. Same with my 4th Graders today — I want to study decimals, and wanted them to shade in 0.7 of a whole square.

This leads me to my final and most important thing to share, which is how to coach kids through sharing pictures of their work through Google Classroom.

How to Coach Kids Through Sharing Pictures

My first obstacle was getting a “student view” of my classes. Google does not have this feature, and my school only allows us to add people with our school accounts to Google Classroom. And they weren’t thrilled by my suggestion that they give me a second email account under the name “fake_pershan.”

So under my personal gmail account I made a class called “Pretend Math” and registered a new personal gmail account (“fakepershan”) and created an assignment for myself.

Then I shared my screen and walked students through the steps.

That solves the problem for most of my students. It’s not so hard to take a picture with a phone and then upload it to the assignment. I walked my older students through these steps.

BUT: my youngest students don’t have phones. And I can’t get their parents in front of the screen to coach them through this.

SO: my brilliant colleague pointed out that there is a simpler way to do this in docs, with a webcam.

In 4th Grade today I created an assignment focused on teaching photo sharing.

I shared my “Pretend Math” screen and showed them how to use the tool. Then, I showed them how to open the assignment, and then every kid went to this doc. (I created a separate doc for each student, which Google Classroom lets you do.)

The big question is whether their webcam pictures would be good enough for me to learn anything from. The answer so far seems to be “sort of.”

To test whether my 4th Graders could actually pull this off with some real math, I sent them to another brief assignment, this time connected to content our class was learning.

This is what I got:

So, not perfect. And some of them were worse than not perfect.

But this is a crucial option that I needed for working with my younger students. And in each of my classes of older students, I have kids whose phones either are not working or they don’t have smart phones. This is a crucial method that kids can do on their own to turn in photos of their own work.

But, of course, this is sort of annoying. I tried to use it for my opening problem in a class or two and that turned out to be a bit of a mistake, as it took far too long for students to submit their work. I would have been better off using one of the other methods.

Feedback

Once students have handed something in to Classroom, things are actually pretty good. Google lets you comment on the work itself via highlighting and commenting, but I’ve found it more useful to give a quick written comment that appears under the assignment itself.

I should say that Desmos has released a tool for adding written comments to students in their activities. There are two reasons why I’m not ready to use this for my classes:

• Desmos is at some other site, rather than my one-stop shop Google Classroom.

But if I had more time and if my students were doing a lot of Desmos, I could see that as being useful. It seems roughly equivalent to what Classroom lets me do with assignments, but it makes it easier for me to assign Desmos “worksheets” for students to do on their own. I’ll need to do that for a lot of graphing activities down the line, probably.

***

Phew! OK, that’s all I’ve got. Good luck, out there, and let me know what you’ve figured out.

I do love the Beast Academy books. My 3rd Graders are working on multiplication, and the Beast Academy books have these crossword puzzles. The kids love them:

(Why do the kids love them? Oh, I don’t know. If interest = “this is new” times “I can do this” then I guess this has enough going on that it feels new. And the puzzle is self-checking, which probably validates that “I can do it” feeling. That’s all I’ve got.)

Every puzzle can both be solved and studied. I’ve made it a habit to encourage my students to ask questions about the puzzles we solve, and I usually do this by sharing a question or two that I have.

My question was, can you make this puzzle using only multiplication?

And my kids’ questions were:

• Can you make one of these crosswords only using subtraction?
• Are the blanks really necessary?
• Could the puzzle be smaller? Larger? Could it be 5 x 4? 8 x 2?
• Could it be shaped like a path?

The next class, I gave students some blank crosswords and asked them to see if they could fill in the blanks in a way that worked.

Which was interesting. But the kids wanted more, so I sat down to make more crosswords in the original style. The original puzzles always include mostly multiplication, and then one addition and one subtraction equation, always on the right and bottom sides.

So I set out to make a few puzzles in this style. I started filling in the boxes, and got stuck. Then I tried again — still not working. I started to get that familiar good/bad feeling that happens with math. It’s the feeling of “oh this is harder than I thought” but also “there might be something here!”

Over lunch, I interrupted two of my colleagues and recruited them into the problem. (I was happy to return the favor after they’ve done the same to me so many times.) We filled out the crossword with variables.

Using these variables, the puzzle is only possible if $ac+bd =ab-cd$. My colleague pointed out that you might factor this a bit and then solve for d:

$d = \frac{a(b - c)}{(b + c)}$

A few things about that equation:

• It means that the whole puzzle is determined by just three of those variables.
• d is a whole number, so (b + c) needs to go into a(b – c).

This is not a ton to work on, but suppose that the sum of b and c is chosen to be a prime number. It clearly won’t go into (b – c). So that means a will have to be a multiple of b + c.

That seems to work!

This leaves me with a bunch of questions, though. Does this characterize all the possible crossword puzzles? I feel like this finds one specific way of getting a crossword that works here, but is it really the only way? Also, I haven’t really thought about whether I could use any multiple of b + c. I think I can, just because it’s worked whenever I’ve tried it so far, but it would be better to understand why.

There’s a math textbook that I like that makes the case that there is significant mathematics that has been developed by teachers, just for the sake of having nice examples to give to students. I always like when that sort of thing happens, a nice mathematical surprise that appears sometimes when you remember to look for it.

I flipped around this Illustrative Math lesson — it went pretty well!

Every year, this lesson has given my students trouble:

Here were the problems I had with the lesson:

• My students didn’t have strategies for making sense of why the faster bug would go on the bottom. Kids would quickly misidentify which line was which bug, and we’d have to back up and talk about that for a while before doing anything else.
• So they didn’t really get a chance to engage with the math. And I wasn’t sure what the math exactly was, beyond this little tricky graph that puts time on the y-axis.
• Also the formatting was tricky, because the ladybug/ant race happens on one page and the graph on the other. You don’t necessarily need cognitive science to tell you that swapping between two very separated images makes learning hard, but it does.

Each of the Illustrative Mathematics lessons has a Summary at the end of the lesson. It’s good, but meant as a reference — it’s not really designed for classroom use.

So here was my redesign idea:

• Turn the summary into sample student work i.e. a worked example
• Pair that with analysis prompts and a follow-up question i.e. an example-problem pair
• Redesign the actual materials so that the graph and the race are next to each other

Here’s what I did:

We used the original “warm up” from the materials. Then shifted into the example-based materials I created. Then the redesigned activity from the lesson itself.

It went well! Here’s how I know that it went well: most kids got to the extension questions, and the students were able to focus their on those.

That seems to me the basic tradeoff. If you leave ideas a bit more implicit, then kids will spend more time uncovering them. That can be good, mathematical thinking, of course. The other choice is to make things more explicit at the outset. Then, maybe you have a better shot of diving into what would otherwise be “advanced” “challenge” problems.

I usually make that second choice, and part of why is because I think good mathematical thinking can happen with the example-based materials I shared. After that warm-up (where I asked kids to notice as many details about the graphs as possible and didn’t really push the “wonder” question) I covered up the “student solution” and showed my students the “problem” I had created. Then, I uncovered the student work. There was a pause — followed by “oh!” and “ah that makes sense.” There’s a little mathematical thrill you get every time you figure something out — a few kids got that when I revealed the work.

Then I asked kids to talk about it with partner, and then to solve a similar problem with neighbors. I listened in on conversations and was able to figure out if kids were understanding the example or not. A few times I inserted myself into conversations to help. And then I led a discussion about the example where kids shared the following ideas:

• That the ratio between the heights of points on each line that are directly atop each other stays equal.
• That you could also compare points that are at the same horizontal.
• That if the axes were swapped the top line would represent the faster racer, because then the top point covers more distance in the same time.

And the kids most eager to share were not the ones who usually solve problems with the most confidence (and therefore least likely to share if all of this had come through problem solving).

And then we did the activity that had given my kids trouble each year of the past, and they were able to be struggle productively i.e. they had the “compare points on the same vertical/horizontal” strategy. And they got to extension problems.

I’m going to keep looking for chances to do make this same trade.

What we’re debating when we debate “misconceptions”

Is ‘misconceptions’ a bad word? I’ve had the conversation about misconceptions a number of times, most recently when I wrote this post. Here is a bit from the conclusion:

We see misconceptions in children because it really is true that there’s stuff that they don’t yet know. Noticing this doesn’t have to be an act of violence — in fact, I don’t think that it usually is. Usually it’s like me playing with my son and noticing there’s stuff he doesn’t yet know how to do, even as my mind is blown because oh my god my son is into puzzles! When did our baby turn into a kid?

Is it good pedagogy to ask people who don’t already see their pedagogy as abusive to forswear from using words that they use all the time? Isn’t this exactly the sort of “intellectual violence” that we’re being urged to refrain from? Shouldn’t we start with the way people actually see the world, rather than asking them to use language that is not their own?

That excerpt did not convince anybody at all, but my goal here isn’t to convince. Really all I want to do is bring up something I learned about the constraints of this argument.

There are a couple people I’ve met who have flirted with the idea of cutting out all evaluative language from discussions of teaching, but it’s largely an unsustainable position. You can’t cut out value from teaching, and the thought that you can is a bad mistake. Even if you don’t talk of “misconception” you’re still in need of language to describe thinking that isn’t yet what it could be. Maybe there are no misconceptions, but there is thinking that is e.g. inflexible, procedural, memorized, additive-but-not-yet-multiplicative, trick-reliant, stage one, whatever it is you want to say.

Plus, the math education community very clearly want to be able to understand problematic language and ideas for what they are. We want to be able to call ideas or patterns of thought racist, sexist, colonialist, etc. That’s very different than the “all thinking is just thinking” position.

And so the discussion is only ever about what is particularly harmful (or not) about the term “misconception” and its popular usage. Though people frequently talk about the issues with evaluative language in general when discussing misconceptions, that argument just confuses things. We need to be able to talk about thinking in terms of what it could, even should ideally be.

So there are really just two questions that are relevant for this discussion. Is the term “misconception” particularly harmful, compared to other evaluative language? And even if the term is intrinsically fine, is it used in particularly harmful ways?

I’ve shared my answers, but I’d make the case that those are the right questions.

YouCubed, Reviewed

This exponents activity is neither original nor at all an interesting version of the idea. It’s no better than what most teachers would make on their own, if they wanted to teach exponent rules inductively.

Better versions of this are readily available in practically any textbook, but Illustrative Math has a totally free and online unit on exponents that does this activity better. It’s less tedious and repetitive and it asks questions to push students towards generalizations, rather than asking kids to churn out rows and notice the structure at the very end (“discovery”).

Yes, it’s at a Grade 8 level, but this lesson is pretty much there too. And if you can wait a few months, you’ll have the high school version available too.

“Equity” is dead, long live equity

By the time organizations — even organizations whose work I really like — start using the language of equity to advertise their work, it’s a sign that we’ve overtaxed the latest bit of edu lingo. “Equity” is at that point in the edu fad life cycle; it’s beginning to mean just about anything.

I don’t know if there’s anything to do about this. I think this is less about education and more about the corporate world — business lingo isn’t much better than edu lingo. People want to signal that they get it, without getting too bogged down in what exactly “getting it” entails.

The thing I try to remind myself is to be specific and to use familiar, boring words whenever possible. In place of stuffing meaning into abstract terms, I try to put it into sentences. And instead of “equity” I try to talk about the particulars: unsafe classrooms, hot schools, bad water, inexperienced teachers, and so on. This is my personal resistance to the educational world’s endless desire for catchy language, as I think it’s really all we’ve got.

Some questions about the problem of teachers leaving the classroom

Is it actually a problem for kids? Would schools be more effective places if more teachers on the margins of leaving were to stay in the classroom? How do we know? Is there a correlation between ambitious and teaching skill? What is the correlation?

Do master teacher programs improve learning for a district?

How much of the stress in education about people leaving the classroom could be explained by how uniquely meaningful working with children is? After all, going into management involves a change at work across professions. (Sales managers don’t go on sales calls; you leave the regular police work to get a desk job; you still do rounds occasionally but mostly you don’t see patients, etc.) How much of the problem is that there is a huge emotional gap between teaching and higher-paying work that keeps teachers in the classroom, marginally?

Would people be more effective at their administrative jobs if they were partly in the classroom? Would they be more influential?

William Carlos Williams was a doctor by day, poet by night. No one suggests that there should be more doctor/poet jobs. How do we decide what sorts of jobs their ought to be?

I find this so confusing. What questions do you have? Comments are open.

Some of my assumptions for communicating about teaching

These all might be wrong, but I think some of them are worth exposing. Maybe you’ll help me see how I’m wrong?

1. When have something I want to say about teaching or learning, there is a temptation to coin a new word that identifies a new concept. I try to avoid this temptation.

Suppose, for example, that I get up at a conference and say “math should be sticky.” There are some risks. First, there’s the risk someone will spend a lot of time puzzling over what I mean by “sticky,” remember the phrase, and have no idea what I meant by it in context. (This happens often — people remember memorable tags but struggle to articulate what they mean.) Probably then I’ll start hearing people say that I believe that you should teach in such-and-such a way because it’s “sticky” when that’s not what I meant. There’s also a risk that my word will have connotations that I didn’t expect. (Oh, you think “sticky” is gross and bad? Oops.)

So as a rule — a writing rule, a speaking rule — I try very hard to only use words that I think everybody pretty much uses in the same way.

This is not easy, because (I associate this thought with Ilana Horn) the meanings people assign to seemingly clear words like “discover” in teaching varies a great deal. I might say “worksheet” and you might imagine “evil packet that kids work on in silence and struggle” and I imagine “a bunch of problems on a page that hit the sweet spot for kids, who are asking questions and talking together about math.”

So it’s not easy, but I do try. It helps to keep an eye out for words (like “worksheet”) that could be misunderstood, and to replace those with context and sentences that make it clearer what’s happening and what I’m imagining.

2. I try to avoid advocating for practices unconditionally. What I mean is that I never say “we should do this in class more!” without suggesting when it might be useful to do that in class. I’m thinking about this right now with worked examples. I think example-based learning is great and cool and fun, but I would never give a talk (I think) calling for greater use of examples in teaching. Instead, though, I would give a talk describing situations that especially call for worked-examples and teaching people how examples can be useful in that context. (Here are two: “examples as feedback” and “examples as models for really complex thinking.”)

Likewise, I try never to talk in general about teaching, or about teaching math in general. I try to stay conditional.

***

These two things, I think, make communication about teaching easier. As a consequence, I think it ensures that nobody thinks that I mean something I don’t mean, and nobody thinks that I have solutions to many of their teaching problems, or a message that would revolutionize math teaching.

And, as a further result of that, what I have to say is less broadly meaningful, polarizing and also less popular. That’s the tradeoff, I think. Clarity for popularity.

Addendum: I have nobody in particular in mind with this post, but it was inspired by a lot of the tweets I saw from the NCTM conference. I’ll say that the “unconditional” thing was inspired by advocacy for a lot of the thinking prompts that don’t call for precise answers — numberless word problems, goal-free problems, estimation problems, notice/wonder, etc.

These are all incredibly useful, but (I think) far more useful when a topic is new to a student. So I think the general direction is that these more open prompts are great ways in, but you sort of want to call for more and more precision in your prompts as the learning progresses.

I was once talking to a friend who felt burned by Estimation180. Why, I asked. Well, she was trying to use it every day to improve her students’ number sense, but it hadn’t worked. She was disillusioned.

I’m not disillusioned. I know that Estimation180 tasks are useful in some situations and less useful in others. I have some thoughts about where and when they’re useful in my teaching. I try to stick to talking about that when I’m talking about teaching and estimation.

What is retrieval practice when you’re learning math?

I’ve never really carefully read the retrieval practice literature, but I think it gets confusing when people talk about retrieval practice when talking about math skills, as opposed to mathematical facts.

Here is the description from @poojaagarwal‘s website, committed to promoting retrieval practice among practitioners:

Retrieval practice is a strategy in which calling information to mind subsequently enhances and boosts learning. Deliberately recalling information forces us to pull our knowledge “out” and examine what we know. For instance, I might have thought that I knew who the fourth U.S. President was, but I can’t be sure unless I try to come up with the answer myself (it was James Madison).

But how does this apply to math skills? Can trying a problem (i.e. practicing the skill) ever count as retrieval practice? Does it make sense to use the metaphor of ‘calling information to mind’ to describe what’s going with skills practice?

I think not. But I also am finding retrieval practice useful in my lesson planning. There is a great deal of knowledge that is useful for students to know when they’re learning something new. This sort of knowledge is the sort of thing that I’d like my students to know (i.e. retrieve from memory), more than I’d like them to derive.

Often, at the beginning of class, the first thing I ask my students to do is to remember some facts that they may (or may not yet) know from memory. Some constraints:

• I don’t ask students to solve a problem and call it retrieval practice — that’s skills practice, not retrieval practice, and tickles other parts of the mind.
• I only ask students questions that I think they could remember, even if it might be difficult to recall these things. Ideally, these would be things that either students could derive if they can’t recall them.
• Because stuff from the last few days of class can often get forgotten really quickly, I often use these prompts to strengthen the memory of what we’ve recently done. (The prompt “Summarize what we did yesterday” is surprisingly difficult!)

Here are some prompts I’ve recently used with students:

“Draw a pair of ramps that are pretty close to being of equal steepness.”

“Write an equation of a quadratic, describe what it would look like.”

“What happens when you use the tan button on the calculator? Give some examples.”

“Write several pairs of decimals, and write the number that is between them.”

The truest ‘retrieval practice’ of these is the one about the tan button. Next in line is the one about the equation of the quadratic, since I’m prompting kids to remember what the features of the graph are (though it’s also skills practice). What made me think about these as retrieval practice is that they were all calling back on the previous day’s class.

Here are some purer examples of retrieval practice prompts in math:

“What’s the Pythagorean Theorem?”

(If a specific procedure is supposed to be known for converting a decimal into a fraction:) “How do you convert a decimal into a fraction?”

etc.

As I’m messing around in graph theory, I’m noticing that there are a lot of things that would be useful to remember — particular proofs that could serve as paradigms, constraints (in the form of inequalities) on possible planar or non-planar graphs, theorems, specific graphs that are useful examples, etc. If I had a teacher of graph theory, I’d want that teacher to prompt me to remember these things so that I could have more of them available as resources when I’m trying to learn something new or do some creative proving or problem solving.

(I should probably bust out some flash cards at some point…)

As an aside, I think that retrieval practice is sometimes mixed up with spaced practice, but I think these are different things. Spaced practice might be a better fit for what people are describing when they talk about intentionally building time-separated practice of skills into their courses and assignments. I think this requires a different sort of finesse than retrieval practice, though, as the problem with spaced practice is making sure students have something productive to do if they’ve actually forgotten the material.


text:
Inclined plane normal force varies by coordinate system?

Tags:
1. Mar 2, 2015

x86

Suppose we have a block on an inclined plane.

If we choose the x-y axis to be parallel and perpendicular to the inclined plane, then we have

Fy = N - mgcos30 = 0

But if we choose our trivial x-y coordinate system, where y is parallel to the force of gravity, then we get:

Fy = Ncos30 - mg = 0

Of course, we get two different normal forces (and thus different frictional forces), for the exact same type of problem.

What is going on?

2. Mar 2, 2015

Staff: Mentor

In the second coordinate system the normal force is no longer given by Fy. There is also an x-component of the normal force in that system.

3. Mar 2, 2015

x86

Right, but can't we take the y component of the normal force (in coordinate system where x is the ground, not the ramp) (Ncos30) and use this to solve for N? I.e. N = mg/cos30

Should this not solve for the normal force?

EDIT: The problem was my vector reflection math was wrong. Also, as it turns out Fy != 0, rather F(normal direction) = 0

Then you get the same normal for both coordinate systems

Last edited: Mar 2, 2015
4. Mar 2, 2015

Staff: Mentor

I think that you are confusing yourself with your notation. The normal force is a vector, not a scalar. So you would not have N=mg/cos30, you would have Ny=mg. However, even that is not sufficient if there is friction as well as the normal force.

You need to break all three forces into components, so in vectors you have N+F+mg=0, and in components you would have Ny+Fy+mgy=0 and Nx+Fx+mgx=0 [where $\mathbf{g}=(g_x,g_y)=(0,-|\mathbf{g}|)$, and $\mathbf{N}=(N_x,N_y)=(-|\mathbf{N}|\sin 30,|\mathbf{N}|\cos 30)$ and $\mathbf{F}=(F_x,F_y)=(-|\mathbf{F}|\cos 30,|\mathbf{F}|\sin 30)$].

One of the reasons to use the ramp coordinate system is that you only need to break one vector (the weight) into components. Using the ground coordinate system you need to break both the normal and the friction force into components, and if there is any acceleration then you have to break that into components also.


text:
Browse Questions
Oscillations

# The damping in a system

a.       Lowers the resonant frequency compared to natural frequency proportionally

b.      Increases the resonant frequency compared to  natural frequency

c.       Lowers the resonant frequency but as a nonlinear function

d.      Does not affect the resonant frequency

$w'= \sqrt {w_n^2-\large\frac{b^2}{4m^2}}$
Where $w_n$ is undamped natural frequency.
Thus $w' < w_n$
Hence 'c' is the correct answer.


text:
# 2017 seminar talk: Computable quotient presentations of models of arithmetic and set theory

Talk held by Michał Tomasz Godziszewski (University of Warsaw, Poland) at the KGRC seminar on 2017-03-23.

### Abstract

A computable quotient presentation of a mathematical structure $\mathbb{A}$ consists of a computable structure on the natural numbers $(\mathbb{N}, \star, \ast, \ldots)$ (meaning that the operations and relations of the structure are computable) and an equivalence relation $E$ on $\mathbb{N}$, not necessarily computable but which is a congruence with respect to this structure, such that the quotient $(\mathbb{N}, \star, \ast, \ldots)_{/E}$ is isomorphic to the given structure $\mathbb{A}$. Thus, one may consider computable quotient presentations of graphs, groups, orders, rings and so on, for any kind of mathematical structure.

In 2016 Bakhadyr Khoussainov discussed some questions and results in this area. Part of this concerns the investigation, for a fixed equivalence relation $E$ or type of equivalence relation, which kind of computable quotient presentations are possible with respect to quotients modulo $E$.

We address two conjectures that Khoussainov had made and prove various extensions of the Tennenbaum phenomenon to the case of computable quotient presentations of models of arithmetic and set theory. Specifically, no nonstandard model of arithmetic has a computable quotient presentation by a c.e. equivalence relation. No $\Sigma_1$-sound nonstandard model of arithmetic has a computable quotient presentation by a co-c.e. equivalence relation. No nonstandard model of arithmetic in the language $\{+, \cdot, \leq\}$ has a computably enumerable quotient presentation by any equivalence relation of any complexity. No model of ZFC or even much weaker set theories has a computable quotient presentation by any equivalence relation of any complexity. And similarly no nonstandard model of finite set theory has a computable quotient presentation. Concluding from that, we indicate how the program of computable quotient presentations has difficulties with purely relational structures (as opposed to algebras).

This is joint work with Joel David Hamkins, GC CUNY.


text:
## Adaptive Deep Learning for Network Intrusion Detection by Risk Analysis

### Introduction

With increasing connectedness, network intrusion has become a critical security concern for modern information systems. Various network intrusion incidents have been reported in recent years and they have brought about considerable financial and societal damage. The purpose of Network Intrusion Detection (NID) is to identify malicious activities in network traffic. As shown in Figure 1, where network activities are represented by records in a table, NID needs to distinguish abnormal network activities from normal ones based on their particular features (e.g., duration, protocol, service and state). In this example, it can be observed that $r_2$ uses the udp protocol and the dns service with fixed-size source-to-destination bytes. Since $r_2$ matches the characteristics of a generic attack, it is considered to be a network intrusion. However, the record of $r_1$ represents a normal activity. NID is usually considered as a binary classification problem tasked with labeling each activity as Normal or Abnormal. Even though NID can be performed based on expert rules and data mining, its state-of-the-art performance has been achieved by deep learning.

Figure 1: A NID running example.

Unfortunately, the efficacy of deep models depends on large quantities of accurately labeled training data, which may not be readily available in real scenarios. In addition, it is common in modern networks that traffic distribution continuously shifts. As a result, even a well trained deep model may still mislabel many activities in real networks. It is noteworthy that limited interpretability of deep models further exacerbates the challenge. Therefore, there is a need for interpretable risk analysis, which can detect the activities mislabeled by a machine classifier and explain why. Toward this end, we propose a novel solution of risk analysis for NID in this paper. Built upon on the recently proposed framework of LearnRisk, the proposed solution first extracts interpretable risk features, then trains a risk model by a learning-to-rank objective and finally applies the learned risk model to rank activities by their misprediction risk. As shown in Figure 2, it consists of three steps: risk feature generation, risk model construction and risk model training. To ensure that risk features are interpretable and discriminative, we construct various risk metrics from different perspectives, including domain knowledge and statistical abnormalcy. Specifically, we leverage the CIA domain knowledge, which stands for Confidentiality, Integrity and Availability, to quantify security compromise. Confidentiality measures information concealment, Integrity measures information trustworthiness and Availability measures the ability to use desired information. In addition, we measure intrusion risk from the perspective of anomaly detection, based on the observation that intrusions are usually statistical outliers among a larger number of normal activities.

Figure 2: The LearnRisk framework for NID.

Furthermore, since risk analysis can measure the misprediction risk of a machine classifier on unlabeled activities, it provides classifier training with a viable way to adapt towards a particular workload. Therefore, we also present a novel solution of adaptive deep training for NID based on risk analysis. As shown in Figure 3, it consists of two phases: the traditional training phase and the following adaptive training phase. In the first phase, it pre-trains a deep model in the traditional way based on labeled training data, while the second phase further fine-tunes the model towards unlabeled target data by minimizing their misprediction risk.

Figure 3: Risk-based adaptive training for NID.

The main contributions of this paper are as follows:

1. We propose a novel solution of interpretable risk analysis for NID. In particular, we present a technique of risk feature generation that can effectively fuse various intrusion risk factors for risk measurement.

2. We present a novel solution of adaptive deep learning for NID, which can effectively tune a deep model towards a target workload by minimizing misprediction risk.

3. We empirically validate the efficacy of the proposed solutions on real benchmark data by a comparative study. Our extensive experiments have shown that the proposed solution of risk analysis can identify the mislabeled activities with considerably higher accuracy than the existing alternatives, and adaptive deep learning can effectively improve the performance of deep models in both offline and online settings.


text:
# Force Systems in Space

Forces in Space (3D Forces)
Magnitude of a force F in space

$F = \sqrt{{F_x}^2 + {F_y}^2 + {F_z}^2}$

Components of a force in space

$F_x = F \cos \theta_x$

$F_y = F \cos \theta_y$

$F_z = F \cos \theta_z$

Direction cosines

$\cos \theta_x = \dfrac{F_x}{F}$

$\cos \theta_y = \dfrac{F_y}{F}$

$\cos \theta_z = \dfrac{F_z}{F}$

Proportion of components

$\dfrac{F_x}{x} = \dfrac{F_y}{y} = \dfrac{F_z}{z} = \dfrac{F}{d}$

Moment of a force about an axis

$M_x = zF_y \pm yF_z$

$M_y = zF_x \pm xF_z$

$M_z = yF_x \pm xF_y$

Resultant of Concurrent Force Systems in Space
Components of the resultant

$R_x = \Sigma F_x$

$R_y = \Sigma F_y$

$R_z = \Sigma F_z$

Magnitude of the resultant

$R = \sqrt{{R_x}^2 + {R_y}^2 + {R_z}^2}$

Equilibrium of Concurrent Space Forces
The resultant of all forces is zero

$\Sigma F_x = 0$

$\Sigma F_y = 0$

$\Sigma F_z = 0$

The sum of moment is zero

$\Sigma M_x = 0$

$\Sigma M_y = 0$

$\Sigma M_z = 0$


text:
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Asked by Pragya Singh | 1 year ago |  62

##### Solution :-

Given:
Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02 m

Glycerine is flowing at the rate of 4.0 × 10-3 kg/s

M = 4.0 × 10-3 kg/s

Density of glycerine, ρ = 1.3 × 103 kg m-3

Viscosity of glycerine, η = 0.83 Pa s

We know, volume of glycerine flowing per sec:

V =$$\dfrac{ M }{ density }= \dfrac{ 4\; \times \;10^{ -3 } }{ 1.3\; \times \;10^{ 3 } }$$

= 3.08 x 10 -6 m3 /s

Using Poiseville’s formula, we get:

V = $$\dfrac{ \pi p^{‘} r^{ 4 }}{ 8\; \eta \;l}p^{‘} = \dfrac{ V \;8\; \eta \;l}{ \pi r^{ 4 }}$$

Where p’ is the pressure difference between the two ends of the pipe.

p’ = $$\dfrac{3.08\; \times \;10^{-6} \;\times \;8\; \times \;0.83\; \times \;2 }{\pi \;\times \;(0.01)^{4} }$$

 = 9.8 x 10Pa

And, we know:
R = $$\dfrac{ 4 \;\rho \;V }{ \pi \;d\; \eta }$$, [Where R = Reynolds’s number]

R = $$\dfrac{4 \times 1.3 \times 10^{3} \times 3.08\times 10^{-6} }{\pi \times 0.83 \times 0.02 }= 0.3$$

Since the Reynolds’s number is 0.3   which is way smaller than 2000 , the flow of glycerine in the pipe is laminar.

Answered by Abhisek | 1 year ago

### Related Questions

#### According to the law of atmospheres density of air decreases with increase in height y

(a) According to the law of atmospheres density of air decreases with increase in height y as $$ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}$$. Where ρ0 =1.25 kg m-3 is the density of air at sea level and yis a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

(b) A zeppelin of volume 1500 m3 is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y0 = 8000 m and ρHe =0.18 kg m-3].

#### Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends.

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).

#### Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius $$2.0 × 10^{–5} m$$ and density $$1.2 × 10^3 kg m^{–3}$$. Take the viscosity of air at the temperature of the experiment to be $$1.8 × 10^{–5} Pa$$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
A plane is in level flight at constant speed and each of its wings has an area of $$25 m^2$$. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be $$1 kg/m^3$$), g = $$9.8 m/s^2$$


text:
Posted on Categories:OS操作系统, 计算机代写

# 计算机代写|OS操作系统代考OS operating system代写|COMP7308 Memory and storage

avatest™

OS操作系统操作系统（OS）是管理计算机硬件、软件资源的系统软件，并为计算机程序提供公共服务。时间共享的操作系统为系统的有效使用安排任务，也可能包括用于处理器时间、大容量存储、打印和其他资源成本分配的会计软件。对于硬件功能，如输入输出和内存分配，操作系统作为程序和计算机硬件之间的中介，尽管应用程序代码通常由硬件直接执行，并经常对操作系统功能进行系统调用或被其打断。操作系统出现在许多包含计算机的设备上–从移动电话和视频游戏机到网络服务器和超级计算机。

avatest.™OS操作系统代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.™， 最高质量的OS操作系统作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此OS操作系统作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

avatest.™ 为您的留学生涯保驾护航 在计算机Computers代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的计算机Computers代写服务。我们的专家在OS操作系统代写方面经验极为丰富，各种OS操作系统相关的作业也就用不着 说。

## 计算机代写|OS操作系统代考OS operating system代写|Memory and storage

While a process is running, most of its data is held in “main memory”, which is usually some kind of random access memory (RAM). On most current computers, main memory is volatile, which means that when the computer shuts down, the contents of main memory are lost. A current typical desktop computer has 1-8 GiB of memory. GiB stands for “gibibyte,” which is $2^{30}$ bytes.

If the process reads and writes files, those files are usually stored on a hard disk drive (HDD) or solid state drive (SSD). These storage devices are nonvolatile, so they are used for long-term storage. Currently a typical desktop computer has a HDD with a capacity of $500 \mathrm{~GB}$ to $2 \mathrm{~TB}$. GB stands for “gigabyte,” which is $10^{9}$ bytes. TB stands for “terabyte,” which is $10^{12}$ bytes.
You might have noticed that I used the binary unit GiB for the size of main memory and the decimal units GB and TB for the size of the HDD. For historical and technical reasons, memory is measured in binary units, and disk drives are measured in decimal units. In this book I will be careful to distinguish binary and decimal units, but you should be aware that the word “gigabyte” and the abbreviation GB are often used ambiguously.
In casual use, the term “memory” is sometimes used for HDDs and SDDs as well as RAM, but the properties of these devices are very different, so we will need to distinguish them. I will use “storage” to refer to HDDs and SDDs.

Each byte in main memory is specified by an integer “physical address.” The set of valid physical addresses is called the physical “address space.” It usually runs from 0 to $N-1$, where $N$ is the size of main memory. On a system with $1 \mathrm{~GB}$ of physical memory, the highest valid address is $2^{30}-1$, which is $1,073,741,823$ in decimal, or 0x03ff ffff in hexadecimal (the prefix 0x indicates a hexadecimal number).

However, most operating systems provide “virtual memory,” which means that programs never deal with physical addresses, and don’t have to know how much physical memory is available.

Instead, programs work with virtual addresses, which are numbered from 0 to $M-1$, where $M$ is the number of valid virtual address. The size of the virtual address space is determined by the operating system and the hardware it runs on.

You have probably heard people talk about 32-bit and 64-bit systems. These terms indicate the size of the registers, which is usually also the size of a virtual address. On a 32-bit system, virtual addresses are 32 bits, which means that the virtual address space runs from 0 to 0xffff ffff. The size of this address space is $2^{32}$ bytes, or $4 \mathrm{GiB}$.

On a 64-bit system, the size of the virtual address space is $2^{64}$ bytes, or 4 . $1024^{6}$ bytes. That’s 16 exbibytes, which is about a billion times bigger than current physical memories. It might seem strange that a virtual address space can be so much bigger than physical memory, but we will see soon how that works.

When a program reads and writes values in memory, it generates virtual addresses. The hardware, with help from the operating system, translates to physical addresses before accessing main memory. This translation is done on a per-process basis, so even if two processes generate the same virtual address, they would map to different locations in physical memory.
Thus, virtual memory is one important way the operating system isolates processes from each other. In general, a process cannot access data belonging to another process, because there is no virtual address it can generate that maps to physical memory allocated to another process.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
# Would pyrokinesis be an effective defense?

I've read comics in which someone with pyrokinetic powers was able to block bullets with a sheet of fire. I have three questions

1. How hot would the fire have to be to melt a bullet passing through it? (fire is about 1 inch/ 2.54 cm thick)
2. Would this be an effective defense, or just mitigate the fact that you still have a small lump of metal flying at you at a high velocity? (Would the aerodynamics change enough to matter) (The "Shield" is about 2 yards/ 1.8288 meters away from you)
3. Would there be a temperature that would make guns unusable?
4. Any other ideas for counter-gun pyrokinetic tricks are welcome.
• One work around which I have seen used effectively, is to consider pyrokinesis to be telekinesis on a molecular scale, mechanically oxidizing burnable substances (including the air) and thus producing flame. When this is the mechanism the implements your pyrokinesis, it is an easy step to shift it from one class of invisible objects (molecules) to another (speeding bullets). As long as the psychic can't see what they are moving with their mind, they don't know that they can't move it psychically. – Henry Taylor Jul 27 '17 at 2:35
• I don't think it would stop a bullet because although the fire could be more than hot enough to melt a bullet, the bullet would pass through an inch of space in a tiny fraction of a second and air only transfers heat to a bullet so fast. Also, the bullet could be molten metal and still end up hitting the person with pyrokinetic powers. What if they used their fire powers to melt a makeshift shield off of their surroundings? – Tom Jul 27 '17 at 2:37
• 3 and 4: Blow up the magazine of the gun by exploding the bullets. – rytan451 Jul 27 '17 at 2:49
• @John It would probably still be lethal, though less so, and it will be really hot. If you've ever stood under a large bucket of water being poured from high heights, you would know that droplets at high speeds can still be painful. And in this case, they would be hot. Really hot. – rytan451 Jul 27 '17 at 3:18
• This question has a huge X Y problem. It is not asking how a pyrokynetic would block bullets, it is asking about a proposed solution. – Mindwin Jul 27 '17 at 10:47

I may be put on a number of watch lists during my researching to answer this. :P

My criteria for making the bullet undamaging is this: if the bullet has been boiled into a vapor, it is no longer capable of damaging your pyrokinetic. This is a bit strict, but easier to calculate. You may want to lower the temperature to compensate.

A Beretta M9 has a muzzle velocity of 381 meters per second. This means that the bullet will be passing through the fire barrier in more than 66.6 microseconds, depending on the distance between the gun and the barrier. That is a very short time (emphasis on very) for the bullet to be rendered ineffective.

Rifles have a greater muzzle velocity, so their bullets would have a much shorter time to be vaporized.

BEGIN EDIT

I am assuming primarily conductive heat transfer over the time span required, and that the energy imparted by the firewall would be distributed constantly over the entire bullet. I admit that is somewhat of a stretch. The bullet starts at around $50 °C$ from the gunshot. Vaporized means that the entire bullet has reached $1749 °C$, which is the boiling point of lead. This means that the bullet must increase around $1700 °C$ in 66 microseconds. Given that amount of time, the lead must increase in temperature at $25757575.\overline{75} °C/s$.

The bullet is around 1 gram of lead (very approximate), so the amount of energy to increase the temperature of the bullet by 1700 °C is 21.76 Joules (ref). That's around 329697 watts of heat transfer needed throughout the 66 microseconds.

That much power requires extreme amounts of heat around the bullet. (I don't actually remember my line of thought after here, but I did calculate this.)

END EDIT

You'll need a temperature of around 3,000,000 Celsius (5,400,000 Fahrenheit) to vaporize an unjacketed lead bullet. That is more than five hundred times the surface temperature of the sun.

Chances are, with that much heat two yards away from your pyrokinetic, getting shot would be the least of his concerns.

A quick message to your pyrokinetic:

Dear pyrokinetic, Please do not try to block bullets with fire. You'll probably burn yourself to death. Perhaps you may consider the saying "The best defense is a good offense"? If your enemies are unable to shoot you, you won't get shot.

Perhaps instead of blocking the bullets with fire, you may explode their cartridges in the magazine, or melt the firing mechanisms of their guns. Just try to keep at a good distance, and be aware of your surroundings.

Don't get shot.

Best regards,
A concerned citizen

• More answers should include safety messages like this. Considering the temperatures involved, wouldn't be easier and not so overheated as to set fire to the shooter (preferably before they fire their weapons)? Good luck with the watch lists! – a4android Jul 27 '17 at 6:07
• I like the use of "around" to describe a temperature of nearly 3 megakelvin ... given to a precision of 10mK. I would have written it as "about 3,000,000 C (5,300,000 F)" – Martin Bonner supports Monica Jul 27 '17 at 10:50
• I'd like to see your math. – Samuel Jul 27 '17 at 16:17
• Note that at that temperature the air would turn to plasma and would start emitting x-ray radiation that would kill everybody. – ZeroOne Jul 28 '17 at 1:12
• @Rytan451 Can you explain what went into your math? As it is right now, you made up a number and are getting a bunch of upvotes for no evidence at all. – BobTheAverage Jul 28 '17 at 14:17

For another trick in using pyrokinesis to block bullets:

Depending on how long the sheet of fire can be maintained for, it could be used simply to obscure the view of the shooter.

It's pretty difficult to see through fire, especially with the addition of smoke, which in an enclosed space would cause the shooter's eyes to suffer as well (though that might also happen to your pyromancer - I would recommend goggles).

Alternatively, depending on how well the heat can be manipulated, it could be possible to conjure a mirage, in which heat bends light rays and creates a different picture than what is really there, which could be used to distort what the shooter sees and cause them to miss.

Whilst these techniques don't necessarily stop bullets in the air, it can stop them from being fired accurately, which is just as useful.

• "though that might also happen to your pyromancer" People wielding super powers are usually somewhat more resistant to effects caused by their own ability, so that's easy to handwave. – Mast Jul 27 '17 at 10:52

Pyrokinetic "sheet of fire" can not be fire at all. What it has to be is a region of space where thermal molecular velocity is increased by the kinetic's power, to the point where oxygen-nitrogen recombination occurs (which means that pyrokinesis is not environmentally friendly, as nitrogen oxides are responsible for acid rains).

The same effect is a defense against most kinds of projectiles: their macroscopic kinetic energy is neutralized, they can be heated or fractured until they disintegrate, and even the shock wave of an explosion - being kinetic energy of air molecules - can be diverted.

The possible limits are where does this energy go - we can somehow transmit it without physical contact, but it's best if we don't repeal the laws of thermodynamics - and how much of it can be transferred by the kinetic, and for how long. Also, how long does it take for the kinetic power to "lock" on the incoming bullet.

Since the power for a sheet of fire is above what can be generated by a human body, the energy must be supplied by something else (alien nanomachines, trans- dimensional demons...). Maybe that something else can take care of monitoring airspace and target acquisition.

As user @rytan451 states, it's not really feasible to flat out vaporize the bullet (though I'd be curious to see the calculations he used to arrive at 3ish Mk). What you could do is super heat the air next to the bullet and use the expansion of the gas to deflect the trajectory slightly. For this to be effective however, your pyrokenetic would have to have superhuman reflexes and precision in order to intercept the bullet as it travels.

• This was the solution I thought made the most sense. However, if you have the ability to deflect the bullet early, it means you have the ability to heat things right near the barrel of the gun substantially. As an easier solution, reach a few inches further, and make a fire which turns the trigger mechanism into a pile of goo. – Cort Ammon Jul 27 '17 at 16:24
• @CortAmmon trigger mechanism, or the person holding the gun ;) (someone had already suggested disabling the gun however) – Aaron Jul 27 '17 at 17:53
• As some point you do transition over from defense to offense =) – Cort Ammon Jul 27 '17 at 18:07
• If the pyrokinetic abilities are psychic in nature, and the "superhuman reflexes and precision" above are precognitive in nature, than this makes perfect sense to me. – Ghotir Jul 27 '17 at 19:45

It would be very effective if you assume one thing:
Your pyrokineter is actually electro/air (apart from Firefly from DC pyros don't use any kind of fuel to burn and/or flamethrowers).
Mostly because it's easier to use nitrogen from the air and add a little of electricity to create plasma.
Plasma has the advantage that it can:

1. cut through metal
2. plasma can be magnetically charged and would push away Pb

So you would not only have to deal with smaller particles, but it would be easier for the field to push them in different directions.

Think of it as a bush that changes the trajectory of a bullet.

Your pyrokinetic will be lucky if his assailant is only using lead bullets. Most ammo today is steel jacketed and there are much more durable materials being used to make armor piercing projectiles.

So, with that in mind, mayhaps we should look for a means of using your pyrokinetic's talents in more imaginative ways. And, as always, the devil is in the details.

For instance, if his assailant happens to be a sniper then your pyrokinetic may never even see it coming, so mounting any kind of defense would become a moot issue. So we must assume you're looking for a way to preserve the life of your pyrokinetic in situations where he has some prior warning of impending doom. Thus he has time to mount a formidable offense and should do so prior to his assailant firing his weapon. In this scenario your pyrokinetic can use his amazing talents to fry his arch nemesis to a crispy thin strip of bacon.

On the other hand, if your pyrokinetic hasn't the stomach for such violence then he could just warm his assailants weapon to a point where holding on to it long enough to take aim is out of the question. That always produces opportunities to lighten the mood for all observers when they see the assailant dancing around trying to find a way to cool his trigger finger.

Now, let us assume the weapon has already been dis-charged in your pyrokinetic's direction. As others have pointed out these projectiles are not walking. If your pyrokinetic hasn't already got a plan ready and begun its initiation, it isn't likely he is going to survive long enough to figure out what to do about the bullet, except maybe figuring out how to get it out from under his skin.

Likewise, if he knows the weapon is about to be discharged in his direction and he has time to flame on, simply trying to generate enough heat to melt the bullet isn't going to be the safest method of ensuring his survival. So let's assume your pyrokinetic has very precise control over his amazing ability and has had enough practice to use it effectively in concentrated bursts. As opposed to a wall of flame, your pyrokinetic should discharge an extremely intense beam of fire about twice the size of a standard bullet in circumference and thrust it in a trajectory that will ensure collision with the bullet at a force much greater than any bullet, irrespective of powder pack or material used, could resist. In this way he simply slows the bullet down very quickly until it drops at his feet, still smoking hot and glowing a bright orange/red, capping off the scene with a loud hiss and sizzle as the ground it has fallen on is damp.

Does that work for you?

The pyrokinetic does not need to destroy the bullet to not get hit.

The pyrokinetic could blow the bullet off course.

Flame is simply air so hot that it is incandescent. When you heat a gas it expands in volume. The faster you heat it, the faster it expands. Rapidly expanding gas is an explosion.

https://www.physicsforums.com/threads/how-much-does-air-expand-with-heat.267530/ PV = nRT; V is volume and T is temperature. So if pressure is stable an increase in temperature causes the same increase in volume. A candle flame is 900C. From 30C to 900C means a volume increase of 30x. An oxyacetylene torch is 3000C; from 30C to 3000C is 100x the volume. If you turn 1 cubic meter of gas into 100 cubic meters of gas in 1 second, that would make a 360 kph wind or 223 mph wind.

A sudden expansion of heated gas (flame!) in the path of a bullet could blow it off course. There is no question that wind can push a bullet off course. The deviation has to do with the mass of the bullet, the speed of the wind, and the flight time of the bullet (or how long the wind gets to push it). There are published tables for shooters. Here is one.

Multiplying the values for 20mph wind by 10, a 200mph wind could push a bullet this size 16 inches over 100 yards and farther over longer distances.

This would not always save the target. The deviation of the bullet would depend on the volume of gas heated, how hot it got, the speed and mass of the bullet and the distance of the shot. The pyrokinetic should practice. But a bullet does not need to miss by much to miss.

• That is a great strategy, and considering he could use subtle applications of heat to create convection to move the air couldn't he technically airbend? – Efialtes Mar 15 '19 at 19:18

I propose another theory in which to 'stop' the bullets with flames or pyrokinesis. Creating a wall of flame is pointless, as other before me have said, it would need to be very hot and very thick to vaporize a bullet.

So instead, create a vacume of air and oxygen within a circle of flames. If possible the barrier should stop the bullets momentum or drag it down lower. And if your pyrotechnic is skilled enough, they should be able to control the vacume of air and fold the flame within to change or divert the bullets.

As my Uncle once said, "Not everything can be solved with fire, but a damn lot can."

• How exactly can a vacuum act as a barrier? Vacuum exercises no drag, so it is even better than air for a moving bullet! – L.Dutch - Reinstate Monica Mar 15 '19 at 6:54
• I think he means that the vacuum would act as space between the air so that the projectile hits the vacuum and speeds up and then suddenly hits a gaint wall of air slowing it down due to the sudden friction. – Efialtes Mar 15 '19 at 19:16
• @Efialtes That's not how it works. The bullet is under constant drag during flight. Taking away some of the drag doesn't speed the bullet up, it just doesn't slow it down. When it hits the air on the other side of the void it just resumes slowing down like it was before. – Corey Mar 15 '19 at 23:58

What is pyrokinesis actually?

Is it a gathering of molecules from the surroundings (air, dust, etc) and then superheating and accelerating them?

If your hero can throw a fireball, could he also generate a mini-fireball that travels at similar speeds as a bullet? Could he generate a fire bullet fast enough and at the correct mass+velocity to counter the energy of the bullet?

You could stop the forward momentum, but the bullet would probably shatter and you would have fragments spreading out from the point of impact. If the fire bullet was a larger mass, then it could envelop the bullet and pretty much vaporize the fragments. That fire bullet would also probably "shatter" and dissipate back into the air after impact.

You need to calculate Heat capacity of air and Heat capacity of lead going through it. Boiling point of lead is 1750 °C. So it is 1730 difference to make. Lets assume that bullet will only gain heat from Air it went through - so basically it has same volume (or very similar).
Air has higher mass heat capacity, but it depends on mass, not volume.
So we take volume heat capacity because of same volume of air and bullet. That is 0.00121 for Air and 1.44 for Lead.
Lead/Air*1730 = 2 058 842°C/K or 3 705 947 degrees Fahrenheit.

You shouldn't care about velocity, unless you think "only 10% of heat would be absorbed by bullet" - in that case you will need 10x higher temperature.
"Small lump of metal flying at you" - Unless you have 2000+ degrees around you, it will change back to metal. But more likely it would stay as hot steam for few seconds - just don't breathe it.

• This answer assumes that the bullet and the air are left in the same space until their temperatures equalize. The bullet will take a microsecond to travel through the flames. – BobTheAverage Jul 29 '17 at 16:18

I'm in the group that think of pyrokinesis as a specialized form of telekinesis where the pyrokinetic energizes molecules to induce heat and ultimately fire. Take this far enough and I don't see why your pyrokinetic couldn't end up generating plasma. CERN has produced a type of plasma that reaches 5.5 TRILLION degrees celcius, which exceeds the above temperatures about 2 million times over. Your pyrokinetic could also have some degree of electro-magnetic influence as well, which would be a bit of a requirement for controlling plasma.

Other answers have focused on the sheer amount of heat needed to vaporize the bullet, but I think (I don't really know -- ask someone who's good at physics) that your first obstacle to get over would be air displacement. It's worth considering that the vortex left in the wake of the bullet's path may effectively extinguish any flame nearby due to starvation.

Depending on your pyrokinetic's origin of ability (e.g. on their person v. at a remote location in their field of vision) combined with their level of detail and speed, the more effective defense (other than obscuring themselves with a smokescreen/the sheet of flames or distracting/frightening the attacker) may be to preempt the shot and cause an early ignition, although this would require considerable attention to detail so that the projectile is not aimed somewhere equally damaging.

Air temperature might also be taken into account -- is there any property of air buffering or flight physics that can be altered by a drastic change in temperature?


text:
# Separable completion of a ring

The completion of a topological ring $A/\bar{\mathfrak{o}}$, where $A$ is a topological ring and $\bar{\mathfrak{o}}$ is the closure in $A$ of the zero ideal $\mathfrak{o}$. The separable completion of a ring is also a topological ring and is usually denoted by $\hat A$. Every continuous homomorphism from $B$ into a complete separable ring $B$ can be uniquely extended to a continuous homomorphism $\hat A \rightarrow B$.

In the most important case where the ring $A$ has a linear topology defined by a fundamental system of ideals $\left(\mathfrak{a}_\lambda\right)_{\lambda \in \Lambda}$, the separable completion $\hat A$ is canonically identified with the projective limit $\lim_{\lambda \in \Lambda} A/\mathfrak{a}_\lambda$ of the discrete rings $A/\mathfrak{a}_\lambda$. The separable completion of a module is achieved in the same way.


text:
Loop space of a category - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:05:36Z http://mathoverflow.net/feeds/question/114553 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114553/loop-space-of-a-category Loop space of a category Mike Shulman 2012-11-26T16:41:59Z 2012-12-06T07:16:23Z
This seems like it should be a "standard" thing, and I think I remember even seeing it somewhere, but I can't remember where.

Let $C$ be a small category. Is there a category $\Lambda C$ whose nerve (or classifying space) is a model for the free loop space of the nerve (or classifying space) of $C$? Probably the objects of $\Lambda C$ should be something like zigzags of morphisms in $C$. A reference would be the best thing to hear.
 http://mathoverflow.net/questions/114553/loop-space-of-a-category/114566#114566 Answer by Karol Szumiło for Loop space of a category Karol Szumiło 2012-11-26T18:30:17Z 2012-11-26T23:35:45Z
My French is not good enough to be sure about it, but it seems that this paper has the definition you are after.

Evrard, Marcel Fibrations de petites catégories. Bull. Soc. Math. France 103 (1975), no. 3, 241–265. (Numdam)
 http://mathoverflow.net/questions/114553/loop-space-of-a-category/114694#114694 Answer by Konrad Waldorf for Loop space of a category Konrad Waldorf 2012-11-27T19:34:27Z 2012-11-27T19:34:27Z
There is a definition of a loop groupoid of a topological groupoid in

Lupercio, Uribe, "LOOP GROUPOIDS AND TWISTED SECTORS"

In Section 4 the concept is reduced to the finite group case, where Proposition 4.2.4 presumably shows the property you were asking for.
 http://mathoverflow.net/questions/114553/loop-space-of-a-category/115502#115502 Answer by Fedotov for Loop space of a category Fedotov 2012-12-05T14:11:55Z 2012-12-06T07:16:23Z
I will try to answer the question. As I said in a comment, the Thomason model structure on $Cat$ is not simplicial model structure. Let $C$ be a small category, we will view it as a topological category. Denote by $C[C^{-1}]$ the topological category where we invert all maps of $C$ such that $C\rightarrow C[C^{-1}]$ is a cofibration of topological categories, then the coherent nerve $N_{\bullet}C\rightarrow N_{\bullet}C[C^{-1}]$ induces a weak equivalence of simplicial sets. Notice that $C[C^{-1}]$ is an infinity groupoid.

Let $C$ be a cofibrant topological category. The mapping space $map(C,D)$ in the model category of topological categories is given by the (standard) nerve of the following $HOM(C,D)$ category :

$\underline{Objects}$ are topological functor $F:C^{op}\times D:\rightarrow Top$ such that for any $c\in C,$ $F(c)$ is equivalent to a representable functor $D(d,-)$ for some $d\in D$.

$\underline{Morphisms}$ in this category are natural transformation $H:F\rightarrow G$ such that $F(c,d)\rightarrow G(c,d)$ is a weak equivalence for all $c\in C$ and $d\in D$.

Let $S^{1}$ a simplicial model for a circle. Let $k: sSet\rightarrow sSet$ the cocontinues Joyal functor which take $\Delta^{n}$ to the nerve of the groupoid with $n+1$ objects and only one isomorphism between any two objects.

Recall that $\mathfrak{C}: sSet\rightarrow Cat_{\Delta}$ is the left quillen adjoint to the coherent $N_{\bullet}$ betwen the joyal model structre on $sSet$ and the Bergner model structure on $Cat_{\Delta}$

Now $k(S^{1})$ is a simplicial set, and the cofibrant topological category $|\mathfrak{C}[k(S^{1})]|$ is an infinity groupoid and its cohenrent nerve is equivalent to $S^{1}$.
 The finial result is that $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is a model for $\Lambda C$, since the nerve of $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is equivalent to $map(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])\sim Map(S^{1},N_{\bullet}C[C^{-1}])\sim Map(S^{1}, N_{\bullet} C)=\Lambda N_{\bullet}C$

N.B. The only point that I did not explained is the construction of $C[C^{-1}]$. Let 1 be the category with two objets a and b and a unique morphisms from $: f:a\rightarrow b$. Let $\widehat{1}=|\mathfrak{C}[k(\Delta^{1})]|$, then $C[C^{-1}]$ is the pushout $colim (\sqcup_{mor C} \widehat{1}\leftarrow \sqcup_{mor C} 1\rightarrow C )$ i.e., for each porphism of $C$ there is a map $1\rightarrow C$.


text:
# Half life problem

amazingphysics2255
Homework Statement:
Yttrium 90 is also radioactive and is a beta particle emitter with a half-life of 2.67 days.

Discuss the amount of Yttrium that would be present in the above example after the
Strontium had been decaying for the first 8 days.
Relevant Equations:
N/a
I believe I use a formula such as $$n\frac{t}{T}$$ but am unsure I've never done one without a given amount of the sample

So if I was to use this I would get 8/(divided by)2.67 = 2.99625468 rounded to either 3 or 2.997 But i don't think this is write because you need to use 1/2 in this formula.

edit: So 8/2.67 is going to equal the number of half-lives has gone by so 2.997 halflives have gone by

(1/2)^2.997= 0.125 rounded so is the answer 0.125g?

Last edited:

Homework Helper
Find a relevant equation ... your belief isn't good enough .

As for the intial amount, pick something and call it ##n_0##. The exercise asks what fraction of that is left over after 8 days.

 misread the (partial ?) problem statement. Sorry ...

Last edited:
amazingphysics2255
So my answer is incorrect?...I don't believe I'm allowed to make up a random value to solve this either.

willem2
Are you starting out with 100 % strontium 90, and then you need to find out how much Yttrium there is after 8 days? Eventually you get an equilibrium where the $$\frac {N_Y} {N_{Sr}} = \frac {T_Y} {T_{Sr}}$$
(N is number of atoms, T is half-life)
Using the equilibrium value isn't too accurate, since there are only 3 half-lifes of yttrium gone, so you'll need to solve a differential equation for the amount of yttrium that's left.

amazingphysics2255
@willem2 The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium 90 is also radioactive and is a beta particle emitter with a half-life of 2.67 days. Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

Perhaps this bit at the start will help?

Last edited:
amazingphysics2255
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%

Homework Helper
Gold Member
2022 Award
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.

Mentor
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%
Yes. That is correct.

Homework Helper
Gold Member
2022 Award
Yes. That is correct.
But it doesn’t help in answering the question. There is not a specific initial quantity of Y90. It is being created and destroyed simultaneously.

Mentor
But it doesn’t help in answering the question. There is not a specific initial quantity of Y90. It is being created and destroyed simultaneously.
Thanks. I misunderstood the question.

Mentor
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.
I think you need to give the OP a more detailed idea of how to proceed.

Homework Helper
Gold Member
2022 Award
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.
I think you need to give the OP a more detailed idea of how to proceed.
I didn't make the problem, so I solved it with what I was given.
Let ##N_Y(t)## and ##N_S(t)## be the number of atoms of Yttrium and Strontium respectively at time t. You know the values at time 0.
If the decay rates are ##\lambda_Y## and ##\lambda_S##, what is the rate of change of each count at time t?

Homework Helper
Gold Member
2022 Award
Is this not 90
90 is the number of nucleons in each atom of the Strontium. It is nothing to do with the number of atoms.
At the start there is no Yttrium, only Strontium
Firstly, isn't it beta decay?
The type of decay is irrelevant.
do you mean t time as in 8 days?
I mean at some arbitrary time t. A rate of decay ##\lambda## means that for each atom there is a probability ##\lambda dt## that it will decay in a short period dt.
If there are N(t) such atoms, how many on average will decay in time dt?
is this problem like mine?
No. The problems at that link only deal with the simplest situation: a decay from one species to a much longer lived one. Your problem involves decay to and from an intermediate species.

amazingphysics2255
90 is the number of nucleons in each atom of the Strontium. It is nothing to do with the number of atoms.
At the start there is no Yttrium, only Strontium
So there is only 1 atom Strontium. Which has 52 neutrons and 38 Protons

Homework Helper
Gold Member
2022 Award
So there is only 1 atom Strontium. Which has 52 neutrons and 38 Protons
There is an unknown quantity of Strontium. Unless there is further information on that, interpret the question as asking what fraction of the total number of atoms will be Yttrium.

amazingphysics2255
All I know is. The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium is a beta particle emitter with a half-life of 2.67 days.

I am then told to
Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

So I need to find out the amount of Yttrium that is in the strontium after decaying for 8 days?

Homework Helper
Gold Member
2022 Award
All I know is. The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium is a beta particle emitter with a half-life of 2.67 days.

I am then told to
Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

So I need to find out the amount of Yttrium that is in the strontium after decaying for 8 days?

amazingphysics2255
I mean at some arbitrary time t. A rate of decay λλ means that for each atom there is a probability λdtλdt that it will decay in a short period dt.
If there are N(t) such atoms, how many on average will decay in time dt?
So I need to find how many on average will decay in the dt? and dt is not 8 days is it?

Homework Helper
Gold Member
2022 Award
So I need to find how many on average will decay in the dt? and dt is not 8 days is it?
dt is some very brief period.
If an event happens at rate x, on average, how many events would you expect in time dt?

amazingphysics2255
I don't know if this help but I figured
For strontium you have the function

$$f(t)=100e^{ \frac{\ln(0.5)}{29 \times 365}}$$

as for your question, I'm not going to lie I don't fully understand it as to why I haven't answered it

Homework Helper
Gold Member
2022 Award
I don't know if this help but I figured
For strontium you have the function

$$f(t)=100e^{ \frac{\ln(0.5)}{29 \times 365}}$$
That's a useful start, but the 100 is only needed if you want the answer as a percentage. That will get confusing here, so let's drop that.
The exponent is missing the factor t; what you have written is not actually a function of time.
The constant in the exponent is (minus) the rate of decay, ##\lambda_S##, in events per day.
So we can write your equation more generally as
##N_S(t)=N_S(0)e^{-\lambda_S t}##
This can be obtained from the differential equation ##\frac{dN_S}{dt}=-\lambda_SN_S##. That is, the rate of loss of Strontium atoms is proportional to the number remaining and the decay rate ##\lambda_S##.

The next step is to obtain the differential equation for the number of Yttrium atoms. It is similar to the equation above, but we now need an extra term to represent the Yttrium atoms being created by decay of the Strontium atoms. Have a go at that.

amazingphysics2255
Homework Helper
If you are like me, the idea of writing down and solving differential equations may be off-putting. There is an approach that dodges that difficulty.

We have a formula for the rate at which Yttrium-90 that is being added to the sample as a function of time.

We can easily derive a formula for the fraction of the Yttrium-90 added at a particular time that will survive until the 8 day interval expires.

We could simply integrate the product of the two to figure out how much Yttrium-90 is added and will survive.

amazingphysics2255
Mentor
The half-life of the Strontium is very large compared to the half-life of the Yttrium (29 years vs 8 days). So during the 8 days, the amount of strontium hardly changes at all, and the rate of production of Yttrium is essentially constant.

Your initial calculation assumed that all the Yttrium that was produced from the Strontium decay suddenly appeared at time zero. In that case, there would be 12.5% of the Yttrium that was produced remaining after 8 days. But, since the Yttrium is being produced at a constant rate (rather than all at once), some of the Yttrium produced does not decay for 8 days. Most of it decays for less than 8 days. For example, the Yttrium produced on day 5 has only 3 days to decay. Based on this, do you think that the fraction of all the Yttrium produced during the 8 days and that still remains after reaching day 8 will be greater of less than 12.5%.

amazingphysics2255
Mentor
@haruspex When I wrote my function for strontium I also wrote this for Yttrium. So for the yttrium, you have to start with the decayed parts of the strontium.
$$f_{2}(t)= (100-f(t))e^{ \frac{\ln(0.5)}{2.7}}$$ But you've since advised me this is not the case.

@Chestermiller I believe it will be less than 12.5%

@jbriggs444 Yes I'd like to see that method?
It will be substantially greater than 12.5 %. I calculate about 42%. This is because all the yttrium produced has been around for less than 8 days. Do you know anything about writing and solving simple ordinary differential equations?

Last edited:
I have been wondering what is the course from which this question was taken? and what is the actual full
statement of the problem?

amazingphysics2255
I just asked it's not meant to be solved. Like this, they just want me to discuss.

Last edited:
Homework Helper
Gold Member
2022 Award
That would work if first the Strontium decayed for 8 days and then the Yttrium produced decayed for eight days. As everybody is trying to explain to you it is complicated by the two processes occurring simultaneously.

The simplest approach is to combine two suggestions that have been made:
Take the quantity of Strontium to be constant, since very little will have decayed in eight days.
Consider the quantity of Yttrium that was created in time dt at time t before the end of the eight days.
Multiply that by the fraction that will remain at the end (i.e. at time t later).
Perform the integral.


text:
# Problem: According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 x 103 m/s, the uncertainty in its position (in m) is at least.66 m 17 m 6.6 x 10-8 m 1.7 x 10-8 m 2.1 x 10-10 m

###### FREE Expert Solution
85% (268 ratings)
###### FREE Expert Solution

We’re being asked to determine the uncertainty in the position of an electron with uncertainty in the speed of 3.5 × 103 m/s.

Recall that Heisenberg’s Uncertainty Principle states that we cannot accurately determine both the position and velocity of an electron. This means we can only know either one at any given time

Mathematically, this is expressed as:

$\overline{){\mathbf{\Delta x}}{\mathbf{·}}{\mathbf{\Delta p}}{\mathbf{\ge }}\frac{\mathbf{h}}{\mathbf{4}\mathbf{\pi }}}$

where:

h = Planck’s constant (6.626 × 10–34 kg  m2/s)

Δx = uncertainty in position (in m)

Δp = uncertainty in momentum (in kg  m/s)

85% (268 ratings)
###### Problem Details

According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 x 103 m/s, the uncertainty in its position (in m) is at least.

1. 66 m

2. 17 m

3. 6.6 x 10-8 m

4. 1.7 x 10-8 m

5. 2.1 x 10-10 m

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Heisenberg Uncertainty Principle concept. You can view video lessons to learn Heisenberg Uncertainty Principle. Or if you need more Heisenberg Uncertainty Principle practice, you can also practice Heisenberg Uncertainty Principle practice problems.

What is the difficulty of this problem?

Our tutors rated the difficulty ofAccording to the Heisenberg uncertainty principle, if the un...as medium difficulty.

How long does this problem take to solve?

Our expert Chemistry tutor, Jules took 5 minutes and 15 seconds to solve this problem. You can follow their steps in the video explanation above.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Costanza's class at USF.