IntMath Newsletter: Subitizing, Jacob, Fractals and Donald Duck
By Murray Bourne, 12 Feb 2013
12 Feb 2013
In this Newsletter:
1. Gong Xi Fa Cai
2. Counting game and Subitizing
3. Jacob's Staff
4. Creating a Google-like classroom climate
5. Math puzzles
6. Friday math movie
7. Final thought - Perfect days
1. Gong Xi Fa Cai!
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Happy Year of the Snake to all those enjoying Chinese New Year festivities this week! Chinese calendars are quite interesting. A "standard" year has around 354 days, and there are leap years containing a leap month, with around 384 days. The calendar is always adjusted so the winter solstice (the shortest day of the year) falls in the 11th month. Months always start on the new Moon (when it is completely in shadow). |
Chinese New Year falls on the second new Moon after the December solstice (unless a leap month messes things up). The earliest possible date is Jan 21st, and the latest is Feb 21st (but that won't happen until 2319).
2. Counting game and subitizing
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Can you count faster than a chimp? Our "number sense" is important for understanding math, and it begins with our ability to count. First, try the counting game. All you have to do is count the dots as quickly as you can. |
Next up is a short article on why I created the game, and some early observations:
Then, for some interesting background, see this article:
3. Jacob's Staff
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Scientists used Jacob's Staff during the Renaissance to find heights and distances, using trigonometry. It was a forerunner to the sextant. Math teachers could design some interesting activities using this, so students would learn some math, and some math history. See: Jacob's Staff |
Here's a similar activity based on historic math: Noon Day Project. This activity re-creates Eratosthenes' famous experiment 2000 years ago where he produced a good approximation of the size of the Earth based on measuring shadow lengths at noon in different parts of Egypt.
4. Creating a Google-like classroom climate
Google is well-known for being one of the best companies in the world to work for. Around 7,000 people per day apply to work there. The company offers excellent benefits and a relaxed, almost playful atmosphere that keeps creative juices flowing.
They put a lot of effort into finding the right people, and because Google doesn't want to lose anyone, they create conditions that encourage success and fulfillment.
The article, Inside Google’s Culture of Success and Employee Happiness, got me thinking about why so many students drop out of school (often because of acute math anxiety, or because they don't have a feeling of belonging, or a host of other reasons).
Good classroom "climate" is very important for effective learning, especially in math!
Here are some of the things Google does that could apply in class:
Hire (and reward) good teachers! I enjoy living in Asia where education is generally held in high esteem, and teachers are, for the most part, regarded as valuable members of society. Good teachers almost always leads to better learning outcomes. Keep the good ones by giving them purpose, recognition, and reward.
Collect data on (and share) what works: I don't mean what works for standardized tests, here. I mean quality class activities, different approaches to explaining things, or different resources. What works? What needs changing? Teachers aren't given enough time to think about what they're doing, nor enough opportunities to learn from each other. (Google collects huge amounts of data to improve its processes, and creates many situations where workers mingle and share ideas.)
Warm greetings make a huge difference: Google found their best managers were warm and accepting of input from their team, leading to a "15% increase in productivity". Sadly, I see a lot of classrooms where there are no greetings at the beginning of the day, and no friendly talk with students about mutual interests. It's often no more than "OK, today we're going to do the quadratic equation...". A sense of belonging is important for learning and for sticking around.
Too much bureaucracy: Google has a "flat" structure where middle managers have quite a bit of responsibility, compared to many other companies. That's not always the situaion with teachers.
Too many distractions: Many classrooms have too many interruptions. Are those announcements about sport buses - or uniforms - really that crucial? There are already too many things trying to grab everyone's attention. Focus is good.
In summary, we need to create the best working & learning environment we can, whether it's in school, or in a company.
5. Math puzzles
The puzzle in the last IntMath Newsletter involved finding the time taken to fill a bath.
Correct answers with reasoning were given by Christopher Buchanan, Christian Mills, Nicos Mavrommatis, S. Nickerson, Thomas A Buckley, Bonnie, Hamid Zandi, Lachezar Borisov, Dineth, Ramesh Babu and Mawanda Ismail.
It was interesting to see the variety of thought processes used to solve the puzzle. Some of you went straight for the algebra, and that's fine. My approach with this kind of puzzle is to assume a reasonable amount of water for the bath (say 100 liters) and then work the rest of it from there. It makes it easier (for me) to check if my "rate in" and "rate out" values make sense.
Please note: As usual in math, the best answers are those that include reasons!
New puzzle: Peter is an enthusiastic student and runs to school each day, but usually gets tired and walks the rest of the way. His running speed is twice his walking speed. He notices on Monday that he walks twice the time he runs and it took 20 minutes to get to school. On Tuesday, he runs twice the time he walks. How long does it take to get to school on Tuesday?
You can leave your responses, with reasoning, here.
6. Friday math movies
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(a) Fractals - Hunting the Hidden Dimension See this PBS documentary about the development of fractals and how they are used in computer graphics. (It's 55 minutes, but good value.) |
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(b) Donald in Mathmagic Land This is a classic late-1950s cartoon where Donald learns where math comes from. |
7. Final thought: A perfect day
I enjoy helping people, and I hope you do, too. I think this sums things up nicely:
You have not lived a perfect day, unless you've done something for someone who will never be able to repay you. [Ruth Smeltzer]
Until next time, enjoy whatever you learn.
See the 13 Comments below.
12 Feb 2013 at 8:50 pm [Comment permalink]
monday :d (runing)=2(t),d(walking)=1(2t)total distance is 4t assuming walkin speed one and running speed 2 tuesday:d(runing)=2(2x) , d (walking)=1(x) so total time on monday is 3t=20 total distance =4t=5x which gives x=16/3 total time on tuseday is 3x=16 answer is 16
13 Feb 2013 at 2:14 am [Comment permalink]
In this Newsletter:4. Creating a Google-like classroom climate .--This part describes some phenomenon of the math class .I think these descriptions are very real.The classrooms lack of inspiration and exchange.Math teacher don't think and find more effective method.
13 Feb 2013 at 4:14 am [Comment permalink]
Let Twa=time spent walking on day a(Monday)
Let Tra=time spent running on day a(Monday)=Twa/2
Let x=walking speed (constant on all days). Running speed is defined as 2x.
Let D=the distance to school (a constant)
Computing the time spent walking on Monday is:
20 = Twa + Twa/2 yielding
Twa=40/3 and consequently Tra=20/3.
The distance (speed*time) is therefore:
D= x*(40/3) + 2x*(20/3)
Solving for walking speed, we get:
x=3*D/80.
Now, let Twb=time spent walking on day b(Tuesday) and let Trb=time spent running on Tuesday, or 2*Twb because he ran twice the time he walked the second day.
The distance equation for the second day is therefore
D = x*Twb + (2x)*2*Twb
But since x (walking speed) is constant both days, and was computed to be 3*D/80, we have:
D = (3*D/80)*Twb + 2*(3*D/80)*2*Twb
The D's cancel and we are left with
Twb=80/15 or 5.333 minutes.
Since we are told he ran twice as much as he walked on Tuesday, Trb=10.667 minutes
Therefore, it took him 5.333 + 10.667 = 16 minutes to get to school on Tuesday.
13 Feb 2013 at 8:11 am [Comment permalink]
Answer, it takes Peter 16 mins to get to school on Tuesday.
Let Distance to school = D, walking speed = s,
For Tuesday, walk time = tw, run time = tr
From distance = speed x time, for Monday
D = walk distance + run distance
D = s (2x20/3) + 2s (20/3) = s(80/3) . . (1)
For Tue, same distance
D = walk distance + run distance
D = tw ( s ) + tr ( 2s)
D = tw ( s ) + 2tw ( 2s) . . Run time is twice walk time.
D = 5tw (s) . . . (2)
As both (2) and (1) = D
5tw (s) = s (80/3) . . Cancel s
tw = 16/3 Mins. . . And Run time is twice walk time
tr = 2 x (16/3) Mins
Total Tuesday time Walk + Run times
= 16/3 + 2 x (16/3) Mins
= 16 mins = Answer
13 Feb 2013 at 6:17 pm [Comment permalink]
The solution is based on the equation S= VT {1}, where V is the velocity , T is the time and S is the distance. Further the distance on monday is equal to the distance on tuesday. On monday we can write : Trun + Twalk = 20 {2}and Twalk = 2Trun. Substituting in {2} : 3Trun = 20 -----> Trun = 20/3 and consequently Twalk = 40/2. Considering {1} for both monday and tuesday : Vwalk.Twalk(m)+Vrun.Trun(m)=Vwalk.Twalk(t)+Vrun.Trun(t)
Substituting Vrun = 2Vwalk and Twalk=40/2 into the equation:
40/2Vwalk(m)+2Vwalk.20/3=Vwalk.Twalk(t)+2Vwalk.Trun
80/3Vwalk = Vwalk{Twalk(t)+2Trun(t)} On tuesday:Trun=2Twalk
80/3=Twalk(t)+4Twalk
Twalk=16/3
Twalk + Trun = 16/3 +32/3 = 48/3 = 16 min
13 Feb 2013 at 10:59 pm [Comment permalink]
Constants: Run speed = 2x
Walk speed = x
Total distance = z
Run time = t
Monday: Walk time = 2t
t + 2t = 20
t = 20/3
z = 2x*t + x*2t
=(80/3)*x
Tuesday: Walk time = t/2
(80/3)*x = 2x*t + x*t/2
(80/3)*x = (5x/2)*t
t = 32/3
Total Time Tuesday = t + (t/2)
= 16 mins
15 Feb 2013 at 11:26 pm [Comment permalink]
let the walking speed be s and t be the time used on Tuesday
then the running speed is 2s
thus distance (d) = (20/3xs + 2x20/3x2s) .........(i)
or (d) = (t/3x2s + 2xt/3xs).........(ii)
equating (20/3xs + 2x20/3x2s)= (t/3x2s + 2xt/3xs)
20/3 +40/3 = 2t/3 +2t/3
60/3 = 4t/3
60 = 4t
t = 60/4
t = 15
It takes 15 minutes to get to school on Tuesday
16 Feb 2013 at 5:20 am [Comment permalink]
Let Vr:running velocity, Vw:walking velocity, Tmw:walking time on Monday, Tmr:running time on Monday, Ttw:walking time on Tuesday, Ttr:running time on Tuesday, Smw & Smr:walking & running distance on Monday, Stw & Str:walking & running distance on Tuesday, So:distance Home-School, Tm:total time on Monday(=20min), Tt:total time on Tuesday.
Generally: (constant)Velocity (V)= Distance (S)/Time (T) or V=S/T
It is given that:Tm=20min, Tmr=2Tmw (1), Ttr=2Ttw (2),Vr=2Vw (3)
We're looking for Tt=?
Tm=20'=Tmr+Tmw=Tmr+2Tmr=3Tmr due to (1)
Thus, Tmr=20/3 min and Tmw=40/3 min (4).
So=Smr+Smw=Vr*Tmr+Vw*Tmw=2Vw*(20/3)+Vw*(40/3) due to (3)&(4)
So=80Vw/3 (5)
On Tuesday: So=Str+Stw=Vr*Ttr+Vw*Ttw=2Vw*2Ttw+Vw*Ttw=5Vw*Ttw due to (2)&(3)
Thus, So=80Vw/3=5Vw*Ttw => Ttw=16/3 min. Due to (2):Ttr=32/3 min
The total time Tt on Tuesday is Tt=Ttr+Ttw=(32+16)/3=48/3=16 min.
It takes 16 minutes to get to school on Tuesday.
18 Feb 2013 at 8:43 am [Comment permalink]
Please correct. You should say Lunar New Year not Chinese New Year. Depending on the longitude of the capitals as well as subtle differences in the way they calculate. In the past, Chinese and Vietnamese may celebrate Lunar New Year on different date.
This is from Computing the Vietnamese lunar calendar
18 Feb 2013 at 8:44 am [Comment permalink]
@Sieu Do: Thanks for the clarification.
19 Feb 2013 at 9:09 pm [Comment permalink]
r=runs w=walk x=path v=velocity X=total path
t=time
we know vr=2*vw
in monday:
tw=2*tr,tw+tr=20min => tr=20/3; tw=40/3
X=xr+xw => X=xr+xw => X=vr*tr+vw*tw => X=vr*20/3+2*vr*40/3
=> X= vr*100/3
in tuesday:
tr=2*tw => X=vr*tr+vw*tw => X=tr*vr*5/4
finnallyn:tr*vr*5/4=vr*100/3=> tr=400/15,tw=200/15=>time=40min
11 Mar 2013 at 6:23 pm [Comment permalink]
I think it took 10 minutes to get to school on tuesday because from the first statement running speed is twice walking speed impling that walking time=twice running time (since speed is inversely proportional to time), on Monday running time=2 times walking time. From the two statements, it implies that whenever his walking time is twice his running time, it takes 20 mins to get to school, therefore, it will take him 10 mins to get to school on tuesday since his running time was twice his walking time and as such less time will be taken to get to school.
24 Mar 2013 at 4:21 am [Comment permalink]
Peter is an enthusiastic student and runs to school each day, but usually gets tired and walks the rest of the way. His running speed is twice his walking speed. He notices on Monday that he walks twice the time he runs and it took 20 minutes to get to school. On Tuesday, he runs twice the time he walks. How long does it take to get to school on Tuesday? If v1=walking speed, v2=running speed=2*v1
On Monday: t1=walking time, t2=running time
On Tuesday: t3=walking time 2*t3=running time
Thus: 2*t1*v1 + t2*2v1=e (distance of school)
2v1(t1+t2)=e : t1+t2=20 : 40*v1=e
On Tuesday: t3*v1+2*t3*2v1=e : v1(t3+4t3)=40*v1
For transitive character: 40= t3+4*t3 : 5*t3=40
On tuesday it take 5 minutes.