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IntMath Newsletter: Forum, pistons, Shodor

By Murray Bourne, 26 Jan 2016

26 Jan 2016

In this Newsletter:

1. New: IntMath Forum
2. Piston curve
3. Resource: Shodor Interactivate
4. Math puzzles
5. Math movies: Misleading statistics
6. Final thought: Resolutions, 1% at a time

1. New: IntMath Forum

A lot of people have asked for a forum over the years. Finally, it's here!

IntMath Forum

I come across too many low quality forums and I'm hoping to avoid that. I also get a lot of mails containing math homework questions where the writer just wants me to provide the answer. This doesn't help him learn much - it just helps his marks. So there are 5 simple principles for the forum:

Principles of the IntMath Forum

You can enter math in the forum using simple plain text (or LaTeX if you prefer). You can also add graphs and images to your post. Details on how to do so can be found here:

How to enter math, graphs and images

Wanted: Volunteer Tutors

The forum needs tutors that embrace the 5 Principles mentioned above. That is, tutors should firstly be confident in answering the question, but help readers figure out problems by themselves, not just do it for them. The best help is given by prompting, reminding, and asking good questions - not simply providing the answer.

To get an idea what I mean, see these exchanges where I'm pointing the user to relevant resources, and asking questions. Eventually, they work it out:

Let me know (by replying to this mail, or in the comments) if you are interested to be a tutor.

2. Does a piston follow a sine curve shape?

Interactive animation of piston curve (almost sinusoidal)

In the Trigonometric Graphs section, I used a piston example for one of the sine curves. A reader pointed out that a piston does not exactly follow a sine curve at all.

Here's a short article with an interactive animation that explores this issue:

Curve Shape for the Movement of a Piston

3. Resource: Shodor's Interactivate

Interactive animation of piston curve (almost sinusoidal

I've been a fan of Shodor's Interactivate for many years. Indeed, it helped to inspire IntMath. Their activities used to be java-based, and so didn't work on mobile devices.

But now, their interactive activities are mobile friendly. See over 300 activities, covering algebra, functions, ciphers, calculus, and probability here:

Shodor's Interactivate

4. Math puzzles

The puzzle in the last IntMath Newsletter asked about the ratio of the areas of Reuleaux Triangles.

Correct answer with explanation was given by Stephen, who was also was one of the few to express the final answer as a ratio, as required.

New math puzzle

The length of a rectangle increases by 10% and the width decreases by 10%. Which of the following is true?

  • The area does not change.
  • The area becomes smaller or larger depending on the original lengths of the sides.
  • The area will always become smaller.
  • The area will always become larger.

You can leave your responses here.

5. Math movie: Misleading statistics

How statistics can be misleading - Mark Liddell

This video outlines Simpson's Paradox, which all budding statisticians need to be aware of. This is part of a TED-Ed lesson.

How statistics can be misleading

Related article: Misleading charts

Around the same time I watched How statistics can be misleading, I came across the following article on Quartz.

I feel it's a vital mathematical skill to be able to properly analyze data, especially when it's in visual form.

I don't agree with all the writer's suggestions, but overall it's quite a good analysis:

The most misleading charts of 2015, fixed

6. Final thought: Resolutions, 1% at a time

You probably made new year resolutions and now that life has taken over, maybe they've been shelved already.

A possibly more effective and realistic way of challenging yourself to improve is to aim for 1% improvement per day (or some other reasonably short time frame). If you want to exercise more, try just walking (or running, or any other exercise) 1% more than the day before. This will mean you'll double your walking distance in just 70 days:

(1.01)70 = 2.007

You could apply a similar small and realistic improvement goal for your studies: say 5% improvement for each weekly test.

Either way, all the best achieving your goals in 2016!

Until next time, enjoy whatever you learn.

See the 17 Comments below.

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Posted in Intmath Newsletters category - 26 Jan 2016 [Permalink]

17 Comments on “IntMath Newsletter: Forum, pistons, Shodor”

  1. Francis Kisner says:
    26 Jan 2016 at 7:51 pm [Comment permalink]

    Original Area = LW
    New Area = (1.1 L)(.9 W) = .99 LW
    The combination of the two changes always results in a decrease in the original area.

  2. Rutwik Tank says:
    26 Jan 2016 at 8:22 pm [Comment permalink]

    Solution of the puzzle:
    The area will always become smaller.
    Explanation:
    Let the length of the rectangle= x
    and the width of the rectangle= y
    Clearly, x>0, y>0
    So the area of the rectangle= x*y
    Now,
    new length= x+(x/10)= (11*x)/10
    new width= y-(y/10)= (9*y)/10
    So the new area= (11x/10)*(9y/10)= 0.99*xy, which is always less than xy.

  3. Kai says:
    27 Jan 2016 at 1:18 am [Comment permalink]

    Rectangular Areas

    Let the area of the original rectangle be

    A = l * w

    If we increase the length by 10%, we have a new length l_1 given by
    l_1 = l + 0.10l = 1.1l

    Similarly, the new width will be
    w_1 = 0.9w

    The new rectangle then will have an area of

    A_1 = (1.1l)(0.9w) = 0.99lw = 0.99A

    So the new area will always be 99% of the old area - a 1% decrease

  4. Dineth says:
    27 Jan 2016 at 1:32 am [Comment permalink]

    Area will always become larger. (As length > width of a rectangle and area = length * width)

  5. Jessica says:
    27 Jan 2016 at 1:48 am [Comment permalink]

    Let L = original length and W = original width. Original area = L*W.
    Length increases by 10%, so new length = 1.1*L
    Width decreases by 10%, so new width = 0.9*W
    New area = (1.1*L)(0.9*W) = 0.99*LW
    Therefore, the new area is always 99% of the old area, so the new area is always smaller than the old area.

  6. Don Miller says:
    27 Jan 2016 at 3:55 am [Comment permalink]

    Always smaller

  7. Tomas Garza says:
    27 Jan 2016 at 9:11 am [Comment permalink]

    The area wll always become smaller. Call w the width and w + k the length (k > 0). Then the area is A = w (w + k). If the length is increased by 10% and the width decreased by 10%, the resulting area is 0.9 w times 1.1 (w + k) = 0.99 w (w + k) = 0.99 A, which is less than A.

  8. Chloe says:
    27 Jan 2016 at 3:30 pm [Comment permalink]

    The areas will become :
    Smaller or larger depending on the original values.

  9. Andrzej Cichy says:
    27 Jan 2016 at 9:26 pm [Comment permalink]

    Consider a rectangle with the length a and width b.
    Area before change {A}_{1}={a}{b}
    The length a increases by 10% so {a}_{2}={1.1}{a}
    The width b decreases by 10% so {b}_{2}={0.9}{b}
    Area after change {A}_{2}={0.99}{a}{b}, so {A}_{2}={0.99}{A}_{1}

    So the area will always become smaller.

  10. Nicos says:
    28 Jan 2016 at 2:31 am [Comment permalink]

    Let the length be a and width b. Then the area E=a*b. If a is increased 10% and b decreased 10%, then a'=1.1*a and b'=0.9*b. The new area E'=a'*b'=1.1*a*0.9*b=0.99*a*b, so E'=0.99*E and the area will always become smaller, the 99% of the initial area.

  11. michael says:
    28 Jan 2016 at 5:40 am [Comment permalink]

    cute

    {A}_{1}={l}_{1}\times{w}_{1}
    {l}_{2}={1.1}\times{l}_{1}
    {w}_{2}={0.9}\times{w}_{1}

    {A}_{2}={l}_{2}\times{w}_{2}
    {A}_{2}={\left({1.1}\times{l}_{1}\right)}\times{\left({0.9}\times{w}_{1}\right)}
    {A}_{2}={0.99}\times{A}_{1}
    The area will always become smaller.

  12. Roger Evans says:
    28 Jan 2016 at 5:26 pm [Comment permalink]

    Hello,
    The math puzzle...
    The area will become smaller.
    Take a rectangle 200 cms long with width 100 cms.
    200 X +10% = 220 cms long, 100 X -10% = 90 cms width.
    Multiply the original size out, subtract from the new size; the new rectangle has a smaller area.
    Kind regards,
    Roger

  13. Guido says:
    28 Jan 2016 at 8:45 pm [Comment permalink]

    The area will always become smaller and exactly by 1% .
    Area = L*B ----> 1.1L*0.9B = 0.99L*B

  14. Nour Hlwani says:
    29 Jan 2016 at 6:26 am [Comment permalink]

    Area of A rectangle = LXl
    Where L --> length and l --> width

    L increased by 10% -> L(1+0.1)=1.1 L
    l decreased by 10% -> l(1-0.1)=0.9 l

    The new area = 1.1×0.9×L×l = 0.99×(old area)
    Since 0.99<1 Area is slightly decreasing

    The right answer is (the area will alway become smaller.

  15. Steve m. says:
    1 Feb 2016 at 3:53 pm [Comment permalink]

    The new area will always be smaller

    i.e. new area= 0.99 * original area

  16. Janet Teeguarden says:
    3 Feb 2016 at 6:27 am [Comment permalink]

    The area will always get smaller, no matter which number is called length and which is width. New area = (L + 0.1L)(W - 0.1W) = LW + 0.1LW - 0.1LW - 0.01LW = LW - 0.01LW. So the new area is always one percent smaller than the original area LW. (This seems too easy...what did I miss here?)

  17. KIPRONO ERIC says:
    16 Feb 2016 at 2:43 pm [Comment permalink]

    The area of the triangle will automatically reduce.the reduction will b an approximation of 1% as compared to the previous one.

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