Graphing calculator pro. – Free app

Until tomorrow (Jan. 5th, 2011), the Graphing Calculator Pro iPod app is free (thanks Josh). It can do a lot, so it might take a minute figure out how to do the graphs and it can be a little fiddly (you can scale both axes separately, for example, which is nice unless you don’t intend to), but I like it. After just a couple of minutes, the student who found it was already playing with slopes.

Explaining algebra: concrete to abstract

Concrete to abstract, or abstract to concrete? Bottom up or top down? Introducing new concepts can be done either way, but which way is best? Students tend to learn better when they’re building on an existing scaffolding. However, some students are more adept at seeing the bigger picture first then analyzing the details, while others favor seeing the smaller details and constructing the bigger picture of the pieces of the puzzle.

In yesterday’s lesson introducing algebra, I started concretely with weights on the scale for variables and the fulcrum as the equal sign. But I worked with them abstractly. Instead of telling students I had 200 g, 100 g and 50 g weights, I said we had three types of weights and called them a, b and c.

The weights were set on the scales as above so we wrote out the equation:
! a + c = b + b + c
and simplified to give:
! a + c = 2b + c

Then we talked about balance. Whatever you do to one side you have to do to the other, so I took the c weights off and showed how it solved for a:

! a + c -c = 2b + c - c
! a = 2b

Now you can solve for b by dividing everything by 2 to get:
! \frac{a}{2} = \frac{2b}{2}
! \frac{a}{2} = b

However, since you can’t exactly divide the 200 g weight into two, you can’t exactly demonstrate this, but at least you can show the math. You can also now substitute in the values for the weights. If b = 100 g.

! a = 2b
! a = 2 (100)
! a = 200

At this point, students could look at the weights and see the numbers.

Concrete to abstract and back again, I like how the lesson turned out, although, today I had to go over how to show their work again.

And Pythagoras Scores!

Locating the center point of the soccer field.

Our school, Lamplighter, has started up a couple soccer teams to play in the local under-8 and under-6 leagues this year. I’m now one of the under-6 coaches, and the curious similarities between them and the middle schoolers is going to have to be the topic of another post; Montessori observed some interesting parallels between the first and third planes of development that are worth getting into. However, since the teams are new, we did not have lines on the practice field. And teaching throw-ins is kinda tricky with imaginary lines.

One of the parents/coaches of the under-8 team, Mr. Surbrook, offered draw out the lines. He also volunteered to give a lesson on geometry and let the middle school (and upper elementary) students help.

Refreshing ourselves on Pythagoras' Theorem.

To prepare the middle schoolers I did a quick review of Pythagoras’ Theorem using the 3×3, 4×4 and 5×5 squares (see above).

! a^2 + b^2 = c^2

The lesson was interesting because the 7th graders had had a more recent exposure to the equation but, unlike the 8th graders, have not had any algebra yet, so there were some puzzled looks when I rearranged the equation.

Lesson on the geometry of rectangles.

That was in the morning. After lunch Mr. Surbrook came in and showed us how to use Pythagoras’ Theorem to make right angles and locate the center of the field. If stretch out six pieces of string, four for the sides and two for the diagonals (calculated with the equation,) at their fullest extent you have a rectangle with decent right angles.

Corner of the soccer field. Note the nice 90 degree angle.

After figuring out the theory inside, we went out to the field and help cut the string and lay out the lines. The kids were a bit disappointed they did not get to actually paint the lines, but we’d run out of time for the day.

Fortunately, they’ll get another chance at surveying when Dr. Houghton brings her class out to map the topography of the campus.

I very much liked how the whole procedure went, with my preparatory lesson first, then Mr. Surbrook practical lesson, and finally the actual practical application. We did something similar when we laid out the greenhouse the first time. That time we threw the kids in without a guide and without the practical lesson. It was a bit of a team-building exercise. It also took quite a bit longer.

Find c.

LaTeX

The easy way to stick mathematical equations into documents would probably be to use Microsoft’s Equation Editor in Word. But that makes it difficult to transfer things from one computer to another, especially if someone does not have MS Word. I prefer to use LaTeX. It’s free, open-source and usually pretty easy to set up on a server. It enables me to put equations inline on the class wiki and now, thanks to the Easy LaTeX plugin I can have them on the blog too.

I’ve been wanting to do this especially since doing the jam algebra post. Then I was lucky that I could, just barely, do everything with text. Now, however, instead of:

(7) 0.4 s / 0.4 = 0.6 j / 0.4

I can do this:
(7) ! \frac{0.4 s}{0.4} = \frac{0.6 j}{0.4}

Which has it’s pluses and minuses. However, before I would not have been able to do this (at least not very easily):
! \sqrt{x} = \sqrt[2]{x^1} = x^{\frac{1}{2}}

Using the LaTeX math markup is not exactly trivial (if you put your cursor over the equation you can see it), but Kocbach (date unknown) and Downes (2002) are great resources.

Jam algebra

I’d like to minimize the sugar content of the jam in order to see the most of the currant’s tartness. According to the FAO, you need to have about 60% sugar concentration in the final jam for good preservation. I’ve squeezed the currants and produced quite a bit of juice. I need to find out how much sugar to add.

To figure out how much sugar we need to add, based on the mass, we need to define our terms. Let’s say the amount of sugar is s, the amount of jam is j and the total mass, the sugar plus the jam, is t.

So the total mass is going to be:

(1) t = s + j

Since the amount of sugar needs to be 60% then:

(2) s = 0.6 t

If we substitute the first equation (1) into the second (2) we’ll have just one equation we can solve to find the amount of sugar:

(3) s = 0.6 (s + j)

Now we solve for the amount of sugar, s. Start by expanding the right hand side of the equation (distributive property):

(4) s = 0.6 s + 0.6 j

Next, isolate s on the left hand side of the equation by subtracting:

(5) s – 0.6 s= 0.6 s + 0.6 j – 0.6 s

Which gives:

(6) 0.4 s= 0.6 j

Now, get rid of the coefficient on the left hand side by dividing through:

(7) 0.4 s / 0.4 = 0.6 j / 0.4

To get:

(8) s = 1.5 j

So to have the right amount of sugar, I need to add one and one half times as much sugar as I have juice. So, if I have 2.0 kg of juice, then:

s = 1.5 (2.0 kg)
s = 3.0 kg

3.0 kg of sugar is a lot of sugar! However, if we boil the jam, some of the water in the juice will evaporate. Therefore, if we know how much sugar we want to add in we should be able to calculate how we need to evaporate to get the right ratio. More evaporation should also lead to a more concentrated flavors in the jam. Hmmm …

Well, I have a scale.