I have been teaching a low literacy class this year and it has taught me the perils of relying on the immersion technique to teach mathematical literacy. After seeing the positive effects of direct instruction of low literacy students, I see immersion as a lazy teaching technique for low ability students - immersion is slow, ineffective and generally detrimental to these students, especially in a large heterogeneous class.
Let me explain. When I teach area, I generally teach students to write a story that I can read. They draw a diagram, label the sides, write a general equation, substitute the values, evaluate the solution and check their answer. Once this technique has been learned I can then easily teach other concepts such as Pythagoras' theorem, trigonometric ratios, surface area and the like.
The explicit teaching of mathematical literacy (requiring specific layout and explicitly explaining the meaning and need for each component of the layout) is the key component in this exercise. By year ten, most can answer the area of a rectangle and write the answer, but cannot abstract the method to a circle. I attribute this to a lack of mathematical literacy and a failure to appreciate the true need for mathematical literacy.
Although modelling has a place in teaching mathematics (a key tool in immersion), we must be mindful that we need to teach literacy explicitly and not assume students will just pick up major concepts by observing a question being completed. The difference between a student answering a question correctly (after being given an example on the board) and being able to identify how to answer a question correctly from a range of tools (without prompting) is considerable.
Mathematics has grammar just like English. If students understand the grammar of mathematics, the meta-language of mathematics and the algebraic/arithmetic/visual representations/tools of mathematics, then their ability to solve problems increases exponentially.
And as mathematics teachers we can appreciate the benefits of exponential growth.
Sunday, May 23, 2010
Teaching of mathematical literacy
Location:Perth, WA, Australia
Perth WA, Australia
Saturday, May 22, 2010
Functions and the Casio Calculator fog(x)
When finding fog(x) the CAS calculator does a good job of simplifying algebraic steps, often to the point of making traditional questions trivial in the calculator section.
I'll start by defining a function f(x) =2x+1 and g(x)=x^2 and attempting to find fof(x), fog(x) and the inverse of (fog(x)).
To define a function f(x) go to the Main window and select
Interactive->Define
A window will appear
Enter the function name (eg. f)
Variable (eg. x)
Expression (eg. 2x+1)
The main window will reply:
Define f(x)=2(x)+1
done
Repeat:
Interactive->Define
Enter the function name (eg. g)
Variable (eg. x)
Expression (eg. x^2)
The main window will reply:
Define g(x)=x^2
done
To find fof(x)
In the main window
Action->Transformation->simplify
f(f(x)) (using the soft keyboard)
press exe
The main window should say
simplify(f(f(x))
4x+3
To find fog(x)
In the main window
transformation->simplify
f(g(x)) (using the soft keyboard)
press exe
The main window should say
simplify(f(g(x))
2x^2+1
To find inverse of fog(x)
In the main window
action->transformation->simplify
action->assistant->invert
action->advanced->solve
y=f(g(x)),x))) (using the soft keyboard= don't forget the "y=")
press exe
The main window should say
simplify(invert(solve(y=f(g(x)),x)))
Here is a link to other CAS calculator posts.
I'll start by defining a function f(x) =2x+1 and g(x)=x^2 and attempting to find fof(x), fog(x) and the inverse of (fog(x)).
To define a function f(x) go to the Main window and select
Interactive->Define
A window will appear
Enter the function name (eg. f)
Variable (eg. x)
Expression (eg. 2x+1)
The main window will reply:
Define f(x)=2(x)+1
done
Repeat:
Interactive->Define
Enter the function name (eg. g)
Variable (eg. x)
Expression (eg. x^2)
The main window will reply:
Define g(x)=x^2
done
To find fof(x)
In the main window
Action->Transformation->simplify
f(f(x)) (using the soft keyboard)
press exe
The main window should say
simplify(f(f(x))
4x+3
To find fog(x)
In the main window
transformation->simplify
f(g(x)) (using the soft keyboard)
press exe
The main window should say
simplify(f(g(x))
2x^2+1
To find inverse of fog(x)
In the main window
action->transformation->simplify
action->assistant->invert
action->advanced->solve
y=f(g(x)),x))) (using the soft keyboard= don't forget the "y=")
press exe
The main window should say
simplify(invert(solve(y=f(g(x)),x)))
Here is a link to other CAS calculator posts.
Inverse Functions and the Casio classpad 330
The 3CD course requires knowledge of domain, range, co-domain and inverse functions. The classpad can handle these in a number of ways.
The most obvious way is in graph mode (Menu-> graph tab). Set up a graph (say y=(x+1)/(x+2), and visually examine it to find the domain and range. To find the inverse, click on the graph and select Inverse (Analysis->Sketch->Inverse). The equation of the inverse can be found by selecting the inverse graph and examining the equation bar at the base of the screen. Be careful with this method, as the resultant inverse graph is not written in "y=" form.
The less obvious way to do it is in the main window. I have not found an "inverse" function yet, but the following is a workaround to find the inverse.
Go
Action->Transformation->Simplify (to simplify the resultant equation)
Action->Assistant->Invert (to swap the x & y variables around)
Action->Advanced->Solve (makes x the subject of the equation)
Enter the function
Press ",x)))" using the soft keyboard (to make x the subject of the new equation)
It would look something like this when finished:
simplify(invert(solve(y=((x+2)/3,x)))
In a way I prefer the main window method as it mirrors the algebraic method. I think students need to really understand the parts of an equation to effectively find the domain and range. I explicitly draw students attention to critical information such as the inability of the function to equal zero or where the function is undefined
a) Look for possible values of x where y=1/0 will occur (eg. {x≠-1} for y=1/(x+1) ).
b) If it is not possible for the numerator to be zero (eg. {y≠0} for y=2/(x+1))
If the range is not obvious it is often easier to examine the domain of the inverse of the function.
Here is a link to other CAS calculator posts.
Purplemath has a good explanation of inverse functions.
The most obvious way is in graph mode (Menu-> graph tab). Set up a graph (say y=(x+1)/(x+2), and visually examine it to find the domain and range. To find the inverse, click on the graph and select Inverse (Analysis->Sketch->Inverse). The equation of the inverse can be found by selecting the inverse graph and examining the equation bar at the base of the screen. Be careful with this method, as the resultant inverse graph is not written in "y=" form.
The less obvious way to do it is in the main window. I have not found an "inverse" function yet, but the following is a workaround to find the inverse.
Go
Action->Transformation->Simplify (to simplify the resultant equation)
Action->Assistant->Invert (to swap the x & y variables around)
Action->Advanced->Solve (makes x the subject of the equation)
Enter the function
Press ",x)))" using the soft keyboard (to make x the subject of the new equation)
It would look something like this when finished:
simplify(invert(solve(y=((x+2)/3,x)))
In a way I prefer the main window method as it mirrors the algebraic method. I think students need to really understand the parts of an equation to effectively find the domain and range. I explicitly draw students attention to critical information such as the inability of the function to equal zero or where the function is undefined
a) Look for possible values of x where y=1/0 will occur (eg. {x≠-1} for y=1/(x+1) ).
b) If it is not possible for the numerator to be zero (eg. {y≠0} for y=2/(x+1))
If the range is not obvious it is often easier to examine the domain of the inverse of the function.
Here is a link to other CAS calculator posts.
Purplemath has a good explanation of inverse functions.
Sunday, May 9, 2010
Composite Area problems
With my year 10 "literacy focus" mathematics class we have been looking at the language of measurement, specifically area and perimeter. The idea has been to get students to clearly understand the importance of units and how to interpret what questions are asking.
When thinking about perimeter we are only thinking about the boundary outside a shape. We then looked at some real world shapes and measured their boundaries, followed by drawing some scale diagrams of shapes and measuring their boundaries.
We then tried to describe how much space was in the shapes. I tried to drive them to describing dimensions, but some students had enough prior knowledge to say calculate the answer by multiplying lengths.
The main issues came as we arrived at composite shapes where composite shapes had to be split into basic geometric shapes. Deriving missing measurements really brought home how much trouble low literacy students can have with basic mathematical concepts such as subtraction. Assuming that a student can see how to find missing sides is in many cases overestimating their ability.
Even at year 10, the majority of low literacy students fail to see f- c = a whereas they are more likely to be able to see a+ c = f.
Because of this to use a scale drawing to assist these students see subtraction in action requires some thought. If a student drew the above diagram with the following measurements they would
not use subtraction to find the missing side on the top rectangle. They would count the squares (if on grid paper) or measure the vertical gap and put the measurement of one on the page.
To encourage students to use subtraction I needed to encourage students to first look at the diagram and do a number of examples with them without using scale diagrams, explicitly doing subtraction sums. We checked our examples with scale diagrams, rather than finding the solution using scale diagrams.
Though a subtle difference, this was far more successful.
Another successful strategy used during these lessons was using formal layout during the early stages with simple examples. By doing this, students were able to see the connection between diagrams, algebraic substitution and the usage of formulae.
For instance:
By learning this and ensuring each example was completed thus, when triangles and circles were introduced it was a trivial case of changing the formula and adding a line to the bottom
Area(Shape) = Area(S1) + Area(S2)
It was interesting to note that students at this stage had now forgotten that perimeter was the outer boundary and were including the S1 - S2 boundary in their perimeter calculations. It had to be explicitly explained that
Perimeter(Shape) != Perimeter(S1) + Perimeter(S2)
and that the intersecting boundary needed to be subtracted. For me, this is learning real mathematical literacy. Students are becoming able to exactly describe their intent on the path to a solution.
When thinking about perimeter we are only thinking about the boundary outside a shape. We then looked at some real world shapes and measured their boundaries, followed by drawing some scale diagrams of shapes and measuring their boundaries.
We then tried to describe how much space was in the shapes. I tried to drive them to describing dimensions, but some students had enough prior knowledge to say calculate the answer by multiplying lengths.
The main issues came as we arrived at composite shapes where composite shapes had to be split into basic geometric shapes. Deriving missing measurements really brought home how much trouble low literacy students can have with basic mathematical concepts such as subtraction. Assuming that a student can see how to find missing sides is in many cases overestimating their ability.
Even at year 10, the majority of low literacy students fail to see f- c = a whereas they are more likely to be able to see a+ c = f.
Because of this to use a scale drawing to assist these students see subtraction in action requires some thought. If a student drew the above diagram with the following measurements they would
not use subtraction to find the missing side on the top rectangle. They would count the squares (if on grid paper) or measure the vertical gap and put the measurement of one on the page.
To encourage students to use subtraction I needed to encourage students to first look at the diagram and do a number of examples with them without using scale diagrams, explicitly doing subtraction sums. We checked our examples with scale diagrams, rather than finding the solution using scale diagrams.
Though a subtle difference, this was far more successful.
Another successful strategy used during these lessons was using formal layout during the early stages with simple examples. By doing this, students were able to see the connection between diagrams, algebraic substitution and the usage of formulae.
For instance:
By learning this and ensuring each example was completed thus, when triangles and circles were introduced it was a trivial case of changing the formula and adding a line to the bottom
Area(Shape) = Area(S1) + Area(S2)
It was interesting to note that students at this stage had now forgotten that perimeter was the outer boundary and were including the S1 - S2 boundary in their perimeter calculations. It had to be explicitly explained that
Perimeter(Shape) != Perimeter(S1) + Perimeter(S2)
and that the intersecting boundary needed to be subtracted. For me, this is learning real mathematical literacy. Students are becoming able to exactly describe their intent on the path to a solution.
Subscribe to:
Posts (Atom)