Friday, December 20, 2013

Task Analysis for Curriculum Design

Two of the first things I do when designing a curriculum is to define the tasks that I want students to be able to do (backwards design), and then perform a task analysis on those tasks. A task analysis basically identifies the steps or sub-tasks needed to perform the task.

As a curriculum designer, I need to know the sub-tasks required to complete a task because I must build opportunities where students can explicitly learn how to perform each of those sub-tasks. I also need to make sure that the instructor is assessing those sub-tasks and knows to and is able to intervene when necessary.

For example, I am developing an area curriculum, and one of the tasks that I want students to perform is to take a polygon, decompose it into rectangles and triangles, find the area of those rectangles and triangles, and then combine those areas to get the area of the entire polygon. One sub-task that students need to perform is to find the length of a side when the length isn't given. Another is to identify the base and height of a triangle. Neither sub-task is trivial, and if a student is unable to do either of them, he or she will be unable to complete the overall task.

When doing a task analysis, I try to break the task down into sub-tasks that are either intuitive or build on sub-tasks students have already mastered. This highlights what students really need to know and be able to do, lowers the entry points into the curriculum (making it more accessible), and enables students to approach tasks as problem solvers rather than procedure followers.

It is very common for curriculum designers to break down a task to the wrong level. When I was working in schools, middle school students were expected to learn how to use prime factorization to find the greatest common factor (GCF) and least common multiple (LCM) of two numbers. (I find it very curious that prime factorization does not show up at all  in the common core standards, although Massachusetts has inserted their own standard for it into grade 6.)

To find the GCF or LCM of 504 and 675, first you would find the prime factorization of each number:

504 = (2^3) × (3^2) × 7
675 = (3^3) × (5^2)

[I apologize for the spreadsheet notation. Unfortunately, Blogger restricts my use of css, so I can't format the exponents the way I'd like.]

Then you'd compare the exponents for each factor.

To find the GCF of 504 and 675, you want to grab the smallest exponent. 504 has three 2's, but 675 has zero, so we go with zero 2's. 675 has three 3's, but 504 only has two, so we go with two 3's. 675 has two 5's, but 504 has zero, so we go with zero 5's. 504 has one 7, but 675 has zero, so we go with zero 7's. Combining zero 2's with two 3's, zero 5's, and zero 7's gives us a GCF of 9.

GCF = 3^2 = 9

To find the LCM of 504 and 675, you want to grab the greatest exponent, which would be three 2's, three 3's, two 5's, and one 7, or an LCM of 37,800.

LCM = (2^3) × (3^3) × (5^2) × 7 = 37,800

Teachers have a number of strategies for helping their students perform these tasks. One is to break out the number of factors in each prime factorization into a table:

Factors
2 3 5 7
504 3 2 0 1
675 0 3 2 0

Another is to teach students how to find the GCF in grade 7 and then the LCM in grade 8 so that they don't confuse the two procedures.

Neither strategy does anything to make the tasks more intuitive or accessible. But by drilling down a little further, we can break these tasks down to sub-tasks that do make sense.

One of the key sub-tasks is using prime factorizations to recognize when one number is a factor of another. If the prime factorization of 675 is (3^3) × (5^2), can you tell if 3 is a factor of 675? How about 5? 15? 45? 35? 45 is a factor of 675 because the prime factorization of 45 is (3^3) × 5, which is a subset of (3^3) × (5^2). 35 is not a factor of 675 because the prime factorization of 35 is 5 × 7, which is not a subset of (3^3) × (5^2).

Once you can use prime factorizations to recognize when one number is a factor of another, you can easily use them to recognize common factors. 3 is a common factor of both 504 and 675. 6 isn't because 2 × 3 is a subset of (2^3) × (3^2) × 7, but not a subset of (3^3) × (5^2). To find the GCF, keep adding factors to a common factor until you can't add anymore.

To find the LCM, you are going to start by finding a common multiple. By definition, a common multiple will have both 504 and 675 as factors, which means that the prime factorization of a common multiple must contain (2^3) × (3^2) × 7 and (3^3) × (5^2) as subsets.

Does (2^10) × (3^10) × (5^10) × (7^10) contain (2^3) × (3^2) × 7 and (3^3) × (5^2) as subsets? Yes, so it is a common multiple of 504 and 675. To find the LCM, we begin by knocking out factors one by one and using our factor test to see if we still have a common multiple. Eventually, when we can't knock out any more factors, we have the LCM, or least common multiple. This process moves rapidly from random trial-and-error to informed trial-and-error to prediction to generalization and a procedure for finding the LCM.

No comments:

Post a Comment