“Procedural” is not enough

B. Abramovitz, M. Berezina,

A. Berman, L. Shvartsman

4th MEDITERRANEAN CONFERENCE ON MATHEMATICS EDUCATION 

University of Palermo - Italy

28 – 30 January 2005

Department of Mathematics,

Ort Braude College,

Karmiel , Israel

Department of Mathematics, Technion, Haifa, Israel

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Teaching mathematics to engineering students has a twofold purpose

We want the students to learn to use mathematics and we want them to understand mathematics. In other words we teach them to compute and we teach them to think.

The procedural part of the teaching deals with computational algorithmic techniques: how to solve a system of linear equations, how to compute the gradient of a function, how to compute an integral.

The conceptual part of the teaching deals with the understanding of the theory behind the computation: why is the theorem correct, what happens if some of the assumptions are changed, why does an algorithm work.

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Many teachers and researchers agree that the procedural part is not enough

We strongly believe that a good engineer should develop mathematical thinking, which will help him to solve problems he has not seen before.

The conceptual thinking develops creativity in problem solving.

The understanding of a proof improves the student’s logical thinking and communication abilities.

Unfortunately, most engineering students are only interested in the computational part of the course and believe that solving many technical problems will be sufficient to succeed at the exam.

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In order to encourage the students to pay more attention to the theoretical part of the course, we developed several problems that we used during the courses and in the exams.

The student cannot solve such problems if you do not explain enough the theoretical material.

We give examples of such problems in Calculus and Linear Algebra. In each example the problem is given in a “procedural form” and in a “conceptual form”.

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An example in Calculus

The concepts addressed in this section are: tangent line, composite function and in particular the Chain Rule. The procedural version can be used to check if the student knows the Chain Rule and how to find the tangent line.

In the conceptual form a technical use of the Chain Rule is not sufficient, and the student has to fully understand the concept of the composite function and the theorem on its derivative. In this version the student also has to be able to use the tangent line to get information on the function.

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The procedural form

Let sin 5 3g x x and 4 12

5

u

f u e

.

Find the tangent line of h x f g x at 0x .

Solution

By the Chain Rule h x f g x g x so that

40 3 0 5 4

5h f g .

Since 0 3 1h f the tangent line is 4 1y x .

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The conceptual form Let 5 3y x be the tangent line of the function

g x at 0x .

Let 4xf g x e for every real number x .

Find the tangent line of the function f x at 3x .

Solution

Since 5 3y x is the tangent line of g x at 0x , it

follows that 0 3, 0 5g g .

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By the Chain Rule the derivative of the composite

function 44 xf g x f g x g x e and

0 3 0 4xf g x f g .Finally

4 4

30 5

fg

. Now 3 0 1f f g such

that the tangent line to f x at 3x is 4

15

y x .

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An example in Linear Algebra

We give two problems related to the well- known theorem on the relation between the solutions of a non homogeneous linear system and the related homogeneous one.

In the procedural form the student has only to know the Gaussian Elimination method and Vector Addition. The aim of this problem is to give the student the main idea of the theorem without formulating and proving the result.

The problem in the conceptual form can be readily solved if the student knows both the theorem and its proof and is able also to apply them.

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The procedural form Solve simultaneously the systems (a) and (b) of linear equations and find out the relation between their general solutions.

(a)

0743

0532

02

321

321

321

xxx

xxx

xxx

(b)

2743

1532

12

321

321

321

xxx

xxx

xxx

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We suppose that the student will solve both systems using the Gaussian Elimination Method and will write down the solutions in the following form:

hpnh XXttttttX ),,()0,1,2(),1,2( .

Here nhX is the general solution of the non-

homogeneous system (b), hX is the general solution of

the homogeneous system (a) and pX is a solution of (b).

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The conceptual form Consider two systems of linear equations: (a) AX = 0 and (b) AX = b with the same 3×3 coefficient matrix A . Assume that the vectors (1,2,1) and (2,1,1) are solutions of the system (b) and that the vector (1,1,1) is a solution of the system (a). Find the general solution of the system (b).

The solution of the problem is based on the following theoretical facts: 1) The difference of every two solutions of (b) is a solution of (a). This fact is a part of the proof of the theorem. This shows that the vector (1,-1,0) is a solution of the system (a) as well. Thus the system (a) has two linearly independent solutions.

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2) The general solution of (a) is a vector space. In our case the student should conclude that the dimension of this vector space is 2 so that the general solution of the system (a) is the linear span of the vectors (1,1,1) and (1,-1,0):

)0,1,1()1,1,1( tsX h

3) The theorem on the structure of the general solution of a non-homogeneous system. Following this result the student obtains the final solution in the form:

)0,1,1()1,1,1()1,2,1( tsXXX hpnh .

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The procedural form

Let ,f x y be a function with continuous derivatives in a

neighborhood of the point 1,1 , such that

1,1 3, 1,1 4f f

x y

. Let : , 0L x y be a

smooth curve in the domain of ,f x y such that

1,1 0 . Suppose that 0

on L . If 1,1 is a

local minimum point of ,f x y on L find the tangent

line to L at the point 1,1 .

An additional example in Calculus

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Solution By the method of Lagrange’s multipliers we define the function

, , , ,x y f x y x y .

Then 0 01,1, 0, 1,1, 0x y

and

0 01,1 1,1 0, 1,1 1,1 0f f

x x y y

.

Thus 0 0

3 41,1 , 1,1

x y

and the tangent

line is

3 1 4 1 0x y .

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The conceptual form

Let ,f x y be a function with continuous derivatives in a

neighborhood of the point 1,1 , such that

1,1 3, 1,1 4f f

x y

. Let

be a smooth curve in the domain of ,f x y such that there exists

a point 0 0,t t with 0 01, 1t t . Suppose that

2 20 on L . If 1,1 is a local minimum point of

,f x y on L find the tangent line to L at the point 1,1 .

: ,x t

L ty t

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S o l u t i o n

D e f i n e t h e f u n c t i o n ,F t f t t . B y t h e d e f i n i t i o n , 0t

i s a m i n i m u m p o i n t f o r F t a n d b y F e r m a t ’ s T h e o r e m

0 0F t . N o w , b y t h e C h a i n R u l e ,

0 0 0 0 0 0 0, ,f f

F t t t t t t tx y

o r

0 01 , 1 1 , 1 0f f

t tx y

a n d f i n a l l y

0 03 4 0t t . S i n c e 0 0

3

4t t i t f o l l o w s

t h a t t h e t a n g e n t l i n e i s 1 1

4 3

x y

.

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The procedural form Compute adjA , if

2 1 1

1 2 1

1 1 2

A

An additional example in Linear Algebra

Solution

3 1 1

1 3 1

1 1 3

adjA

.

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The conceptual form Given a matrix B , find a matrix A such that adjA B , where

2 1 1

1 2 1

1 1 2

B

.

Solution

Since B is nonsingular, A is nonsingular and 1

det

BA

A .

Thus 1A kB where k is a nonzero scalar.

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1

3 1 11

1 3 14

1 1 3

B

so

3 1 1

1 3 1

1 1 3

A c