4. Fieldwork and results

In this section we start considering the sheep problem in its original version, posed in a direct form. After that we show some of the most creative reformulations proposed by the prospective teachers who participate in the fieldwork at UGR. Finally, we conclude with some general remarks about the productions of the prospective teachers in this experience.

It is important to mention that the following results represent part of the general research about task enrichment by prospective teachers (see, for instance, [14]). In this opportunity our work is focused on the mathematical content of the proposals. Other aspects of the didactic analysis, like the cognitive and the instructional dimensions of the enriched tasks, will be part of further research.

#### 4.1 The sheep problem

In this problem, a sheep is grazing in a square field with side length L. The sheep is tied at the point (L/2, 0), and the rope attached to the sheep has a length R as shown in Figure 5.

In Figure 5, A represents the area of the sector where the sheep may graze, is the ratio of the rope length to field side length, and represents the fraction of the total area accessible for the sheep. It can be observed that f is a function of the ratio r that can be obtained by integration techniques.

Typical exercises consist in supplying students with this figure and asking them to obtain f corresponding to one or more values for r. However, a more interesting approach is to ask the students to draw a diagram hoping they realize that the problem can be solved as an intersection of circles and squares. Four different situations may happen:

• When , the sheep does not reach the lateral edges of the field.

Figure 5. Part of the field accessible for the sheep.

Inverse Modeling Problems in Task Enrichment for STEM Courses DOI: http://dx.doi.org/10.5772/intechopen.89109


• When , the sheep can graze all around the field.

This problem requires modeling and integral calculation, and it can be easily converted into an inverse problem. It is obvious that for every value of r ≥ 0 there exists a unique value of A, but more challenging is to ask the question from another angle. For instance, for any value of A, does a corresponding unique value of r exist or not? To solve this problem, the function f(r) must be studied in terms of continuity and growth with r ≥ 0, in order to ensure its invertibility.

#### 4.2 The new fieldwork design

in the year 2017 fieldwork, the participants were not asked to solve their proposed problems, so this was an aspect that needs to be improved in further research. The new results showed interesting differences and few similarities which are

In this section we start considering the sheep problem in its original version, posed in a direct form. After that we show some of the most creative reformulations proposed by the prospective teachers who participate in the fieldwork at UGR. Finally, we conclude with some general remarks about the productions of the

It is important to mention that the following results represent part of the general research about task enrichment by prospective teachers (see, for instance, [14]). In this opportunity our work is focused on the mathematical content of the proposals. Other aspects of the didactic analysis, like the cognitive and the instructional

In this problem, a sheep is grazing in a square field with side length L. The sheep is tied at the point (L/2, 0), and the rope attached to the sheep has a length R as

Typical exercises consist in supplying students with this figure and asking them to obtain f corresponding to one or more values for r. However, a more interesting approach is to ask the students to draw a diagram hoping they realize that the problem can be solved as an intersection of circles and squares. Four different

In Figure 5, A represents the area of the sector where the sheep may graze, is the ratio of the rope length to field side length, and represents the fraction of the total area accessible for the sheep. It can be observed that f is a function of the ratio r

• When , the sheep does not reach the lateral edges of the field.

dimensions of the enriched tasks, will be part of further research.

analyzed in the next sections.

Theorizing STEM Education in the 21st Century

4. Fieldwork and results

4.1 The sheep problem

shown in Figure 5.

situations may happen:

Figure 5.

34

Part of the field accessible for the sheep.

prospective teachers in this experience.

that can be obtained by integration techniques.

As it was mentioned, the new fieldwork was designed in order to avoid, or at least attenuate, the difficulties observed in the previous experience, carried out during year 2017. In particular, both experimental designs had three main differences:


This new design produced different responses that cannot be included in the previous nine groups observed during the year 2017 experience. Some of the most creative and new proposals are analyzed in the following subsection.

#### 4.3 Some of the most creative reformulations for the sheep problem

As already stated, some of the prospective teachers' productions cannot be classified into the nine groups observed in the previous fieldwork. The following examples illustrate this situation.

Example 1: An unusual specification problem.

One of the prospective teachers solved the direct problem by integration, observing that the circumference equation can be written as <sup>L</sup> <sup>2</sup> � <sup>x</sup> � �<sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>¼</sup> <sup>R</sup><sup>2</sup> and then the area accessible for the sheep is <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup> 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup> <sup>2</sup> � <sup>x</sup> � �<sup>2</sup> <sup>q</sup> dx. After that, he solves the integral by using the change of variables <sup>x</sup> <sup>¼</sup> <sup>L</sup> <sup>2</sup> � R sin t and several wellknown trigonometric formulas to obtain the following long formula:

<sup>A</sup> ¼ � <sup>R</sup><sup>2</sup> <sup>2</sup> arcsin �<sup>L</sup> 2R � � � arcsin <sup>L</sup> 2R � � <sup>þ</sup> <sup>1</sup> <sup>2</sup> sin 2arcsin �<sup>L</sup> 2R � � � � � sin 2arcsin <sup>L</sup> 2R � � � � � � � � It is easy to observe that this formula can be simplified, but the prospective teacher leaves it in the long version, as shown above.

After this classical solution, his inverse reformulation proposes to get the solution in a geometrical way and compare the final result with the one obtained by integration. A ¼

1 2

8 >>>>>><

>>>>>>:

the comparison between R and <sup>L</sup>

final cost as a function of variable R.

function, since he claims that <sup>L</sup> <sup>¼</sup> dA

multiplying by the cost per meter.

Example 3: An inversion by intervals.

accessible area can be obtained as <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup>=<sup>2</sup>

percentages possible.

37

After some algebraic manipulations, he arrives at f rð Þ¼ <sup>1</sup>

than L are not considered.

<sup>R</sup><sup>2</sup> <sup>L</sup> 2R

DOI: http://dx.doi.org/10.5772/intechopen.89109

@

πR<sup>2</sup> if 0≤R≤

Inverse Modeling Problems in Task Enrichment for STEM Courses

solution required, at least for this proposed reformulation.

stration or justification—is not true for every region in R<sup>2</sup>

considered as the unique weak connection between both proposals.

One of the prospective teachers proposed another interesting inverse

s

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � <sup>L</sup> 2R � �<sup>2</sup>

� � <sup>0</sup>

L 2

<sup>þ</sup> arcsin <sup>L</sup>

So, as it can be observed, he considered two different situations, depending on

In the inverse reformulation, he proposes to give this piecewise function as part of the data. He informs that the shepherd decides to eliminate the rope and, instead of it, he wants to build a circular fence like in Figure 5, i.e., the same as in the original problem. This fence costs 15 €/m, and the prospective teacher asks for the

As it can be easily observed, the problem could be solved in a direct way, by using the arc length formula and then calculating the corresponding integral and lastly multiplying by the cost per meter. However, this solution does not use the given area function—the input of this problem—so it cannot be considered as the

The prospective teacher solves his own problem by differentiating the given

this case, since it is composed of a circular sector and triangles. A more detailed discussion about when the derivative of the area gives the perimeter can be read at [35]. Finally, after obtaining L Rð Þ by differentiation, the price is easily obtained

In our previous experience, in 2017, only one prospective teacher proposed a reformulation involving arc length in a problem where two sheep were running, one along a straight line and the other along the circumference. In that problem, the accessible area did not appear as an input, so the presence of the arc length can be

As it was commented, the participant solution is not general, though it is simple, short, elegant, and accurate for this particular situation. As in the previous example, this proposal cannot be included in one of the nine groups observed in 2017.

reformulation that cannot be classified in the nine groups previously obtained in 2017. Firstly, he solves the direct problem, changing the axis position such that the new origin is located in the point where the sheep is tied. Due to this change, the

> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> dx <sup>¼</sup> <sup>x</sup>

2

q

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> <sup>þ</sup> <sup>R</sup><sup>2</sup>

> ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4

<sup>2</sup> arcsin <sup>x</sup> R

<sup>þ</sup> <sup>r</sup>2arcsin <sup>1</sup>

�L=2 .

2r � �.

� � h i<sup>L</sup>=<sup>2</sup>

�L=2

In his reformulation, the prospective teacher proposes that between 60 and 70% of the field is needed for the sheep to graze and between 30 and 40% is needed for a

In the solution of his own inverse problem, he makes mistakes in the derivative of f rð Þ; nevertheless, his conclusion about the growth of this function is obviously

Later, he will use this formula for solving the corresponding inverse problem.

pumpkin plantation, and he asks for at least one value of r that makes both

right. For this reason he tries with particular values, like fð Þffi 1 0:956 and

2R

1

<sup>A</sup> if <sup>L</sup>

<sup>2</sup>. It is important to mention that other radii greater

dR. This statement—given without any demon-

<sup>2</sup> <sup>≤</sup>R≤<sup>L</sup>

, although it is correct in

:

This is a very interesting specification problem, since the data and final result are known and he asks for another procedure in order to get the desired result.

When the prospective teacher solves his own reformulation, he divides the area accessible for the sheep into three parts, a circular sector and two triangles, as it can be observed in Figure 6.

In Figure 6 both triangles have a height h that can be easily obtained by Pythagoras' theorem, giving h ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4 q and then the area of each triangle is <sup>1</sup> 2 L 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4 q . After that, the angle α in both triangles is determined by using trigonometric concepts, for instance, <sup>α</sup> <sup>¼</sup> arctan <sup>2</sup><sup>h</sup> L � � <sup>¼</sup> arctan <sup>2</sup> L ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4 � � q , and so, the angle of the circular sector, π � 2α, is easily obtained. Then, the accessible area can be

written as

$$A = \frac{L}{2} \sqrt{R^2 - \frac{L^2}{4}} + \pi R^2 \frac{\pi - 2 \arctan\left(\frac{2}{L} \sqrt{R^2 - \frac{L^2}{4}}\right)}{2\pi}.$$

The prospective teacher shows that both formulas give the same results for particular values like <sup>R</sup> <sup>¼</sup> <sup>L</sup> <sup>2</sup> and <sup>R</sup> <sup>¼</sup> <sup>L</sup>ffiffi 2 <sup>p</sup> . The participant ends his work observing that "as it was expected, both methods gave the same results."

It is important to remark that in the previous experience, carried out in 2017, all the reformulations (i.e., the nine groups and their variants) corresponded to causation inverse problems. None of them proposed an inverse specification problem as in this creative production.

Example 2: An arc length inverse problem.

Another prospective teacher solved the direct problem by using integrals and the same change of variables showed above. Nevertheless, he used other trigonometric formulas, and he takes advantage of symmetry arguments to get a different formula:

Figure 6. The accessible area divided into three parts.

Inverse Modeling Problems in Task Enrichment for STEM Courses DOI: http://dx.doi.org/10.5772/intechopen.89109

<sup>A</sup> ¼ � <sup>R</sup><sup>2</sup>

<sup>2</sup> arcsin �<sup>L</sup>

be observed in Figure 6.

oras' theorem, giving h ¼

particular values like <sup>R</sup> <sup>¼</sup> <sup>L</sup>

in this creative production.

written as

formula:

Figure 6.

36

The accessible area divided into three parts.

concepts, for instance, <sup>α</sup> <sup>¼</sup> arctan <sup>2</sup><sup>h</sup>

<sup>A</sup> <sup>¼</sup> <sup>L</sup> 2

2R

Theorizing STEM Education in the 21st Century

leaves it in the long version, as shown above.

� � � arcsin <sup>L</sup>

2R � � <sup>þ</sup> <sup>1</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4

<sup>2</sup> and <sup>R</sup> <sup>¼</sup> <sup>L</sup>ffiffi

that "as it was expected, both methods gave the same results."

s

Example 2: An arc length inverse problem.

q

is easy to observe that this formula can be simplified, but the prospective teacher

known and he asks for another procedure in order to get the desired result.

After that, the angle α in both triangles is determined by using trigonometric

the circular sector, π � 2α, is easily obtained. Then, the accessible area can be

<sup>þ</sup> <sup>π</sup>R<sup>2</sup>

� � <sup>¼</sup> arctan <sup>2</sup>

The prospective teacher shows that both formulas give the same results for

It is important to remark that in the previous experience, carried out in 2017, all the reformulations (i.e., the nine groups and their variants) corresponded to causation inverse problems. None of them proposed an inverse specification problem as

Another prospective teacher solved the direct problem by using integrals and the same change of variables showed above. Nevertheless, he used other trigonometric formulas, and he takes advantage of symmetry arguments to get a different

2

L

After this classical solution, his inverse reformulation proposes to get the solution in a geometrical way and compare the final result with the one obtained by integration. This is a very interesting specification problem, since the data and final result are

When the prospective teacher solves his own reformulation, he divides the area accessible for the sheep into three parts, a circular sector and two triangles, as it can

In Figure 6 both triangles have a height h that can be easily obtained by Pythag-

<sup>2</sup> sin 2arcsin �<sup>L</sup>

� � � � � � � � It

2R

and then the area of each triangle is <sup>1</sup>

� � q

L

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4

� � q

<sup>2</sup><sup>π</sup> :

<sup>p</sup> . The participant ends his work observing

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4

L

<sup>π</sup> � <sup>2</sup>arctan <sup>2</sup>

� � � � � sin 2arcsin <sup>L</sup>

2R

2 L 2

, and so, the angle of

q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup><sup>2</sup> 4

.

$$A = \begin{cases} \frac{1}{2}\pi R^2 & \text{if } \quad 0 \le R \le \frac{L}{2} \\\\ R^2 \left(\frac{L}{2R}\sqrt{1 - \left(\frac{L}{2R}\right)^2} + \arcsin\left(\frac{L}{2R}\right)\right) & \text{if } \quad \frac{L}{2} \le R \le L \end{cases}$$

:

So, as it can be observed, he considered two different situations, depending on the comparison between R and <sup>L</sup> <sup>2</sup>. It is important to mention that other radii greater than L are not considered.

In the inverse reformulation, he proposes to give this piecewise function as part of the data. He informs that the shepherd decides to eliminate the rope and, instead of it, he wants to build a circular fence like in Figure 5, i.e., the same as in the original problem. This fence costs 15 €/m, and the prospective teacher asks for the final cost as a function of variable R.

As it can be easily observed, the problem could be solved in a direct way, by using the arc length formula and then calculating the corresponding integral and lastly multiplying by the cost per meter. However, this solution does not use the given area function—the input of this problem—so it cannot be considered as the solution required, at least for this proposed reformulation.

The prospective teacher solves his own problem by differentiating the given function, since he claims that <sup>L</sup> <sup>¼</sup> dA dR. This statement—given without any demonstration or justification—is not true for every region in R<sup>2</sup> , although it is correct in this case, since it is composed of a circular sector and triangles. A more detailed discussion about when the derivative of the area gives the perimeter can be read at [35]. Finally, after obtaining L Rð Þ by differentiation, the price is easily obtained multiplying by the cost per meter.

In our previous experience, in 2017, only one prospective teacher proposed a reformulation involving arc length in a problem where two sheep were running, one along a straight line and the other along the circumference. In that problem, the accessible area did not appear as an input, so the presence of the arc length can be considered as the unique weak connection between both proposals.

As it was commented, the participant solution is not general, though it is simple, short, elegant, and accurate for this particular situation. As in the previous example, this proposal cannot be included in one of the nine groups observed in 2017.

Example 3: An inversion by intervals.

One of the prospective teachers proposed another interesting inverse reformulation that cannot be classified in the nine groups previously obtained in 2017.

Firstly, he solves the direct problem, changing the axis position such that the new origin is located in the point where the sheep is tied. Due to this change, the accessible area can be obtained as <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup>=<sup>2</sup> �L=2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> dx <sup>¼</sup> <sup>x</sup> 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> <sup>þ</sup> <sup>R</sup><sup>2</sup> <sup>2</sup> arcsin <sup>x</sup> R � � h i<sup>L</sup>=<sup>2</sup> �L=2 . After some algebraic manipulations, he arrives at f rð Þ¼ <sup>1</sup> 2 ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4 q <sup>þ</sup> <sup>r</sup>2arcsin <sup>1</sup> 2r � �.

Later, he will use this formula for solving the corresponding inverse problem.

In his reformulation, the prospective teacher proposes that between 60 and 70% of the field is needed for the sheep to graze and between 30 and 40% is needed for a pumpkin plantation, and he asks for at least one value of r that makes both percentages possible.

In the solution of his own inverse problem, he makes mistakes in the derivative of f rð Þ; nevertheless, his conclusion about the growth of this function is obviously right. For this reason he tries with particular values, like fð Þffi 1 0:956 and

fð Þffi 0:5 0:3926, and after some iterations, he concludes that fð Þffi 0:7 0:625, so r ¼ 0:7 is a possible answer.

The proposal itself seems to be not as creative as the previous ones (Examples 1 and 2), but it is the only one that asked for the pre-image of an interval, an interesting topic related to continuity, monotony, and derivatives, among other important calculus concepts.

Example 4: General inversion of a vector function.

Another prospective teacher tries to solve the direct problem by integration. For this purpose he puts the area accessible for the sheep as <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup> 0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup> <sup>2</sup> � <sup>x</sup> � �<sup>2</sup> <sup>q</sup> dx, and he proposes the change of variables <sup>x</sup> <sup>¼</sup> <sup>L</sup> <sup>2</sup> � R sin t. Unfortunately some mistakes when using trigonometric formulas led him to a wrong result f rð Þ¼ <sup>5</sup> <sup>4</sup> <sup>r</sup>2arcsin <sup>L</sup><sup>2</sup> 2 r � �. He does not use this function in the inverse problem proposal, but it appears again in the corresponding solution.

Example 6: Interpretation problems.

ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4

Two possible cases for the accessible area.

q

5. Discussion

39

f rð Þ¼ <sup>1</sup> 2

Figure 7.

following function as a data for the inverse problem:

Inverse Modeling Problems in Task Enrichment for STEM Courses

DOI: http://dx.doi.org/10.5772/intechopen.89109

2r

<sup>þ</sup> <sup>2</sup>r2arcsin <sup>1</sup>

problem did not appear in the year 2017.

innovative and authentic STEM content…".

of mathematical concepts and processes.

A couple of prospective teachers proposed a different kind of inverse problem, where the elements of Figure 5 are explained. After that, one of them gives the

between the variables and the problem consists in determining the meaning of both. In the last one, the situation proposed is almost the same, but there is a mistake

in the given function f; so, the correct answer will be that r has no meaning. As it happened in the other examples commented above, this kind of inverse

Problem posing has an important role to play in the STEM classroom. For instance, Beal and Cohen [36] suggested "… significant changes to the existing model of education, in which students would move from passive consumers of educational resources that have been developed by others to creators of rich,

Moreover, problem posing has a positive influence on students' ability to solve word problems and provided a chance to gain insight into students' understanding

Some researchers have found evidences that students' experience with problem posing enhances their perception of the subject; causes excitement and motivation; improves students' thinking, problem solving skills, attitudes, and confidence in mathematics; and contributes to a broader understanding of mathematical concepts [37]. Some of these capabilities were observed in our fieldwork with prospective teachers, particularly in the selected examples, where some responses were very creative. This is an important aspect not only for teaching but also as a prominent feature of mathematical activity. As Poincaré said "Mathematicians may solve some problems that have been posed for them by others or may work on problems that have been identified as important problems in the literature, but it is more common for them to formulate their own problems, based on their personal experience and interests" [38]. In the same way, Hadamard [39] identified the ability to find key research questions as an indicator of exceptional mathematical talent. For instance, a challenging question appears in the solution of Example 2 of the previous section, where the arc length is obtained as the derivative of the area accessible to the sheep. The question about the accuracy of this procedure led us to an interesting research

question whose answer is not trivial (see [35] for a general discussion).

� � h i. She only mentions that <sup>f</sup> and <sup>r</sup> express relations

The reformulation considers the same situation schematized in Figure 5, like in the original problem, and he asks to obtain L and R for given values of f and r. In other words, the input is the vector ð Þ L, R , and the expected output is another vector ð Þ f,r ; then, it corresponds to a vector function inversion.

The solution is wrong and easier than it should be, since he considers the function previously obtained f rð Þ¼ <sup>5</sup> <sup>4</sup> <sup>r</sup>2arcsin <sup>L</sup><sup>2</sup> 2 r � �, which is simpler than the correct one: f rð Þ¼ <sup>1</sup> 2 ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4 q <sup>þ</sup> <sup>r</sup>2arcsin <sup>1</sup> 2r � �. If the wrong f rð Þ is utilized, it follows that 4f <sup>5</sup>r<sup>2</sup> <sup>¼</sup> arcsin <sup>L</sup><sup>2</sup> 2 r � �, and then <sup>L</sup> can be obtained as <sup>L</sup> <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 <sup>r</sup> sin <sup>4</sup><sup>f</sup> 5r<sup>2</sup> <sup>r</sup> � �. Finally, this <sup>L</sup> can be multiplied by r to obtain R.

It is obvious that the inverse problem was unwittingly simplified; however, it is an interesting proposal and the only one of this type in both years 2017 and 2019. Another important characteristic is that it requires a general inversion, since the input vector ð Þ <sup>L</sup>, <sup>R</sup> is a generic vector of the vector space <sup>R</sup><sup>2</sup> .

Example 5: Problems that ask for a sketch of the region.

Four prospective teachers (two in each group) proposed inverse problems that ask for a sketch of the region, with different variants.

In the first one, the axis are chosen such that the sheep is tied at the origin of the coordinate system, and so, the accessible area can be obtained as <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup>=<sup>2</sup> �L=2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> dx. In her reformulation, the prospective teacher gives the integral, and the problem consists in doing a sketch of the region, including an

identification of its elements.

The second one is similar, but there are no axis changes, and the accessible area is given by the formula <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> arcsinð Þ <sup>L</sup>=2<sup>R</sup> �arcsinð Þ <sup>L</sup>=2<sup>R</sup> <sup>R</sup><sup>2</sup> cos <sup>θ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � sin <sup>2</sup> <sup>θ</sup> <sup>p</sup> <sup>d</sup>θ. The prospective teacher informs that this formula was obtained after performing a change of variable <sup>x</sup> <sup>¼</sup> <sup>R</sup> sin <sup>θ</sup> <sup>þ</sup> <sup>L</sup> <sup>2</sup>, and, like in the previous case, she asks for the region involved.

The last two cases are different since in both of them the student is asked to propose a criteria that allow to choose between regions (a) and (b), for any given value of the accessible area A (see Figure 7). The solution can be obtained by considering the limit case, i.e., <sup>R</sup> <sup>¼</sup> <sup>L</sup> <sup>2</sup> and then <sup>A</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> <sup>π</sup>R<sup>2</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> π <sup>L</sup> 2 � �<sup>2</sup> <sup>¼</sup> <sup>1</sup> <sup>8</sup> <sup>π</sup>L<sup>2</sup> . So the requested criteria is very simple: if A> <sup>1</sup> <sup>8</sup> <sup>π</sup>L<sup>2</sup> , the region is like (b), whereas if A < <sup>1</sup> <sup>8</sup> <sup>π</sup>L<sup>2</sup> , the region is like (a).

It can be noted that in those cases, the problem has a weak connection with the sheep problem, since the context about the sheep, the rope, etc. can be eliminated from the proposals, and the solution remains unchanged.

Inverse Modeling Problems in Task Enrichment for STEM Courses DOI: http://dx.doi.org/10.5772/intechopen.89109

Figure 7. Two possible cases for the accessible area.

fð Þffi 0:5 0:3926, and after some iterations, he concludes that fð Þffi 0:7 0:625, so r ¼

and 2), but it is the only one that asked for the pre-image of an interval, an interesting topic related to continuity, monotony, and derivatives, among other

Example 4: General inversion of a vector function.

and he proposes the change of variables <sup>x</sup> <sup>¼</sup> <sup>L</sup>

Theorizing STEM Education in the 21st Century

2 r � �

function previously obtained f rð Þ¼ <sup>5</sup>

q

ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4

2

2 r � �

be multiplied by r to obtain R.

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

identification of its elements.

able <sup>x</sup> <sup>¼</sup> <sup>R</sup> sin <sup>θ</sup> <sup>þ</sup> <sup>L</sup>

is given by the formula <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> arcsinð Þ <sup>L</sup>=2<sup>R</sup>

considering the limit case, i.e., <sup>R</sup> <sup>¼</sup> <sup>L</sup>

requested criteria is very simple: if A> <sup>1</sup>

, the region is like (a).

from the proposals, and the solution remains unchanged.

this purpose he puts the area accessible for the sheep as <sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup>

proposal, but it appears again in the corresponding solution.

ð Þ f,r ; then, it corresponds to a vector function inversion.

<sup>þ</sup> <sup>r</sup>2arcsin <sup>1</sup>

input vector ð Þ <sup>L</sup>, <sup>R</sup> is a generic vector of the vector space <sup>R</sup><sup>2</sup>

ask for a sketch of the region, with different variants.

Example 5: Problems that ask for a sketch of the region.

coordinate system, and so, the accessible area can be obtained as

, and then L can be obtained as L ¼

mistakes when using trigonometric formulas led him to a wrong result

The proposal itself seems to be not as creative as the previous ones (Examples 1

Another prospective teacher tries to solve the direct problem by integration. For

The reformulation considers the same situation schematized in Figure 5, like in the original problem, and he asks to obtain L and R for given values of f and r. In other words, the input is the vector ð Þ L, R , and the expected output is another vector

The solution is wrong and easier than it should be, since he considers the

2r

<sup>4</sup> <sup>r</sup>2arcsin <sup>L</sup><sup>2</sup>

It is obvious that the inverse problem was unwittingly simplified; however, it is an interesting proposal and the only one of this type in both years 2017 and 2019. Another important characteristic is that it requires a general inversion, since the

Four prospective teachers (two in each group) proposed inverse problems that

In the first one, the axis are chosen such that the sheep is tied at the origin of the

<sup>R</sup><sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>p</sup> dx. In her reformulation, the prospective teacher gives the

The second one is similar, but there are no axis changes, and the accessible area

teacher informs that this formula was obtained after performing a change of vari-

The last two cases are different since in both of them the student is asked to propose a criteria that allow to choose between regions (a) and (b), for any given value of the accessible area A (see Figure 7). The solution can be obtained by

<sup>8</sup> <sup>π</sup>L<sup>2</sup>

�arcsinð Þ <sup>L</sup>=2<sup>R</sup> <sup>R</sup><sup>2</sup> cos <sup>θ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>2</sup> and then <sup>A</sup> <sup>¼</sup> <sup>1</sup>

It can be noted that in those cases, the problem has a weak connection with the sheep problem, since the context about the sheep, the rope, etc. can be eliminated

<sup>1</sup> � sin <sup>2</sup>

<sup>2</sup> <sup>π</sup>R<sup>2</sup> <sup>¼</sup> <sup>1</sup>

<sup>2</sup>, and, like in the previous case, she asks for the region involved.

integral, and the problem consists in doing a sketch of the region, including an

. He does not use this function in the inverse problem

2 r � � 0

, which is simpler than the cor-

. Finally, this L can

<sup>2</sup> � R sin t. Unfortunately some

� �. If the wrong f rð Þ is utilized, it follows that

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

.

<sup>θ</sup> <sup>p</sup> <sup>d</sup>θ. The prospective

<sup>2</sup> π <sup>L</sup> 2 � �<sup>2</sup> <sup>¼</sup> <sup>1</sup>

, the region is like (b), whereas if

<sup>8</sup> <sup>π</sup>L<sup>2</sup>

. So the

2 <sup>r</sup> sin <sup>4</sup><sup>f</sup> 5r<sup>2</sup> r � �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>R</sup><sup>2</sup> � <sup>L</sup> <sup>2</sup> � <sup>x</sup> � �<sup>2</sup> <sup>q</sup>

dx,

0:7 is a possible answer.

f rð Þ¼ <sup>5</sup>

important calculus concepts.

<sup>4</sup> <sup>r</sup>2arcsin <sup>L</sup><sup>2</sup>

rect one: f rð Þ¼ <sup>1</sup>

<sup>5</sup>r<sup>2</sup> <sup>¼</sup> arcsin <sup>L</sup><sup>2</sup>

<sup>A</sup> <sup>¼</sup> <sup>Ð</sup> <sup>L</sup>=<sup>2</sup> �L=2

A < <sup>1</sup> <sup>8</sup> <sup>π</sup>L<sup>2</sup>

38

4f

Example 6: Interpretation problems.

A couple of prospective teachers proposed a different kind of inverse problem, where the elements of Figure 5 are explained. After that, one of them gives the following function as a data for the inverse problem:

f rð Þ¼ <sup>1</sup> 2 ffiffiffiffiffiffiffiffiffiffiffiffi <sup>r</sup><sup>2</sup> � <sup>1</sup> 4 q <sup>þ</sup> <sup>2</sup>r2arcsin <sup>1</sup> 2r � � h i. She only mentions that <sup>f</sup> and <sup>r</sup> express relations between the variables and the problem consists in determining the meaning of both.

In the last one, the situation proposed is almost the same, but there is a mistake in the given function f; so, the correct answer will be that r has no meaning.

As it happened in the other examples commented above, this kind of inverse problem did not appear in the year 2017.
