POSSIBLE SOLUTION(S)

First, identify more expressions, having identified the length as x in our revision of the original question. We let T(x) be the time necessary to build a straight tunnel of length x in uniform, moist, fine sand into the side of a vertical wall. And we now know that we are seeking a precise description for T(x) .

At this point several students may make conjectures as to the nature of T(x) . ``I know. It is some constant times x .'' ``Why?'' is our response. It sure sounds good - the longer the tunnel the more time it takes . We all make a note of this reasonable observation as a characteristic which any solution for T(x) must possess.

We do not offer any reasons why the linear model (constant times x ) may not be appropriate. We do point out that stating the model is linear without some support may not be ``the solution'' to determining just how long it takes an ant to build a tunnel of length x .

Moreover, we ask, ``Is it reasonable that if we double the tunnel length the time to construct the longer tunnel is merely twice as long? After all, isn't this what this linear model would predict? Does this linear model take into account the fact that the ant will have to drag material from the deeper second half of the tunnel all the way through the first half?'' We point out that perhaps we had better take a closer look at the situation.

Perhaps from knowing the time it takes to build a small section of the tunnel we can build up to a larger section of the tunnel from the assumptions more directly rather than just jumping on a function offered ``by the Class,'' e.g. TC(x) = k x .

We ask the question, ``How long does it take to dig out the section from distance x to distance x+h ?'' Quickly the students respond, ``The time it takes to dig to x+h less the time to dig to x . That's T(x+h) - T(x) .''

Thus we have the time to extend the tunnel from distance x to x+h as T(x+h) - T(x) . And what might we be able to say about this time?

Since the ant must bring all the sand in this little x to x + h section out a distance x and since we assume that the ant travels at a constant rate - with or without baggage - we can assume that the time to cart out all the stuff from the x to x+h section is proportional to x , i.e.

T(x+h) - T(x) = b x .

Based on our previous discussion and the proposed linear model for T(x) itself which was offered earlier we postulate that the time it takes to extend the tunnel h units (for small h ) is also proportional to h , for the longer the section, the longer the time needed to dig out the section. In fact, we make it reasonable by saying, ``If h is 10 little granules high by 1 granule deep, then the ant has to make 10 trips back out the tunnel. If h is 10 little granules high by 2 granules deep, then the ant has to make 20 trips back out the tunnel - all over the same trip distance, x .'' Thus it seems reasonable that T(x+h) - T(x) = c h .

Since this tunnel is into the side of the hill, the ant merely drops the sand over the edge at the tunnel opening and this does not affect the time for digging.

Now we point out that we have T(x+h) - T(x)= c h and we also have T(x+h) - T(x) = b x . This means c h = b x for all x and all h . The only way this can happen is if c = a x and b = a h , i.e. a is a constant of proportionality.

Thus, we produce the equation T(x+h) - T(x) = a x h .

At this point, if the class is studying the derivative, we ask if they see the makings of a derivative in this equation. In all cases we point out that our objective is to get a full description of T(x) , perhaps to solve for T(x) .

We again check our assumptions. Have we used them? Have we used them well?

We mention the students' developing skills for differentiation which demand the form (f(x+h)-f(x))/h . In this case we get the difference expression (T(x+h) - T(x))/h = a x .

We recall that we wish information about T(x) , not T(x) and T(x+h) , and recalling the definition of derivative if the students do not mention it (and someone usually does) we proceed to obtain

T'(x) =Limit as h goes to 0 of (T(x+h) - T(x))/h = a x .

Thus to find T(x) we need to find a function T(x) which has a derivative equal to a x for some constant a.

In the classes in which the formal antiderivative or indefinite integral concept has been covered we know that the answer is T(x) = a/2 x^2 + k .

And in those classes in which we have not covered the indefinite integral we ask them to think of a function which has x as a derivative, then one which has a x as a derivative. They usually get the quadratic part, and upon being prompted, ``Could we add, say 4, or 5, or k to your answer?'' they will see the general form.

Now we are getting somewhere. And the group usually feels that way, for they have a candidate for T(x) which was built upon the simple assumptions and not just thrown out to us. But there is concern about the two constants a and k . We ask the students how they might determine them.

We sometimes need to prompt them with such questions as, ``How long does it take the ant to build a tunnel of length x = 0 ?'' And will this knowledge give us information about these constants? Sure enough, 0 = T(0) = a/2 (0)^2 + k = k . Thus we have a simpler model and one less constant T(x) =a/2 x^2 .

But what about the constant a? If someone knew some other data point, e.g., how long it takes to build a tunnel of length, say x = 1 , then we could determine T(1) = a/2(1)^2 and thus a = 2 T(1). But we do not have any more realistic, known data.

Thus we are comfortable with the constant a and our model (a one parameter model), for T(x), the time necessary for an ant to build a straight tunnel of length x into the side of a vertical wall.

We look at T(x) =a/2 x^2 and ask, ``Is this a reasonable model?'' and ``Why is it more reasonable than the `Class offered' model TC(x) = kx ?''

Perhaps we can answer both questions with the same issue, namely, what if we double the length of the tunnel from x to 2x ? How does that effect the time to build the tunnel?

With the model TC(x) = kx , we see that doubling the length from x to 2x has the following effect:

TC(2x) = k(2x) = 2(kx) = 2TC(x) .

Thus the class model predicts that it takes twice as long to build a tunnel of length 2x as it does to build a tunnel of length x .

This sounds good -- but wait. This says it takes T(x) time to build the initial tunnel of length x and it takes the same amount of time to build the next x units of the tunnel. This is unreasonable for the ant has to carry all the material from the deeper section of the tunnel through the outer section of tunnel in addition to the distance in the deeper section. So, surely the time to build say the first half of the tunnel in cannot be the same as the time to build the second half in. With this, we finally dismiss the linear model. (We could have done this earlier, even before we arrived at our answer, and, indeed, we have done this in some instances.)

Now let us consider our derived model, T(x) = a/2x^2 , and ask the questions, ``What if we double the length of the tunnel from x to 2x ? How does that effect the time to build the tunnel?''

T(2x) = a/2 (2x)^2 = a/2 4x^2 = 4 a/2 x^2 = 4T(x) .

Thus it takes four times as long to build a tunnel of length 2x as it does to build a tunnel of length x . We suggest that this IS interesting and it is reasonable now that we believe doubling the length of the tunnel from x to 2x should more than double the time it takes, T(x), to build a tunnel of length x .

These discussions make the model appear reasonable as it seems to fit well with our perceptions of reality. We interpret the model as only a rough estimation, for unless we have more data, we cannot confirm or deny the model. But when we teach calculus for engineering students this analysis suggests issues for the civil engineering students who will eventually be making bids on projects - namely doubling the length of a tunnel will not merely double the cost!