In my Business Calculus class (U.S. college-level), we discuss three aspects of cost: Total Cost $C(q)$, Marginal Cost $MC(q)$, and Average Cost $A(q)$ where $q$ is quantity produced. The defining relationships are: $$\frac{dC(q)}{dq} = MC(q)$$ $$A(q) = \frac{C(q)}{q}$$ $$C(q) = C(0) + \int_0^q MC(q)\ dq$$ where $C(0)$ is referred to as the Fixed Costs.

A question on a recent quiz was:

Given the cost function $C(q) = q^2+50$, what is the average cost when producing 10 units?

This is a pretty straightforward calculation: $\frac{10^2+50}{10} =15$.

For the exam I'd like to ask (likely a bonus question now since I'm obviously struggling enough to post here) essentially the same question but in a position-velocity scenario instead.

The main problem I'm running into is how to ask the question since I can't figure out the position-analogous thing to Average Cost.

At first, considering units in each scenario, I thought it should be something like "Average Velocity":

Given the position function $s(t) = t^2+50$, what is the average velocity after 10 seconds?

But that doesn't match up quite right because the solution to this is $\frac{(10^2+50)-(0^2+50)}{10} = 10$.

It seems like the obstruction to forming the concept I want is that, in the Cost scenario, the Fixed Costs get incorporated into the averaging (whereas in the average velocity they get subtracted out). I'm not sure how the Position question should be phrased to treat the initial position the same as Fixed Costs.

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    $\begingroup$ In your first example, you give the cost function, C(t), and then you find the average cost. In your proposed problem, you give the position function, p(t), and then ask for the average velocity. If you want something analogous, and you want to ask for the average velocity, why not just give the velocity function, v(t)? $\endgroup$
    – Nick C
    Nov 19 '21 at 1:43
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    $\begingroup$ I'm not understanding why you want the question so much like your business question. $\endgroup$
    – Sue VanHattum
    Nov 19 '21 at 2:27
  • $\begingroup$ Fun fact: for constant acceleration, $v = v_o+at$ and if we let $v(t_1)=v_1$ and $v(t_2) = t_2$ then the average velocity $\frac{1}{t_2-t_1}\int_{t_1}^{t_2} v(t) dt = \frac{\triangle x}{\triangle t}$ is equal to the average of the velocities $\frac{v_1+v_2}{2}$. Woe to the freshman Physics student who has need of this understanding for such problems are evil. $\endgroup$ Nov 19 '21 at 12:26
  1. These two averages are of different types.

    Given the position function $s(t) = t^2+50$, what is the average velocity after 10 seconds?

    $\frac{(10^2+50)-(0^2+50)}{10} = 10$.

    The average velocity considers an interval on the independent axis (so, your sample question is in fact wrongly phrased: you should be asking for the average velocity over $10$ seconds or, alternatively, what the average velocity has been after $10$ seconds).

    • Given the position function $s(t) = t^2+50$, what is the average displacement and average velocity over 10 seconds?

      average displacement $=\displaystyle\frac{\int_0^{10}s\,\mathrm dt}{10-0},$

      average velocity $=\displaystyle\frac{s(10)-s(0)}{10-0}.$

    Given the cost function $C(q) = q^2+50$, what is the average cost when producing 10 units?


    The average cost depends on only one point on the independent axis.

    Unlike the instantaneous position/displacement/velocity functions, the function $C(q)$ tracks total cost, not the cost of producing the $q$th unit.

    The cost of producing $10$ units is $C(10),$ rather than $C(1)+C(2)+\ldots+C(10).$

    The production cost of the third unit out of $10$ total units differs from the production cost of the third unit out of $7$ total units, because it is determined by the total production ($10$ versus $7$).

  2. The analogous sample question that you want is this (it's a cheat though):

    • Given the distance-travelled function $d(t) = t^2$, what is the average speed over 10 seconds?

      average speed $=\displaystyle\frac{d(10)}{10}.$

      P.S. Do not replace write “distance-travelled” with just “distance”; the latter is ambiguous and might wrongly suggest distance from the origin, i.e., the absolute value of the position function, instead of $\displaystyle\int_0^{t}\left|\frac{\mathrm ds}{\mathrm dt}\right|\,\mathrm dt,$ i.e., distance travelled.

  3. Based on the definition

    satiation is the feeling of satisfaction and fullness that occurs during a meal and halts eating; satiety is a feeling of fullness after a meal

    the following comparison parallels the above:

    • [average displacement] average satiation (total satiation divided by number of pizza slices) versus
    • [average cost/speed] average satiety (satiety divided by number of pizza slices).

Addendum (reply to the third comment below)

Rereading this, it's worth noting that C(10) = C(0) + MC(1)+MC(2)+...+MC(10). So MC plays the role of instantaneous velocity.

Precisely; on the other hand, neither absement $\int_0^{10}s\,\mathrm dt$ nor resultant displacement $s(10)-s(0)$involves summing up instantaneous velocities, so both aren't analogous to total cost.

So it's still unclear to me what the motion measurement analogous to Average Cost would be

$$\text{average cost}=\frac{\text{total cost}}{\text{quantity}}$$ and $$\text{average speed}=\frac{\text{distance travelled}}{\text{time}}$$ have similar computation mechanics; besides, observe that both total cost and distance travelled start from zero, are non-decreasing, and are each functions of a single input.

The similarity is superficial, but it's the most appropriate analogy from the world of kinematics.

  • $\begingroup$ I totally agree and already understood these aspects of the average. And I really like the satiety/satiation analogy. But it doesn't answer my question: What is the motion/position concept corresponding to total cost (or satiety)? And maybe, in light of what others have said too, there is no such sensical thing. $\endgroup$
    – Aeryk
    Nov 19 '21 at 14:50
  • $\begingroup$ @Aeryk I've updated the Answer in response to your comment. Additionally, this is a related previous post: units of average. $\endgroup$
    – ryang
    Nov 22 '21 at 14:40
  • $\begingroup$ Rereading this, it's worth noting that C(10) = C(0) + MC(1)+MC(2)+...+MC(10). So MC plays the role of instantaneous velocity. So it's still unclear to me what the motion measurement analogous to A would be (if it exists/makes sense). $\endgroup$
    – Aeryk
    Nov 29 '21 at 15:43

I think the key thing making these examples feel different is the level of measurement of the two quantities. The short version of the story is

If there is a well-defined, meaningful zero point for your data, it might be ratio data, so it could make sense to divide such quantities.

If there is no well-defined, meaningful zero-point, then the data is at best called interval data; dividing these does not make any mathematical sense.

Quantities produced are ratio data. Zero is zero; special and well-defined.

Positions are interval data. Zero is arbitrary; it can be moved around.

So when you divided the total cost by 10 in your economics example, you got a meaningful answer, while you noticed that you couldn't reasonably divide your position by 10 -- the only way to convert your positions into ratio data was to convert them into displacement (which is ratio level!) by finding $(10^2+50)−(0^2+50)$.

For a physics example that parallels the one you want, you will need to use something that has a meaningful physical zero value so that you have ratio data; degrees Kelvin, barometric pressure, molarity of a solution, and so on. Position in meters, temperature in degrees Celsius, and other interval data will give you the same awkwardness you found in your question.


After reading the responses and thinking about this a lot since posting, I have decided to answer my own post (certainly not my intent when I first posted).

IMHO the analogous motion question would be the following

Suppose the position of Car C at time $t$ seconds is given by $C(t) = t^2+50$ (meters). Car B, starting at the origin at $t=0$, wants to travel at a constant speed such that, after 10 seconds, it has caught up to Car C. What is that constant speed?

Answer: Car B needs to travel at a constant speed of $\frac{C(10)}{10} = 15 \text{ m/s}$.

There's a couple of reasons why this is what I was originally looking for:

  1. The solution has the correct form, $\frac{C(T)}{T}$, without introducing any subtractions. This means that, like Average Cost, it approaches $\infty$ as $T \to 0^+$. This is often a key conceptual point in distinguishing between Average Cost and Marginal Cost.
  2. There is no calculus required to understand and/or find the solution. You do not have to understand or consider instantaneous velocity. Again, this mirrors Average Cost which students usually understand quicker precisely because it doesn't really require any calculus knowledge. (In fact, all the extra calculus surrounding the topic of Average Cost usually works against them as they try to shoe-horn it into some sort of Calculus idea.)

I'm a physicist by training. What struck me immediately as awkward or ill-defined about your physics question is that it assumes there's a natural starting point at $t=0$. This is a totally wrong/alien way of thinking to a physicist. There is nothing special about $t=0$. So to phrase the physics question in a way that is a natural fit to the subject and the way the subject is conceptualized by workers in the field, it should really be something like "Given ..., what is the average velocity over the time interval from 0 to 10 seconds?"

This just isn't perfectly analogous to your business example, where producing zero units really is special in some sense. I suppose in a businessy context, C(0) represents the fact that you have fixed costs such as property taxes, which accrue regardless of whether you produce anything. Then I suppose C(q)-C(0) might be referred to as something like an incremental cost, relative to keeping your factory in mothballs.

Please also keep in mind that average velocity simply isn't of any interest in physics. The only reason it's discussed in the first chapter of freshman physics texts is to disabuse students of the notion that its definition, which is what they learned in grade school, would be of interest.

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    $\begingroup$ Average velocity isn't of any interest in physics? The only velocity which can actually be measured is average velocity! Instantaneous velocity can only be approximated by average velocities (or computed in a model). $\endgroup$ Nov 19 '21 at 11:03
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    $\begingroup$ I gave this -1 because it mostly tells the questioner to go away instead of helping them answer their question. $\endgroup$
    – Chris Cunningham
    Nov 19 '21 at 16:43

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