# Calculus: why do we define rate of change as $dy/dx$?

I'm just starting to learn Calculus using Morris Klines' awesome book, "Calculus, an intuitive and physical approach." I really like it so far.

I'm just at the beginning, and after learning how to differentiate I was wondering why rate of change is defined exactly the same for every function. Allow me to explain.

If we deal, for example, with functions the describe distance traveled over time, and we search for the exact speed at a specific time along this distance, then I totally understand why we define the rate of change as $\frac{dy}{dx}$ - it follows perfectly the physical way speed is defined and being calculated: speed=distance/time. Here, $dy$=difference in distance='a distance' and $dx$=difference in time='an amount of time'. So it makes sense to me. All that is left to do is make $dx$ (time) approach 0 and calculate the result.

My confusion comes when we deal with other kinds of physical quantities. Physical quantities whose very physical definition\calculation has nothing to do with division. As an example, let us view the area of a rectangle: $A=a*b$. Allow me to differentiate it, please, so you'll see what I mean.

(Do forgive me as I do not know how to write subscripts on this forum.)

Let us assume a is a constant and b is the independant variable. It follows then that for $b=b_1$ we get:

$a_1 = a\cdot b_1.$

$a_2 = a\cdot (b_1+db)=a\cdot b_1+a\cdot d_b.$

$da = a_2-a_1=a\cdot b_1+a\cdot db-a\cdot b_1 = a\cdot db.$

So far, so good. But then, in the book, for some odd and strange reason, we simpy divide both sides of the equation by $db$.

As mentioned above, the area of a rectangle is defined by MULTIPLYING two adjacent sides. It has nothing to do with division. So finding the rate of change of the rectangle area should also have nothing to do with divison. (In my opinion, of course, and I'll sooon explain why.)

You may tell me that 'the rate of change of the rectangles' area' is just half the sentence - it needs to be in relation to something - and that is where the divison comes from. When you look at the relation between two things mathematically - you divide them. Hence, relation is a quotient by definition. But I disagree, and here is why.

IMO, right where we stopped when we found the derivative of the rectangles' area IS the definition of 'rate of change in the area of the rectangle with relation to one of its sides' - it is right there in the last equation - it is the difference in areas ($dA$) between a rectangle whose side is $b_1$ and another whose side is slightly longer, $b_1+db$. I see here three variables: $dA$, $b_1$ and $db$. To me, that equation is also a mathematical relation between them that explains how they change with relation to one another.

Following this logic - all we need to do now is let db get smaller and smaller until it reaches $0$ to find the exact change in area at $b_1$. But when we do so, the entire right side equals $0$. (Which stands to reason, by the way, because what it actually means is that we subtract the areas of two identical rectangles - so it should indeed zero and cancel out.)

To me, this seems like the right way to calculate the rate of change IN THIS PARTICULAR SITUATION - an area of a rectangle with one side fixed as the other varies, as compared to dividing $dA$ by $db$.

It seems to me that at times, using $\frac{dy}{dx}$ really is the right and logical choice, and in others, we use it to kind of "cheat" because it is an algebraic trick that yields us a solution other than 0.

So, why is it that we define the rate of change EXACTLY the same for every function?

• Because $dy/dx\approx\Delta y/\Delta x$. Commented Feb 9, 2017 at 15:28
• If change is the change in $y$ and is the rate of change of $y$ in terms of $x$ multiplied by the change in $x$, then the average rate of change is the change in $y$ divided by the change in $x$, and the instantaneous rate of change is the limit of the average rate of change as the change in $x$ becomes arbitrarily small Commented Feb 9, 2017 at 15:29
• The reason that a rate of change with respect to one variable with another can be seen as "related" to division is the chain rule. If you can accept that $dy/dx$ = $du/dx * dy/du$ then it might become more clear why we can algebraically manipulate rates. Have you got this far in the book? Commented Feb 9, 2017 at 15:42
• @SimplyBeautifulArt why '~' and not '='? As for as I understand, dy/dx is exactly the same as (delta y)/(delta x). Commented Feb 9, 2017 at 15:57
• @Henry You just summed up the method of increments if I understood you correctly. Unfortunately, I can't see how this explains why we always divide by the increment. Commented Feb 9, 2017 at 16:01

Besides thinking of derivatives as rates of change one can think about it as "the best linear approximation". Given any function $f$ depending on a variable $x$ we may inquire what the best linear approximation of the function around some fixed value $x_0$ is. The answer is that $f (x_0+\epsilon) \approx f (x_0)+\epsilon f'(x_0)$ where $f'(x_0)$ is the derrivative of $f$ evaluated at $x_0$, that is the slope of the function at that point. This interpretation generalizes easily to functions of several variables.

When thinking of rates of change, imagine a rectangle whose one side has a fixed length $l$ and the other depends on time. Suppose the other side depends on time via $b (t)=ct$ where $c$ has units of velocity, that is to say: The $b$ side of the rectangle moves with velocity $c$ making the area of the rectangle larger over time. The area as a function of time is $A (t)=lb (t)=lct$. The rate of change has units area/time. It is $\frac {dA}{dt}(t)=lc$. What this means is that if you have an area $A_0$ at some time $t_0$ and wait a very small amount of time $\delta t$, your area then increases to a very good approximation (which gets better the smaller $\delta t$ is) to become the value $A(t_0+\delta t)\approx A_0+ \delta t\cdot \frac {dA}{dt}(t_0)$.

Surely you find this intuitive as it is basically the same as velocity. The above example justifies the identification of "absolute change of a function due to small change of the independant variable" and "rate of change times small change of independent variable". These two things are almost equal and the difference between them becomes smaller if we make the change in the variable smaller. The diference is also small if the function "looks linear" at the initial value of the variable as opposed to "fluctuating vildly". In fact the area of a rectangle is a linear function of obe side's length and the approximation is exactly true in this case. This is the same as "the rate of change being constant", or equivalently "the acceleration being zero".

Now consider what it means to say how much the area of a rectangle changes if we change one of the sides a little bit. The initial area being $A_0$ and increasing one side by $\delta b$ the area increases by a small rectangle $\delta b\cdot l$. Compare the total area after the increase $A (b+\delta b)=A_0+ \delta b\cdot l=A (b)+\delta b \cdot \frac {dA}{db}(b)$ to the formulas above and maybe you will be convinced that the definition of rate of change as a ratio is correct. The approximation is exactly true here because area is a linear function as discussed above.

• Thank you for the lengthy reply. :) It is quite late at night right now and my brain is very tired. I hope I understand "linear approximation" the same as the actual meaning of the word. If it does, I got some experiments I'd like to do tomorrow on a piece of paper following what you said. Maybe it'll help things click in. Thank you once again. Commented Feb 10, 2017 at 21:24

Multiplication and division are not as different as you seem to think - they're so closely interrelated that there really isn't any difference between them. Objecting to using division in a problem about multiplication is like objecting to using subtraction to solve the equation $x + 7 = 10$.

The reasoning you present would argue that in every situation, the instantaneous rate of change of any quantity is always zero. No matter what situation you consider, you will always end up with something of the form $dy = f(x)dx$. As $dx$ is taken to zero, the right-hand side goes to zero.

What's important to understand is that this is nonsense. When considering a rectangle's area as one of the sides changes, we certainly believe that the area is changing. If the rate of change is $0$, the area is not changing - that's part of what we mean by "rate of change". The key idea is how the area changes as compared to the change in the side length.

For example: say $a = 2$. $A = ab = 2b$. $dA = 2db$, by the reasoning you used. This says that when $b$ changes by a small amount, $A$ changes by twice that amount - so $A$ is changing twice as fast as $b$. We want to be able to say, then, that the rate of change of $A$ is twice that of $b$. To formalize that, we want to say something along the lines of "$A$ changes by two units per unit of change in $b$". Note that the word "per" means you're using division - if I have $8$ pieces of cake and $4$ people, there are $8 / 4 = 2$ pieces of cake per person. So think of $dA$ as "the number of units of change in $A$" and $db$ as "the number of units of change in $b$". Then what we want is "the number of units of change in $A$ per unit of change in $b$"; that exactly means $dA/db$.

• I totally agree with everything you wrote - except with your definition of the word 'per' and how it gets translated mathematically, into equations. I gave this example above and I'd like to give it here for you convinence: Saying that dy/dx is the only way to measure rate-of-change is like saying the number 3 is the only way to represent the physical quantity three. It's not right. How about 2+1? how about 5-2? how about 27/9? There are infinite other ways. Each should yeild the same result - and it does indeed. but not so with calculus :\ at least the way I see things. Commented Feb 10, 2017 at 21:32
• @AryeSegal That seems like a very convincing reason not to say "dy/dx is the only way to measure rate of change" - which is why I never said that. What I said amounts to "dy/dx is a correct way to measure rate of change". Analogously, it's true that there are many ways to represent the quantity "three"; but how would you demonstrate that $2+2$ is not one of them? You would observe that $2+2=4$, which gives a different value than the other ones. Likewise, this is a correct way to measure rate of change, and therefore any other approach which yields a different result can't be correct. Commented Feb 10, 2017 at 22:16
• @AryeSegal To elaborate a little further, in calculus it is the case that every correct way of measuring the same quantity (for example, rate of change) gives the same answer. It's just that the notion of "rate of change" that you outlined with the rectangle example isn't the same concept as "rate of change" as referred to in calculus. What I'm arguing in my answer above is that "rate of change" as referred to in calculus really is what we mean when we say "rate of change" in ordinary conversation, so it's (in some sense) the "right" definition of rate of change. Commented Feb 10, 2017 at 22:19

Good question. You are starting out well. I think I understand why the rectangle example puzzles you.

The calculus of derivatives was invented to study rates. You clearly understand that when the rate is velocity - that is, the rate of change of distance with respect to time - you have to "divide $dy$ by $dx$" to get the instantaneous rate, because the value of that rate depends on $x$.

But when the velocity $v$ is constant you don't need calculus. The ratio of the change in distance to the time elapsed is the constant velocity. There's no need to worry about the limit when the time interval is small - no need to divide by the time interval. In time $t$ you always cover distance $vt$.

If you look at a rectangle with one side length $a$ fixed and think about the area as a function of the second side then in fact the rate at which the area changes in constant, just like uniform velocity. A change of $h$ units in the other side length causes the area to change by $ah$ square units, so the rate of change is the constant $a$. So for this particular rate of change you really don't need to think about what happens when the change $h$ is small. You don't need calculus and you are essentially right when you say

To me, this seems like the right way to calculate the rate of change IN THIS PARTICULAR SITUATION - an area of a rectangle with one side fixed as the other varies, as compared to dividing dA by db.

Whenever some quantity $q$ depends on another quantity $r$ you can calculate the change in $q$ when $r$ changes by $h$: it's $q(r+h) - q(r)$. Then the average rate of change over that small interval is $$\frac{q(r+h) - q(r)}{h} .$$ If that fraction happens to be a constant $c$ independent of $h$ then you don't need calculus: $$q(r+h) - q(r) = ch$$ and $c$ is the rate.

I hope I have addressed this:

So, why is it that we define the rate of change EXACTLY the same for every function?

(There are some problems in your calculus vocabulary, because you're new to the subject. Soon you'll have to be more careful about using "$dx$" for the change in $x$ .)

• Thank you very much for the lengthy answer and the kind and cheering words at the beginning, much needed ;) Took your note about the vocabulary and will do my best, reading some of the comments here really did show that my terms aren't quite spot-on. I believe it'll grow on me as I'll gain more experience in Calculus :) Thanks once more. Commented Feb 10, 2017 at 21:39
• In regards to the post itself, I can definitely see what you mean when you talk about the area of a rectangle, that we don't really need calculus for that, and that's right - just like the example you gave with constant speed. Still, I can't fathom why calculus yeilds different results for dA=a*db and dA/db=a. They are practically the exact same equation, yet calculus views them as a totally different thing. :\ (I hope I'm explaining myself okay. it's tough to explain exactly not in my mother tongue.) Commented Feb 10, 2017 at 21:53
• @AryeSegal They are different ways of saying "the same thing". Sometimes one is more useful or insightful than the other. The ratio is a number: the value of the derivative, the rate of change. The product formulation expresses how a small change in b affects A. Commented Feb 10, 2017 at 22:01
• "The product formulation expresses how a small change in b affects A." Sounds exactly like the rate of change I'm looking for, and the exact definition of what dA/db tells me. So, both equations tell me the exact same thing, have the exact same variables, are mathematically equal - yet each one eventually gives me a different result? That's what I'm having trouble understanding. It's like saying 9=2+7, but when you subtract two you get: 9-2=3. That just doesn't make sense to me :\ Commented Feb 10, 2017 at 22:15
• They don't give you different results. You can't simply "let db go to 0" without taking into account the fact that dA is going to 0 too, but at a different rate, determined by a. What you seem to be saying is that an equation like $2y=x$ says only that $0=0$. It does say that when $x=0$ but it says much more when $x$ is not, even if it's very small. That equation and $y/x = 2$ say the same thing everywhere except when $x=0$. Commented Feb 10, 2017 at 22:33

I think your confusion comes from misunderstanding what we mean when we say 'rate'. A rate is a ratio between two quantities. This ratio is meant to describe how those two quantities are related.

Velocity $\frac{\mathrm d y}{\mathrm d x}$ is the ratio between displacement and time -- it tells me (approximately) how much displacement ($y$) changes for every unit change in time ($x$).

In the same way, it is not quite correct to think about the rate at which the area changes as just the absolute change in the area, as you do in your post. Implicit in speaking about the rate of change of area is that we are taking that rate with respect to the change in something else (for instance, width).

• Thank you for the reply. I can't disagree with the first two paragraphs, but Im having trouble with the last one and I believe you touched exactly the thing I have a problem with: that I can't rate the change in area as the absolute change in area. I would agree with this sentece very much if indeed, the equation of calculating the absolute change in area didn't have any reference to db, but it does. it's right there: dA=adb. I can't see why dA/db tells me something about the rate-of-cahnge that dA=adb doesn't. They both say, and should say, the exact same thing. Commented Feb 10, 2017 at 21:44
• I don't quite understand what you mean in the second half of your comment (the part starting 'I would agree with this...'). However, re the first half: a rate is, by definition, a ratio. One may very well be interested in the absolute change in area, but that absolute change is not a rate. A rate is taken with respect to something else. I guess one might take issue with why rates are defined this way, but that's really just what we mean when we use the term 'rate'. Other quantities may well be of interest, but those other quantities are called something else. Commented Feb 10, 2017 at 21:51