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?
 A: 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.
A: 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 $x$ .)
A: 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 $dA/db$.
A: 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).
