Combined Variation/Proportionality Question

[dtmowns]

Can someone try to explain to me why you write something like "x is jointly proportional to y and z and inversely proportional to w" as x = Kyz/w instead of writing two expressions like this: x=kyz , x=k/w? The second implies two constants, which is incorrect right? But I'm not sure exactly why this is incorrect. Can someone attempt to give me more information?

Thanks,

Dylan

 

[JeffM]

Can someone try to explain to me why you write something like "x is jointly proportional to y and z and inversely proportional to w" as x = Kyz/w instead of writing two expressions like this: x=kyz , x=k/w? The second implies two constants, which is incorrect right? But I'm not sure exactly why this is incorrect. Can someone attempt to give me more information?

Thanks,

Dylan

Sure Dylan. I'll make an attempt at answering a very reasonable question. If I fail, not that I mean to, ask again. Someone else will help.

\(\displaystyle x = ayz\ and\ x = \dfrac{b}{w}\ do\ not\ mean\ the\ same\ thing\ as\ x = \dfrac{kyz}{w},\ a,\ b,\ and\ k\ constants.\)

The easiest way to see this is that

\(\displaystyle ayz = x = \dfrac{b}{w} \implies b = awyz\) so b is NOT a constant. Rather b is a variable dependent on w, y, and z.

What ARE two equivalent ways to say the same thing are

\(\displaystyle x = ayz * \dfrac{b}{w}\ and\ x = \dfrac{kyz}{w}.\)

k is simply the product of a and b. But the product of two constants is also a constant. For example

\(\displaystyle x = 2yz * \dfrac{3}{w} = \dfrac{6yz}{w}.\)

The two expressions mean exactly the same thing, but one is considerably more concise.

 

[dtmowns]

I think I've almost got it now. The only thing is I'm not sure of the reasoning behind interpreting the "and" as multiplication. For some reason the logic behind more complex Variation problems eludes me.

 

[JeffM]

I think I've almost got it now. The only thing is I'm not sure of the reasoning behind interpreting the "and" as multiplication. For some reason the logic behind more complex Variation problems eludes me.

The problem you are having is that natural languages like English are not so precise as the artificial languages created by mathematicians. Even the very simple word "and" contains ambiguities. It is very tricky to translate between math speak and a natural language. That is why students find word problems hard.

Let's go back to your original statement (slightly modified).

x = Kyz/w instead of writing two expressions like this: \(\displaystyle x=k_1yz,\ and\ x=k_2/w?\)

What is implied by your alternative is that x is simultaneously equal to two functions, one a function of two variables like g(y, z) and the other a function of one variable like h(w). Both k₁ and k₂ are considered to be constants.

It is certainly possible that some values of x may be equal to g(y, z) and h(w). Let's take an example.

x = 3yz and x = 72 / w. If y = 8, z = 1.5, and w = 2, it is true that 3 * 8 * 1.5 = 36 = 72 / 2. But it is not generally true for all values.

If y = 7, z = 2, and w = 3, 3yz = 42 and 72 / w = 24. It is not true that x is simultaneously equal to 42 and 24.

What was described is a multiplicative function of THREE variables, and it is YOU who have said that "and" describes how that function of three variables is related to functions of one and two variables. In my very first sentence, I stated that "and" as you used it does NOT translate the mathematical statement at issue. You are in the position of saying "In my opinion, a@b means a and b, but I do not understand how "and" means the same thing as @." Well, no one said it did except you. You are saying that you do not understand your own explanation, which is usually a good clue that there is a defect somewhere in your explanation. (Don't worry: it happens to me all the time. It just means you need to think a bit longer.)

Now here is where I think you are going off the rails.

The statement x is directly proportional to y means this: x is a constant multiple of the independent variable y, PROVIDED that any other independent variables are held constant.

The statement x is inversely proportional to y means this: x is a constant multiple of the multiplicative inverse (or reciprocal) of the independent variable y, PROVIDED that any other independent variables are held constant.

The statement x is jointly proportional to w, y, and z means this: x is a multiple of the PRODUCT of the independent variables w, y, and z OR some or all of their multiplicative inverses.

I am not sure I can explain it better than this, but if you still are confused about ANYTHING, please explain where.

 

[dtmowns]

I'm missing some fundamental knowledge about this I think. When you have combined variation like x = Ky/z, am I right in saying that the expression is telling you the relationship between x and y when z is held constant, and the relationship between x and z when y is held constant? For instance, the fact that Ky is being divided by z doesn't change the Direct Proportional relationship between x and y as long as z is held constant?

When you manipulate it into zx/y = K, do we have a "new" constant? What's really going on here? I think part of what is tripping me up is I read proportionality expressions as normal algebraic equations instead of realizing the message underneath.

Thank you for your patience in answering these questions. I like these kinds of problems but, annoyingly, I am having a hard time truly understanding them. I want to get to the point where they make as much sense to me as 3x + 1 = 4 where x is 1.

 

[JeffM]

I'm missing some fundamental knowledge about this I think. When you have combined variation like x = Ky/z, am I right in saying that the expression is telling you the relationship between x and y when z is held constant, and the relationship between x and z when y is held constant? For instance, the fact that Ky is being divided by z doesn't change the Direct Proportional relationship between x and y as long as z is held constant?

When you manipulate it into zx/y = K, do we have a "new" constant? What's really going on here? I think part of what is tripping me up is I read proportionality expressions as normal algebraic equations instead of realizing the message underneath.

Thank you for your patience in answering these questions. I like these kinds of problems but, annoyingly, I am having a hard time truly understanding them. I want to get to the point where they make as much sense to me as 3x + 1 = 4 where x is 1.

I am having trouble answering you in part because I do not know what you know. In particular, do you have an intuitive understanding of functional notation?

Let's take your new (and simpler) example of x = ky / z. We can say that x = f(y, z). That is, x is dependent on two independent variables.

It is a common technique in calculus (which I am pretty sure you have not studied yet) to think about a multivariate function such as f(y, z) in terms of what happens when just one independent variable varies.

So yes, it makes perfect sense to think about x = ky / z as meaning that x is directly proportional to y if z is held constant. And all that means is that x is a constant multiple of y (if z does not change). Similarly, it makes perfect sense to think about x = ky / z as meaning that x is inversely proportional to z if y is held constant. And all that means is that x is a constant multiple of the multiplicative inverse (or reciprocal) of z (if y does not change). This is just looking at something from different perspectives. So far so good. But of course neither y nor z is constrained. Both are independent variables, and both are free to be varied at the same time. It just simplifies things to think that only one changes at a time.

So why use the language of proportionality at all? Because x IS proportional to the product of y and z's multiplicative inverse. Let's define a new variable u = y / z. Then x = ky / z = k(y / z) = ku.

These are regular algebraic equations of a particular kind, namely where the dependent variable is a constant multiple of the product of the independent variables or a constant multiple of the product of the multiplicative inverses of the independent variables or a constant multiple of the product of some combination of some independent variables and multiplicative inverses of other independent variables.

This is ordinary algebra, with a special vocabulary to talk about this family of very simple relationships.

 

[dtmowns]

I have experience with simple functions like f(x) = 3x + 2 for a line but not much experience with multivariate functions.

Two final questions:

When you say "x is a constant multiple of y," you simply mean that x equals a constant times y right? wrong?

In your experience, what is the best way to learn mathematical concepts outside of the classroom?

 

[JeffM]

I have experience with simple functions like f(x) = 3x + 2 for a line but not much experience with multivariate functions.

Two final questions:

When you say "x is a constant multiple of y," you simply mean that x equals a constant times y right? wrong? Right. Sorry not be clear.

In your experience, what is the best way to learn mathematical concepts outside of the classroom?

It is easy to go from the concept of a function of one variable to a function of two variables because the geometric analogue is easy to fathom.

Geometrically, z = f(x) means that z is a line or curve in the x, z plane such that no point along the x axis corresponds to more than one point in the curve. Geometrically, z = g(x, y) means that z is a plane or surface in the x, y, z volume or space such that no point in the x-y plane corresponds to more than one point in the surface. We just go from two dimensions to three dimensions, and it is easy to visualize (although not so easy to draw) a three dimensional space because it is the space of our sense experience. Functions of three or more variables are not easy to imagine geometrically (I at least find it impossible).

As for for how best to learn mathematical concepts, I am not the right person to ask. My method has been to fuss at them until the light dawns and to do problems, lots of problems. Probably not very efficient, but I am not a mathematician, never studied math beyond the normal college courses, and know virtually no math beyond what was standard material about 1850. But you have what I believe is the right idea. To be able to use the math that you do learn, what is much more important than mechanics is to understand the concepts behind the mechanics. If you remember the concepts, which in part requires applying them dozens of times in carefully designed problems, you can quickly relearn whatever mechanics you forget. You can come to this site to ask conceptual questions. There are lots of people here who can answer them well.