Expert: Josh - 10/4/2004

Question
Given that y varies inversely as x and x varies directly as z.If z is doubled then y is halved.Why is that true.Please explain.I'm having trouble understanding the different types of variation

Answer
Hi,

Please bear with me, as I am going to give quite a lengthy description to get you to see what is going on. By the end of this, you will hopefully have developed a much better understanding, knowing that variation is closely related to the idea of slope (also called, gradient or derivative) which measures the rate of change between two variables.
Here goes....

Let us introduce the notation ~ in our discussion.
The concept of "x varies in direct proportion to z" is expressed as
x~z     ...[1]
The concept of "y varies inversely as x" is expressed as
y~(1/x) ...[2]

That is to say, the relationship for x and z is given by a straight line equation, x=az, with some proportionality constant "a". Whereas, y changes inversely with x. We can write this in one of two ways. They both describe what is called a "hyperbolic" relationship, which derives from the noun, hyperbola. We can write this as y=b/x or xy=b; once again, "b" is nothing more than a constant (some number).

So, how is this hyperbolic equation (consider y=1/x for instance) different from a straight line equation with a negative slope like "\"? (consider y=-x+10 for example.)

Answer:

SIMILARITIES- In both cases, the value of y (the dependent variable) depends on x (the independent variable). That is x determines what the value of y should be. As x increases, in both cases, y decreases.

DIFFERENCES- In the case of y=-x+10, the rate of change of "y" is constant with "x". Best way to see this, is to draw a graph.
It doesn't matter what the value of x is, the slope[*see footnote] (or rise over run) always remains constant (for y=-x+10). This is obvious if you plot a straight line.

In contrast, the rate of change between y and x increases progressively as you move from larger to smaller x values.

Suggestion: Draw a graph to see this.
Consider the following set of values for y=1/x for instance. The simplest form of the hyperbola.
Plot (x=4,y=1/4),(x=2,y=1/2),(x=1,y=1),(x=1/2,y=2),(x=1/4,y=4) and (x=1/8,y=8)**.........[A]

Note:* For our purpose, you can think of the slope as dy/dx. Where dy and dx represent small changes in the y and x direction respectively. eg., if you were given two points on the curve y=1/x, for instance, (xo=1,yo=1),(x1=xo+dx=3/2,y1=yo+dy=2/3), then the slope is approximately given by m(xo,yo) = (y1-yo)/(x1-x0)=dy/dx=(2/3-1)/(3/2-1)=(-1/3)/(1/2)=1/6.
The interpretation of the slope is that at that point, (xo=1,yo=1), the change of y with respect to x is approximately 1/6 unit of y, every unit of x (in the positive direction). My claim is that this slope value, dy/dx changes depending on x. With smaller values (gentler slope) as you get further away towards larger positive x values. It gets much larger (extremely steep slope) as you walk closer towards x=0. Consider another two points for example, this time, (Xo=1/3,Yo=3),(X1=Xo+dX=1/2,Y1=Yo+dY=2), the slope m(Xo,Yo)=(2-3)/(1/2-1/3)=-1/(1/6)=-6. Clearly, the slope at x=1/3 is much steeper than x=1. [Refer to the graph that you have drawn]

Technically, the derivative is defined as dy/dx in the limit as dx tends toward zero, so that instead of joining two separate points (xo,yo) and (x1,y1) on the curve, we actually find the gradient corresponding to a tangent that passes (touches) the curve at one single point (xo,yo). This happens, when the width dx gets smaller and smaller. There is of course more convenient ways of finding this derivative, dy/dx. But this geometric approach gives you a good mental picture.

Note: ** The trend near the y-axis, i.e., enormous increase in y as you get closer to x=0 from the right hand side of the x-axis is referred to as "asymptotic behavior".

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Recapping what we have earlier, suppose that
y=b/x (inverse relationship) ...[#1]
x=az  (direct proportion)    ...[#2]

If y becomes y/2, to balance the equation (maintain the inverse relationship specified in [#1]), we need x to double.
Starting from xy=b, if y is replaced by y/2, x must be replaced by 2x, so that the RHS remains the same.
i.e., (2x)(y/2)=b
So, y->y/2 induces x->2x.
According to [#2], If x doubles, z must also double.
Writing [#2] slightly differently, as
x/z=a
We clearly need x->2x and z->2z to maintain the ratio,
(2x)/(2z)=a.

P.S. I would be happy to explain further if you have any question. But that has got to wait until Sunday, since I have a conference to attend for the rest of the week.