Units for Inverse Quantity

Date: 9/10/96 at 16:6:23
From: Anonymous
Subject: Units for Inverse Quantity

We just started doing inverse-proportion graphs. There is a data 
table with time vs distance. Distance is in centimeters, time is in 
seconds. For my situation, the two are inversely proportional. After 
creating the graph of their values, we are supposed to add a third 
column to the data table, "1/distance." It shows the reciprocals of 
the values in the distance column. 

When we plot the graph of time vs. 1/distance, it makes for a nice 
best-fit line graph. What units should be used when referring to the 
1/distance values? My teacher has suggested that it might be cm^-1, 
but he isn't sure. Can you tell me?

Date: 9/10/96 at 18:41:45
From: Doctor Tom
Subject: Re: Units for Inverse Quantity

Your teacher is right.  If you invert the number, you have to invert
the units.  cm^-1 for the distance, sec^-1 for time.

You can read these as "per centimeter" or "per second".

To get (perhaps) a better feel for why this is true, let me use a
hypothetical situation of my own.  Suppose a bell rings every 1/4 of a 
second.  So the time between rings is 1/4 second, right?  We also 
often say that the bell rings at a rate of "4 per second" - that's 
what you get if you invert "1/4 second".

Or here's a more interesting example.  Suppose you work out some
physics problem, and the answer is a velocity divided by an 
acceleration.  What kind of a "thing" is it?  I'll just use D to 
represent "distance" units (you can use centimeters, feet, furlongs, 
or light-years; I don't care), and I'll use T for the time units 
(seconds, hours, fortnights, ... you pick!)

Velocities always have units D/T and accelerations have units of 
D/(T^2). So a velocity/acceleration has units of (D/T)/((D/(T^2)) = T, 
so you had better be calculating a time in your formula. You can 
always use the units to check your results this way.  It doesn't prove 
that your answer is right, but it can certainly prove that it's wrong!

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/