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/
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