
Variation Equations (page 3 of 3)
Remembering that "weight" is a force, let the weight be designated by "F". The distance of a body from the center of the earth is "d". Then the formula is the following: F = k / d^{2} Plug in the given data point of (d, F) = (4000, 200) and solve for k: 200
= k / (4000)^{2} (Hey; there's nothing that says that k has to be small!) Since the distance is always measured from the center of the earth, if the guy is in orbit a thousand miles up, then his distance is the 4000 miles from the center to the surface plus the 1000 miles from the surface to his ship. That is, d = 5000. Plug this in, and solve for F: F = (3,200,000,000) / (5000)^{2} Then the guy weighs 128 pounds.
Your first instinct is to say, "What the heck?", and your second is to say, "But they didn't give us any data points! We've got no numbers!". Here's a tip: when you have no idea what to do, try playing around with what they gave you, and see if anything useful happens.
At the very least, I can translate the formula from English into math: T = k d^{4} n^{2} Now what? Well, whatever the diameter used to be, my new diameter is now half the old diameter. And whatever the number of revolutions used to be, the new number is twice that value. So I'll plug in "( ^{1}/_{2} )d = d / 2" where "d" used to be, and plug in "2n" where "n" used to be, and see if I can "find" the original "thrust" expression, k d^{4} n^{2}, within the results: new T = k( d / 2 )^{4}( 2n )^{2} In other words, when I make the changes they said to make, my new thrust is onefourth of the old thrust, which means that the thrust has been decreased by threefourths, or 75%.
First, translate the English into math: h = kx / m Plug in the given data point and solve for k: 4 = k ( 12 ) / ( 4 ) Now plug in the new information, and solve for the answer they want: h = ( ^{4}/_{3} ) x / m Remember that they didn't ask for the value of the variable m. I have to answer the question that they did ask: "They will need six men." As an aside, note that the language of variation is often used by pundits and politicians when they're trying to sound smart, but they don't really know what the terms mean, so they use them incorrectly. They will say "this is directly proportional to that", when all they mean is that there is a positive statistical correlation between the two things (increasing one thing makes the other thing increase, too). They will say "this is inversely proportional to that", when all they mean is that there is a negative statistical correlation between the two things (increase one thing makes the other thing decrease). When listening to people, keep in the back of your head what is the actual definition of the terms (in case you're listening to, say, a scientist, who will use the terms properly) and also the common misuse of the terms (in case you're listening to, say, the evening news). << Previous Top  1  2  3  Return to Index


MATHHELP LESSONS
This lesson may be printed out for your personal use.

Copyright © 19992014 Elizabeth Stapel  About  Terms of Use  Linking  Site Licensing 




