Variation-equation word problems can be more complex, either because they involve more things that are varying with respect to each other, or because the exercise itself seems vague or complex. But the same techniques still work.
This exercise has some variation that's direct and some variation that's inverse, so this is a combined-variation problem.
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The symbol for electrical resistance, in ohms, is Ω (pronounced oh-MAY-guh, being the upper-case Greek letter for "O"). So they've given me that Ω varies directly with the length L and inversely with the square of the diameter d. This tells me that L will go on top of my fraction, and d ^{2} will go underneath. So my equation is:
Ω = (k L) / d ^{2}
They've given me a data point. I'll plug those values into my equation, and solve for the value of the constant:
16 = (640k)(16)
1 = 640k
Now that I've got the value of my constant, I can find the value they want:
My answer won't be complete without units, so my answer is:
2 ohms
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This exercise uses the combined gas law, which combines Charles' Law, Boyles' Law, and Guy-Lussac's Law. I remember this from when I took chemistry, but I don't remember the exact equation. That's okay; they've given me enough information to figure things out.
The volume V varies directly as the temperature T (so T is on top) and inversely as the pressure P (so P goes underneath). Then my equation is:
Plugging in the values they've given me, I get:
I'll leave the value of my variation constant in exact fractional form, since using a decimal approximation could introduce round-off error. If I round at all, I'll wait until the very end.
Now that I have my variation constant, I can find what they've asked:
This is the numeric portion of my answer. Looking back at the problem statement, I see that volume V is given in terms of cubic centimeters, so my complete answer is:
138 cm^{3}
As it happens, the fraction from the variation constant cancelled out, giving me a nice neat whole number. I'm glad I didn't try using the decimal approximation in my working, because that could have given me the wrong answer.
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The force F varies jointly with the radius and the mass (so r and m are on top) and inversely with the square of the time it takes to go once around (so t ^{2} goes underneath). (Note: In this exercise, "grams" stands for the mass, not the weight.)
So my equation is:
Plugging in the values they gave me, I get:
5,000 = 50k
100 = k
Now that I have my variation constant, I can find my answer:
Of course, this is only the numerical portion of my answer. Looking back, I see that the units for "force" are "dynes", so my complete answer is:
43,750 dynes
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First, I'll translate the English into math:
I'll plug the values from given data point into my equation, and solve for the value of the variation constant k:
Now that I have the value of my variation constant, I'll plug in the new information, and solve for the answer they're wanting:
I need to remember that they didn't ask for the value of the variable m. To get full points, I have to answer the question that they did ask, which was how many men were needed for the task:
6 men
My first instinct is to say, "What the heck?", and my second is to say, "But they didn't give me any data points! I'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 formulaic relation from English into math:
original thrust equation:
T_{orig} = k d^{ 4} n^{2}
Now what? Well, whatever the diameter used to be, my new diameter is now half of the old diameter.
And whatever the number of revolutions used to be, the new number is twice that value (or, in other words, two times of that value).
So I'll plug in " " where d used to be, and plug in "2n" where n used to be, and see what happens.
new thrust T_{new}:
I notice that the remaining parenthetical (that is, the expression within the parentheses) is the original expression for the thrust. (Go back and look, if you're not sure.) So the new thrust is the original thrust, multiplied by one-fourth.
In other words, when I make the changes they said to make, my new thrust is one-fourth of the old thrust. This means that the original thrust has been decreased by three-fourths, or reduced by 75%
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In case you're wondering, the "answer" to the above question is the steps that show T_{new} = ¼T_{orig}. If you like you can end the exercise with a statement along the lines of "Thus, the statement is proved", but the part that'll be graded is lightly-highlighted steps above, that end with the new thrust being one-fourth of the original.
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 an increasing relation (linear, logarithmic, whatever) or a positive statistical correlation between the two things (so that 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).
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