It's one thing to be able to take the words for a variation equation (such as "y varies directly as the square of x and inversely as the cube root of z") and turn this into an equation that you can solve or use. It's another thing to extract the words from a word problem. But, because the lingo for variation equations is so specific, it's not really that hard. Just look for the keywords, and you're nearly home and dry.
The only other keywords (or "key-phrases", really) you might need to know are "is proportional to" which, in the strictly-mathematical sense, means "varies directly as"; and "is inversely proportional to" which means "varies inversely with".
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Translating the above from the English into algebra, I see the key-phrase "inversely proportional to", which means "varies indirectly as". In practical terms, it means that the variable part that does the varying is going to be in the denominator. So I get the formula:
Plugging in the data point they gave me, I can solve for the value of k:
Now that I have found the value of the variation constant, I can plug in the x-value they gave me, and find the value of y when x = 4:
Then my answer is:
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Most word problems, of course, are not nearly as simple as the above example (or the ones on the previous page). Instead, you have to figure out which values go where, what the equation is, and how to interpret it. Fortunately, the keywords and key-phrases should generally be fairly clear, telling you exactly what format to use.
"Is proportional to" means "varies directly with", so the formula for Hooke's Law is:
F = kd
...where "F" is the force and "d" is the distance that the spring is stretched.
Note: In physics, "weight" is a force. These Hooke's Law word problems, among other types, are often stated in terms of weight, and the weight they list is the force they mean.
First I have to solve for the value of k. They've given me the data point (d, F) = (5, 50), so I'll plug this in to the formula:
F = kd
50 = k (5)
10 = k
Now I know that the formula for this particular spring is:
F = 10d
(Hooke's Law doesn't change, but each spring is different, so each spring will have its own "k".)
Once I know the formula, I can answer their question: "How much will the spring be stretched by a force of 120 pounds?" I'll plug the value they've given me for the force into the equation I've found:
F = 10d
120 = 10d
12 = d
Note that they did not ask "What is the value of 'd'?". They asked me for a distance. I need to be sure to answer the question that they actually asked. That final answer is the distance that the spring is stretched, including the units (which are "inches", in this case):
12 inches
Note: If you give the above answer as being only "12", the grader will be perfectly correct to count your answer as being at least partly wrong. The answer is not a number, but is a number of units.
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This one is a bit different from the previous exercise. Normally, I've given given a relation in terms of variation with a plain old variable, like y. In this case, though, the variation relationship is between the square of the time and the cube of the distance. This means that the left-hand side of my equation will have a squared variable!
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My variation formula is:
t^{2} = k d^{3}
If I take "d = 1" to mean "the distance is one AU", an AU being an "astronomical unit" (the distance of earth from the sun), then the distance for Mars is 1.5 AU. Also, I will take "t = 1" to stand for "one earth year". Then, in terms of the planet Earth, I get:
(1)^{2} = k (1)^{3}
1 = k
Then the formula, in terms of Earth, is:
t^{2} = d^{3}
Now I'll plug in the information for Mars (in comparison to earth):
d = 1.5
t^{2} = (1.5)^{3}
This is one of those times when a calculator's decimal approximation is probably going to be a little more useful in answering the question. I'll show the exact answer in my working, but I'll use a sensible approximation in my final answer. The decimal expansion starts as:
In other words, the Martian year is approximately the length of:
1.837 earth years
By the way, you can make the above answer more intuitive by finding the number of (Earth) days, approximately, represented by that "0.837" part of the answer above. Since the average Earth year, technically, has about 365.25 days, then the 0.837 of an Earth year is:
0.837 × 365.25 = 305.71425
Letting the "average" month have 365.25 ÷ 12 = 30.4375 days, then the above number of (Earth) days is:
305.71425 ÷ 30.4375 = 10.044
In other words, the Martian year is almost exactly one Earth year and ten Earth months long.
If you were writing for an audience (like a fellow student, as you'll be required to do in some class projects or essay questions), this "one year and ten months" form would probably be the best way to go.
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Remembering that "weight" is a force, I'll let the weight be designated by "F". The distance of a body from the center of the earth is "d ". Then my variation formula is the following:
I'll plug the given data point of (d, F) = (4000, 200) to this equation, and then I'll solve for k:
(200)(16,000,000) = k
3,200,000,000 = k
(Hey; there's nothing that says that the value of the variation constant k has to be small!)
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The distance is always measured from the center of the earth. If the guy is in orbit a thousand miles up (from the surface of the planet), then his distance (from the center of the planet) is the 4000 miles from the center to the surface plus the 1000 miles from the surface to his ship. That is, d = 5000. I'll plug this in to my equation, and solve for the value of the force (which is here called "weight") F:
F = 128
Remembering my units, my answer is that, up in his spacecraft, the guy weighs:
128 pounds
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