Variation Word Problems

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:

"Is proportional to" means "varies directly with", so the formula for Hooke's Law is:

...where "F" is the force and "d" is the distance that the spring is stretched.

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:

Now I know that the formula for this particular spring is:

(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:

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):

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!

My variation formula is:

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:

Then the formula, in terms of Earth, is:

Now I'll plug in the information for Mars (in comparison to earth):

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:

0.837 × 365.25 = 305.71425

305.71425 ÷ 30.4375 = 10.044

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:

(Hey; there's nothing that says that the value of the variation constant k has to be small!)

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:

Remembering my units, my answer is that, up in his spacecraft, the guy weighs: