Direct Variation Problem Solving

Direct Variation Problem Solving-78
Instead, you have to figure out which values go where, what the equation is, and how to interpret it.

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If \(k\) is negative, as one variable goes up, the other goes down.

(\(k\ne 0\)) Think of linear direct variation as a “\(y=mx\)” line, where the ratio of \(y\) to \(x\) is the slope (\(m\)).

Let’s suppose you are comparing how fast you are driving (average speed) to how fast you get to your school.

You might have measured the following speeds and times: (Note that \(\approx \) means “approximately equal to”).

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: Most word problems, of course, are not nearly as simple as the above example (or the ones on the previous page).

(I’m assuming in these examples that direct variation is linear; sometime I see it where it’s not, like in a Direct Square Variation where \(y=k\).

There is a word problem example of this here.) We can also set up direct variation problems in a ratio, as long as we have the same variable in either the top or bottom of the ratio, or on the same side. Don’t let this scare you; the subscripts just refer to the either the first set of variables \((,)\), or the second \((,)\).

Plug in the first numbers we have for \(x\) and \(y\) to see that \(\displaystyle k=\frac\). We plug the new \(x\), which is \(\displaystyle \begin\frac&=\frac\\frac&=\frac\y=\frac&=\frac\end\) We can set up a proportion with the \(y\)’s on top, and \(\)’s on the bottom.

We can plug in the numbers we have, and then cross multiply to get the new \(y\). Inverse or Indirect Variation refers to relationships of two variables that go in the opposite direction (their product is a constant, \(k\)).


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