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Working with Formulas

Working with formulas. Sometimes the test will present us with a formula and then they’ll ask about it. The formula is usually a real-world formula from science or business or other areas. The test will always give the formula and explain what the quantities in it mean. You never need to have outside knowledge of the formula or the subject area to answer the question.

Typically, the question, the test will give you the formula, then ask one of three question types. There are questions that will give you some numbers and ask you to solve for others. So it’s actually plug in and solve. There are also questions that ask you to solve algebraically for one of the variables in the formula.

And then there are proportional reasoning questions. So these are typic, the, the three most common questions, three most common tasks that they will give you when they present you with a formula. Here’s a plug in number question. Pause the video and then we’ll talk about this. Okay, when a car, initially moving at a speed of v, decelerates with a constant acceleration a, its minimum stopping distance d is related to v and a by the following formula.

Incidentally, this, this question is throwing in an idea from physics that deceleration is a kind of acceleration, that the word acceleration is actually broad, and it means speeding up and slowing down. That’s some, some outside knowledge that you don’t need, but that’s what’s going on in the terminology of the question. So, if v equals 30 and a equals 10, find the stopping distance.

So all we have to do is take that formula and plug in the numbers that we’re given. And then we’re gonna leave d as a variable. We’re gonna simply solve for that. Well, 30 squared, of course, that’s 900. I’m gonna divide both sides by 10. So I get 90 equals 2d or d equals 45.

And so, that must be the distance. We’re not given any units, so we don’t have to worry about units. That’s the distance. That’s the answer. Second question type, solving for a variable.

So it might give that same explanatory paragraph and then give us this equation and ask us to solve v squared equals 2ad for v. And of course, what we’re gonna do is simply take a square root. So v equals the square root of 2ad. And we can assume everything is positive, so we don’t have to worry about the plus or minus.

For example, we might have to solve h equals one-half gt squared plus vt for g. And again, the formula, they may or may not give us an explanation of what this formula is. So we’re gonna multiply all the terms by 2. Then we’re gonna solve that one term that has a g in it.

And we’re gonna divide by t squared, so we could write the answer like that, or we could split up the fractions, cancel a factor of t in second fraction and write it like this. Either one would be correct. This is a tricky one.

Solve 1 over p plus 1 over q equals 1 over f. Solve this equation for q. Pause the video and then we’ll talk about this. So, a very common mistake here would be to think that we could just take 1 over everything and p, we’d get p plus q equals f.

And so that q would equal f minus p. That is 100% incorrect. We cannot do that. So, the first thing we need to do is solve for 1 over q. And then on that right side, 1 over f minus 1 over p, we need to find a common denominator.

We need to make this into a single fraction by finding a common denominator. So the first fraction, 1 over f, we multiply by p over p. The second fraction, 1 over p, we multiply that by f over f. We get this. We combine the fractions to a single fraction. Well, now we have a single fraction on both sides of the equation, 1 over q equals the single fraction p minus f over pf.

Well, now we can take the reciprocal of both sides and we get q equals pf over p minus f. And that is the correct answer. Now we can talk about proportional reasoning questions. We’ll go back to that same formula for the stopping distance, v squared equals 2ad.

Here’s a typical question, if v doubles and a stays the same, then t is multiplied by what? So the question loves this kind of quest, question, where some of the quantities, they’re multiplied by this or they’re multiplied by this, they’re doubled or they’re halved, and then what happens to this other quantity? They love this.

So how do we approach this? Here’s a very easy way to approach this. Pick ridiculously easy numbers that satisfy the equation. So, for example, we could just pick, start out by saying that everything equals 1. And we could even set the constant equal to 1. Now, why would we set, doesn’t make sense to set 2 equal to 1, but think about what we are doing.

It’s like we have two different scenarios. We have an original scenario where we plug in number,s and then a second scenario when we plug in the numbers again. And really, what we’re doing is we’re taking a ratio of the two equations. Well, that factor of 2 is gonna be the same both times, and so the 2 over 2 is just gonna equal 1 when we take a ratio.

And so, it is easier just to set it equal to 1 right away because we are not gonna have to worry about it. So here, we would set v equal to 1, 2 equal to 1, a equal to 1 and d equal to 1. And then we’ll change whatever values need to be changed, and leave the quantity in the question unknown and solve for it. So, for example, in this question, I will change everything to 1.

I’ll even change that 2 to a 1. And let all three variables equal 1 for my start. That equation works. That would just be 1 squared equals 1 times 1 times 1. So that, that’s a working equation. Now, double v, and leave d as an unknown, so everything else is still 1.

And I get 2 squared equals 1 times 1 times d. And so, of course what we get 2 squared is 4. We get 4 equals d or d equals 4. So in other words, the distance is multiplied by 4 when v doubles and a stays the same. Sometimes, more than one quantity changes at once.

In v squared equals 2ad, if v triples and a doubles, then d is multiplied by what? So pause the video, try to do this on your own, and then we’ll talk about it. Okay, set everything equal to 1 for the start. Even the 2 is gonna get set equal to 1. So that equation works.

Now we’re gonna replace the v with 3 because it triples. We’re gonna replace the a with 2, and we’re gonna leave the d as a variable. So 3 squared equals 1 times 2 times d, or in other words, 9 equals 2d, and d equals 9 over 2. So the distance is multiplied by 9 over 2, or 4.5. Here’s a practice problem.

Pause the video and then we’ll talk about this. Okay. So, this is Newton’s formula for gravitation. The test may or may not tell you that. So we get this formula and we want to know if P triples and R doubles, F is multiplied by what?

So we’re gonna assume everything else stays the same. So we’ll start with everything equal to 1. 1 equals 1 times 1 times 1 over 1 squared. That works. Now we’re gonna change the P to a 3. We’re gonna change the R to a 2, so we get a 2 squared in the denominator.

We’re gonna leave the F as a variable. And what we’re gonna get is three-halves. So the force is multiplied by three-quarters when we make these changes. Here’s another practice problem. Pause the video and then we’ll talk about this.

So the test may or may not tell you that this is Kepler’s Third Law of Planetary Motion. That doesn’t really matter for solving this. So, what we’re going to do is set everything equal to 1 originally. 1 e, 1 squared equals 1 times 1 cubed. That works.

Then we’re gonna replace T with 5 and leave R as a variable. So, 5 squared, which is 25, equals 1 times R cube. So in other words, R cube equals 25. R equals the cube root of 25. We might also write this as the cube root of 5 squared, or 5 to the two-thirds. Any one of those could be a correct form of the answer.

The test will always give you the formula and explain what you need to know about it. The test may ask you to plug in numbers and solve, or it may ask you to solve for a variable in terms of the other variables. In proportional reasoning question, it usually, it’s usually easiest to make each number and the constant equal to 1 in the starting case, and then change what needs to be changed.