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# Proportionality

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- Let's talk about proportionality.
- All this means, if we say that two variables are
- proportional, it just means that they are constant
- multiples of each other.
- So if we say that y is equal to 3 times x, we can say that
- y is-- and we can be a little bit more specific-- we could
- say y is directly proportional.
- y is directly proportional to x.
- And of course, we could divide both sides of this equation by
- 3, and we would get 1/3 y is equal to x.
- So once again, x is a constant multiple of y.
- We could write this as x is equal to 1/3 y.
- So we could also write that x is directly proportional to y.
- If y is directly proportional to x, then x is also directly
- proportional to y, by a different
- proportionality constant.
- So this term right here, this is the constant of
- proportionality between y and x, or between x and y.
- This is the constant of proportionality.
- So in general, if I say that a is directly proportional--
- sometimes you'll just see it written as a is proportional--
- a is directly proportional to b.
- In general, from this statement, we could write that
- a is equal to some constant times b, where this constant
- is equivalent to that constant or that constant.
- It's equal to some number.
- And you could also say that b is directly proportional to a,
- where if you divide both sides of this equation by k, you
- would get b is equal to 1/k a.
- This is direct proportionality.
- Now another thing you might see-- and we'll see it later
- in this video-- is being inversely proportional.
- So let me draw a little line here.
- So let me give you a few examples of inverse
- proportionality.
- Let's say that y is equal to 5 times 1/x.
- Notice what's going on here.
- In this situation, y is proportional to the
- inverse of x, right?
- Or you could say that y is inversely proportional-- I'll
- abbreviate it-- proportional to x.
- It is directly proportional to the inverse of x.
- So here is our proportionality constant, but it is directly
- proportional to the inverse of x.
- Here's another example, and I'm just playing
- with letters here.
- I could write a is equal to 10/b.
- Once again, a is inversely proportional to b.
- You could rewrite this as a is equal to 10 times 1/b.
- a is proportional to the inverse of b, or it's
- inversely proportional to b.
- So now that we have that out of the way, let's do some
- examples using our newly-found knowledge of direct
- proportionality and inverse proportionality.
- So here I have a problem.
- It says that x is proportional to y.
- So that means that x is equal to some constant times y.
- And just so you're familiar with other notation that you
- might see when you're talking about proportionality.
- Other ways to write this statement here. x is
- proportional to y.
- You could write it like this. x is equal to some
- constant times y.
- Or you could write it as x is proportional to y.
- That little, I don't know, left facing fish looking thing
- means proportional.
- This and this mean the same thing.
- This means that x is equal to some constant times y.
- Or sometimes you'll even see it x squiggly line y. x is
- equal to some constant times y.
- This, this, and this, and that are all equivalent statements.
- But this is the most useful because you say, oh, there's
- some constant, maybe we can solve for that constant.
- Then the next statement here says, x is proportional to y
- and inversely proportional to z.
- So that second statement says that x is equal to k.
- It's going to be a different constant, not necessarily the
- same constant.
- So let me put a different-- well, I'll call it k2, I'll
- call this k1-- is equal to some other constant-- not
- necessarily some other, but I'm guessing it's going to be
- some other constant-- times the inverse of z, times 1/z.
- That's what inversely proportional to z means.
- And other ways we could write this is, x is proportional to
- the inverse of z.
- x is inversely proportional to z, or x is inversely
- proportional to z.
- These things in orange are all equivalent.
- I just wanted to make you familiar with it in case you
- ever see it.
- And then they tell us, OK, if x is proportional to y,
- inversely proportional to z-- we wrote that already-- and x
- is equal to 2 when y is equal to 10.
- So let's do that.
- x is equal to 2 when y is equal to 10.
- So x is equal to 2 when k1 is multiplied by y.
- That's what that statement tells us, and y is 10 when k1
- is multiplied by 10.
- x is 2 when y is 10.
- I just put those numbers in there.
- And then we can actually use this now to solve for k1.
- But before that, let's see what this next
- statement tells us.
- It says x is 2 when y is 10 and z is 25.
- So x is 2 when k2 is multiplied by 1/z or 1/25.
- So that's what that second statement tells us.
- So they say find x when y is equal to 8 and
- z is equal to 35.
- So essentially what they're asking us, hey, why don't you
- use this information that we gave you in the green and the
- red, solve for the different proportionality constants, and
- then use that to write, I guess, the specific equations
- for the proportionality or the inverse proportionality, and
- then solve for y or z.
- Let's just do it.
- So here, up here in green, let me rewrite it over here.
- We know that x is 2 when k1 is multiplied by y where y is 10.
- Divide both sides of this equation by 10.
- You get 2 over 10 is 1/5.
- You get k1 is equal to 1/5.
- So that tells us that this first equation right here--
- I'll write it in that original color-- is x is
- equal to 1/5 y.
- Now, that second statement in red, we know that 2 is equal
- to k2 times 1 over 25.
- Let's multiply both sides of this equation by 25.
- This cancels out.
- And we're left with k2-- k2 is equal to 50.
- k2 is equal to 50, so we can write this equation right here
- that x is equal to 50 times 1/z or is equal to 50/z.
- So we've now solved for the two constants of
- proportionality for these two equations, so now we can
- answer the second part of this question.
- Find x when y is equal to 8 and z is equal to 35.
- So the situation when y is equal to 8-- we just go right
- here-- x is equal to 1/5 times 8, which is equal to 8/5.
- So that's what x is equal to when y is equal to 8.
- When z is equal to 35, once again, x is equal to 50/z, is
- equal to 50/35.
- Now let's see, we can divide the numerator and the
- denominator here by 5, so we get 10/7.
- So when z is equal to 35, we get 10/7.
- When y is equal to 8, x is equal to 8/5.
- So those are our two answers, right there.
- Let's do another one.
- Ohm's Law.
- Ohm's Law states that current flowing in a wire is inversely
- proportional to resistance of the wire.
- So current flowing in a wire is inversely proportional to
- resistance of a wire.
- So let's use I for current.
- I is equal to current.
- And we'll use R for resistance.
- R is equal to resistance.
- And you might be wondering why I picked I, but later on when
- you start studying electricity and maybe you become an
- electrical engineer, you'll see that I is the conventional
- letter used for current.
- And I won't go into the details of why, just yet.
- But it tells us that current, that I, is inversely
- proportional to resistance.
- So that first statement I underlined in yellow says that
- current is inversely proportional.
- It's equal to some constant times the inverse of
- resistance.
- It's inversely proportional to resistance of the wire.
- If the current is 2.5 amperes when the resistance is 20
- ohms. So the current is 2.5 when k is multiplied by 1 over
- a resistance of 20 ohms, 1/20 ohms. And I'm assuming,
- because they're going to keep the units in amperes and ohms
- later on, so we could write the units here if we wanted
- to, but I'll keep it simple and not write the units.
- They ask, find the resistance when the current is 5 amperes.
- So this first statement, right here.
- The current is 2.5 amperes when the resistance is 20
- ohms. That's this equation, right here.
- So we can use this to solve for k.
- You multiply both sides of this by 20.
- These cancel out, and you're left with k is equal to--
- what's 20 times 2.5?
- 20 times 2 is 40.
- 20 times 0.5 is 10.
- So it's going to be 50, 40 plus 10.
- So k is equal to 50.
- So in this situation, this equation is I is equal to 50
- times 1/R or 50/R.
- That's our relationship, right there.
- And then they ask, find the resistance when the
- current is 5 amperes.
- So when our current is 5 amperes, so 5 is equal to 50
- over the resistance.
- Let's multiply both sides of this equation by the
- resistance, and you get 5R is equal to 50.
- I just multiplied both sides of this equation by R.
- These canceled out, so I got 5R is equal to 50.
- Divide both sides of this equation by 5.
- We get R is equal to 50 divided by 5 is 10.
- So the answer is, when the current is 5 amperes, the
- resistance will be equal to 10 ohms. And we could write it
- like this, 10 ohms, just like that.
- Let's do one more.
- The intensity of light is inversely proportional to
- the-- now let's be careful here-- to the square of the
- distance between the light source and the object being
- illuminated.
- So let's just say, well, we could use I again.
- Let's say I is equal to intensity of light.
- And let's say that D-- I'm going to do it in a different
- color-- D is equal to the distance between the light
- source and object being illuminated.
- So that's what?
- D, the distance between the light source and the object
- being illuminated.
- Now what does this first sentence tell us?
- The intensity of the light is inversely proportional.
- So the intensity of the light, I, is inversely proportional.
- So it's going to be some proportionality constant times
- the inverse.
- But notice, the intensity of light is inversely
- proportional to the square of the distance.
- Not just to the distance.
- So to the square of the distance.
- So to distance squared.
- If it just said to the distance, we would
- just have a D here.
- But it says the intensity of light is inversely
- proportional.
- So 1 over the square of the distance between the light
- source and the object being illuminated.
- So that's what this first statement is going to give us,
- this equation right here.
- Now, they tell us, a light meter that is 10 meters-- so
- they're saying, when distance is equal to 10 meters-- a
- light meter that is 10 meters from a light source
- registers 35 lux.
- So when the distance is equal to 10 meters, they're telling
- us that the intensity-- the lux is a measure of
- intensity-- the intensity is equal to 35 lux.
- So what intensity would register 25 meters from the
- light source?
- So this first statement they gave us, that I wrote down the
- information in this purple color, we can use that to
- solve for the proportionality constant.
- And then once we have that constant, we can solve for I
- or D, given one or the other.
- So that statement told us, they told us, that when D is
- equal to 10, I is equal to 35.
- So I is equal to 35, so 35 is equal to K times 1 over D
- squared, 1 over 10 squared.
- Or we could say that 35 is equal to k times 1/100,
- multiply-- 10 squared is just 100-- multiply both sides of
- this equation by 100.
- These cancel out.
- And we're left with k is equal to-- what is this?
- 3,500.
- So we've solved, we've used this statement here where they
- gave us some values to solve for our
- proportionality constant.
- So now we know that the equation is-- the intensity is
- equal to 3,500-- that's k times 1 over D squared.
- Or you can just write over D squared.
- Now, they say, what intensity would it register 25 meters
- from the light source?
- So they're saying D is 25.
- So intensity is equal to 3,500 divided by 25 squared.
- Our distance is 25 meters.
- So the intensity is going to be equal to 3,500 divided by--
- well, actually, let's just keep it simple.
- Let me see, well, this is divided by 25 times 25.
- It's 625.
- Let's just get the calculator out.
- So we get, it's going to be 3,500 divided by 25 times 25,
- which is 625, which is equal to 5.6.
- It's equal to 5.6 lux, which is the
- unit for light intensity.
- Hopefully you found that useful.

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