Proportionality

Proportionality


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.

13 thoughts to “Proportionality”

  1. You should include in the name of the title, "Direct Variation", and "Inverse Variation", i have a hard time finding this video.

  2. Your amazing sal if Yoruba teacher you should be the best one and I'm in by math but I still had trouble with this thx and your smart and good at what you so

  3. uhhhh great video if only he didn't repeat himself so much. It made it very hard to concentrate on all the maths. But thats just me I suppose!

  4. I have a doubt…
    In a series circuit , current(I) is a constant
    So,
    P=IV
    Let ~ be the directly proportional symbol…
    P~V ( since I is a constant)
    P~IR ( According to Ohm's law, V=IR)
    P~1/R
    Please say whether above mentioned things are correct. If wrong, please give correct explanation…..

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