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.

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

thank God i found your videos, i have a quiz tomorrow. partial fractions, binomials and this. now im confident that ill pass.

hey, you are the best teacher ever!

i like this one…thanks sal.

this deserves way more views!

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

THANKS SAL! YOU'RE THE BEST!

hi

Is this an example of the 7th grade version

خريا

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!

Uggggghhhhhhhh I can’t believe I have to look up

MATHon YouTube there goes my streak of memesI 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…..