BAM!
Mr. Tarrou.
In this video, we are going to expand our knowledge of variation problems whether direct
or inverse variation and introduce some more complexity of the problem cause in real life,
things are always just a relationship between just two variables.
There's a lot of times especially in like physics problems and electrical problems where
there's more going on than just an x and a y.
So joint variation is when y varies jointly as x and z if there is some non-zero constant
'k' such that y equals k times x times z so it's kind of like just an expanded form of
your direct variation.
It's just that there's more than two variables being directly related to each other.
Now this is kind of a special case of the more general idea of a combined variation
and that’s when one variable or quantity varies either directly and/or can be a combination
of two, direct and inverse variation, or inversely as two or more other variables or quantities.
You don’t have to just have three quantities or variables in your equation other than a
constant.
It can be more complicated than that.
Quantities that vary directly are in the numerator.
This is sort of like a footnote for the combined variation.
Quantities that vary directly are in the numerator and quantities that vary inversely are going
to be found in the denominator cause remember direct variation is y=kx and inverse variation
when you have the equations set up for equal to y, it is going to be equal to k over x.
So direct variation, those variables are going to be found in the numerator, and the inverse
variation, those variables will be found in the denominator.
So let’s get to our first example.
We are going to do four in this video, I just don’t want to do a lot of them, they tend
to be a little long.
This will be the first of four.
Suppose y varies jointly as x and z.
Find y when x equals 8 and z equals 3 if y equals 105 when x equals 5 and z is equal
to 6.
Even though our problems are getting a little bit more complex, they still have to give
you enough information to find the constant of change, that’s basically never given
to you.
SO we are going to see that last little bit here where they gave us y, x and z to find
out what k is.
Now if you don’t have the basic structure of the equation, you are not going to be able
to find anything so as we start of the problem, it says y varies jointly, okay so it’s going
to be y is equal to, and joint variation is y equals k times x times z so it’s basically
a direct variation but just with more than one variable.
so y equals k times x times z.
so we got the basic structure of our joint variation written down as soon as I read this,
I stop and write this out and now we are going to go to the part of the problem where they
gave us all the information except for the constant so when y equals 105, we don’t
know the constant of change but we know that x=5 and z=6.
Okay so 5 times 6 is 30.
So we have 105 equals k times 30 or you are going to write 30k.
We need to undo that multiplication by dividing both sides by 30.
Now if you do it with your calculator you have your answer but let’s just do this
a little bit in our head.
3o, 60, 90, 120, okay so well 120 is bigger than 105 so 30, 60, 90.
So there's three 30's in 105.
Now if we look at the 90, 105-90=15 and 15 is remainder over 30 is going to reduce to
15/30 is going to reduce to 1/2 so this is 3 and one half or I want to write that as
an improper fraction because we don’t, I don’t ever like to work with mixed fractions
unless I am writing the final answers, so three times, or two times 3 is equal to 6
plus 1 is 7 so k is equal to 7/2 at least in the format that we are going to work in
through the rest of the problem.
Now we are going to come up here, write the actual equation out, we have y equals 7/2
cause that's k, times x times z.
Okay!
Now let’s go to the other part of the problem where they don’t, they ask us to find one
of these three variables x, y or z.
we got the constants, we got the equation.
The actual problem says find y when x is equal to 8 and z is equal to 3.
So we are going to do y is equal to 7/2, x again is 8 z is 3.
1/2 of 8 is 4 cause we have a division of 2 and a multiplication of 8.
You know what, you might want to when you are working with fractions cause a lot of
my students, I see a lot of students struggling with them.
If your problem has one fraction, it might be helpful to just write every number as a
fraction so you can see the common denominators you might need or you can see this cancellation
opportunity that we have.
2/2 is 1 and 8/2 is 4.
So we really just have with the 8/2 is 4, we have 7 times 4 which is 28.
And that times 3, well 3 times 20 is 60 and 3 times 8 is 24 so 60 and 24 is 84.
And let’s just check my notes here.
That’s exactly the same I got in my notes as right now!
so there you go, there’s our first example dealing with joint variation.
Let’s get to the second one.
BAM!
Our second example is again joint variation.
It's just going to be a touch harder because I am bringing in this idea of the cube root
of y, so it’s not going to be all exponents of 1.
And before I start this example, the previous one where we set up the work to find out what
the constant was and took that extra step of actually writing the joint variation equation
out.
I thought that was important to see the actual equation that you have for the problem then
using it to finish the problem up.
Some textbooks will show the previous example I just worked out by setting up two equal
ratios and doing a cross multiplication process.
It's of course going to be correct, maybe just a tiny tiny bit quicker than how I did
it.
I personally feel it’s important that you can see the actual joint variation equation
before you actually use it to finish the problem up.
So it’s just a style of teaching, some of your books might show the work the same as
I did, some might have done cross multiplication.
If z varies jointly as x and the cube root of y, blah blah blah.
I am not going to even look the number here, because the first part of the sentence here
has shown me or tell me how to set up the problem.
z varies.
so the way it’s worded, if it said y varies directly or jointly or whatever it would be
y equals, but its z varies jointly or combined with or inversely or whatever so they are
telling me that z is by itself and it varies jointly which is again sort of this extended
direct variation set up here z varies, so it’s going to be k and it varies jointly
as x so it’s going to be k times x and the cube root of y so that’s going to be the
basic structure of our problem.
now this is joint variation and my last example was joint variation and I just gave you a
little bit, introduce little bit of a twist here with the problem so I am going to pause
and see if you can work out the problem and I am just going to come back in a few seconds
with the solution.
You should be able to do this now after seeing the previous example, just this little extra
difficulty with the cube root and the fact that I threw in another fraction that you'll
have to deal with but take a second and see if you can work this quick little problem
on your own!
Hmm!
Nananana!
I hope this is what you got.
I plugged in, again with the first part of the problem where they said what z, x and
y were.
I plugged all those values in and found that my constant value, my constant variaton was
with my joint variation formula and then went onto the rest of the problem where they said
find z when x equals 9 and y equals 27/8 and worked out the rest of the problem and got
a final answer of 189/2.
Let’s get to that next example.
It will be a combined variation problem and the last one will be a word problem.
And example number 3 says f varies directly as g and inversely as h.
f varies, something's going on with f and it equal to and a bunch of other stuff going
on so, so f equals to, f varies directly as g, so it’s going to be basically like g
times k, that’s direct variation, right?
f equals another variable times a constant, like well kx.
And inversely as h.
Well inverse variation is y equals k over x, so again that inverse variation you are
going to find that in the denominator.
so inversely as h, that means we have f is equal to, started writing f there again, f
varies directly with g and inversely with h so this is the structure of the equation
we are going to do for the rest of the problem.
Find g when, so this is the part of the sentence where you are being told what you need to
find so that’s not the part of the problem that is going to tell us how to find, or give
us enough information to find out what k is.
This will though.
If g=54 when h=2 and f=9.
So they are giving us g and h and f and that means that they are giving us everything except
for the constant so this part of the question again is where we need to start our work.
So f=9, g=54 times our constant of change over h which is equal to 2.
So we have, let's see here we have 54/2 is 27.
So we have 9 equals 27k.
We need to move that 27 away from the k, they are touching that means multiplication.
So I am going to divide both sides by 27 and last time I checked 9/27 reduced to 1/3.
So here is our constant of change.
That means we are ready to write the equation now.
So, bringing this up, we get f is equal to g times, or over h times our constant.
Now, I am going to rewrite this initial equation as, instead of g times k over h, let that
k be over 1 and pull it off to the side and write it as g/h times k/1.
Right?
If I write this k as k/1 and multiply my two fractions, I'd have g times k which is gk
over h times 1 which would be h so this is good.
This is a different way of looking at the same equation.
I am just pulling the numerator off and the reason why is cause I don’t want to have
a fraction of 1/3 in my numerator.
I don’t want to build up my complex fraction if I don’t need to.
So that k, I am going to pull off to the side as k/1 or just k and these are equivalent
expressions so I take that value of k and plug it in, I get f= 1/3, that’s my value
of k times g over, pretty ugly g, here we go, g/h.
f=1/3 times g/h.
Let’s go to the part of the problem where we have to actually find out what g is.
Find g when f=3, so we have 3 equals and h=4 and h is in our denominator so we have f which
is 3 equals 1/3 times g well that’s what we are looking for, so g over h and find g
when f=3 and h=4.
Okay so we have 3.
Let’s go ahead and multiply all of this together.
1 times g is g. 3 times 4=12 and we are trying to solve for g well g has a division of 12
with it, we are going to undo that division by multiplying
and here the multiplication of 12 and the division of 12 cancel out and we get g=3 times
10 is 30, 3 times 2 is 6 so 36!
BAM!
Alright that’s our first combined variation problem.
We have got one more that’s going to be set up as a word problem and the work itself
is not going to actually look any different.
Thank you for watching by the way.
The stretch in a wire under a given tension varies directly as the length and inversely
as the square of its diameter.
Stop reading and write off the equation!
What do we have here?
We have, The stretch in a wire, so the stretch, under a given tension varies directly so that’s
going to be y=kx, it’s going to be some kind of multiplication or the direct variation
is found in the numerator, varies directly as the length.
Okay-dokey!
And inversely as the square of the diameter and then somewhere in there we need a constant
of multiplication.
So that’s going to be the structure of our combined variation problem.
Again the stretch in a wire under a given tension varies directly and inversely as the
square of its diameter.
A wire having a length of 5m and a diameter of 1.7 millimeters stretches 2.4 millimeters.
If a second wire made of the same material has a length of 7 meters and a diameter of
1.5 mm, how much will it stretch?
Well again, even though it’s a fancier looking word problem, they need to give us enough
information to find the constant of change.
So that’s going to be here we have a length of 5 and a diameter of 1.7 and with those
conditions the stretch is 2.4 mm.
So we have stretch of 2.4 is equal to the length of 5 and the diameter of 1.7 squared
and we are going to then finish that equation and solve for k and though I tried using some
stuff in my head, I don’t think you want to watch me do division and multiplication
of the decimals, so that being said, right of the top off my head, oops forgot the square
there, we have 2.4 equal to 5 divided by 1.7 squared is 1.73 times k.
Divide both sides by 1.73 and you get k, your constant of change, is equal to, rounded off
values of course, approximately 1.39.
So now that we have got that, we can write our final or actual equation which is stretch
is equal to length over diameter squared times our constant which is 1.39 now.
Okay!
So the rest of the problem says, if a second wire made of the same material has a length
of 7, okay so the length here is 7 and a diameter of 1.5, how long is it going to stretch?
Well our constant here, we have a length of 7 meters and a diameter of 1.5 mm, it’s
going to be squared and that times 1.39 is going to give us a stretch that is equal to
6.49 mm and that is the end of discussion of joint and combined variation.
I am Mr. Tarrou.
Go do your homework!