no... how did you the p^2 in the middle?
we had
6p^2+3p(-4t)+t(2p)-4t^2
parentheses look cleaner btw, you should use them here
what is
3p(-4t)
and
t(2p)
???
they are like terms in the middle
-12pt+2pt=-10pt
this goes for any combination of variables
a^2bc+3a^2bc=4a^2bc
wr^4dt^3+3wr^2dt+6wr^4dt^3=7wr^4dt^3+3wr^2dt
notice in this last example we couldn't combine all the terms because one had different exponents. We can only combine when they are the exact same in exponents and number of variables.
no, just add the coefficients
factoring is one way to see it...
factor out 'pt' from the middle two terms:
\[6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2\]
not necessarily. You can't multiply things unless the equation tells you too, or you do a trick like multiply both sides
if you have
-10pt^2+7pt
you cannot simplify that...
@Inopeki
We just showed that\[(a+b)(a-b)=a^2-b^2\]That relation is called the 'Difference of Squares' formula.
It is used all the time, as you will see, as a quick way to factor anything of the form\[a^2-b^2\]
Yes, it's always true for that form.
This is handy as you will see because if you remember how factoring helped us solve other equations, it will do the same for us here:
For instance\[x^2-4=0\]we can factor this..\[(x-2)(x+2)=0\]which as I mentioned befor by the 'zero factor property' means that we need to solve\[x-2=0\]and\[x+2=0\]to find the zeros of our polynomial
so that's not always zero, is it?
if x=1 then f(x)=-3, not zero
what the expression f(x) is equal to here depends on what we put for x.
There are in fact only two values for x that will make f(x)=0
Which should make sense by what I said yesterday:
The order of the equation is 2, so the total multiplicity of its roots is 2
(fundamental theorem of algebra)
No you are, perhaps I haven't been clear enough
whenever I write
x^2-4
If I want to know where this graph hits the x-axis it is useful to call this a function f(x)
here we have
f(x)=x^2-4
Think about a graph: When the graph is toughing the x-axis it is at y=0, correct?
well, wait now, if you have a y in the function then it is not f(x)
here's one thing that I bet is confusing you:
f(x)=y
when we write a line equation
y=mx+b
we are saying something about the graph:
that the y coordinate is proportional to the x coordinate in this way.
However this is rather simplistic. It is really more that we have a function
f(x)=mx+b
and we have chosen to graph it by letting the y-coordinate of our graph be f(x)
they are effectively the same here, but a function can be graphed in many different types of coordinates, so y would not be used unless we are in Cartesian coordinates as we are now.
there are other coordinate systems like polar, spherical, and cylindrical coordinates
If we tried to graph our function in polar coordinates, y would have no meaning.
however the zeros of the function are independent of how we choose to graph it. The math will be the same.
Hence when we want to know the zeros of some polynomial we call it f(x), and that allows us to put numbers into it and see what happens
f(x)=x^2-4
f(0)=0-4=-4
f(1)=1-4=-3
f(2)=4-4=0
so we found a zero simply by calling this a function and putting numbers into it.
lol, why /don't/ we want to do that?!
Almost all things in physics can be represented by function. If you want to know when the water pressure of a system will become too high, what trajectory to shoot your spaceship, or where a quantum particle will be at a certain time, you must analyse functions in this way.
Really, though of course actual physics problems are usually (though not always) a bit trickier than just plugging in a number
but yes, functions are instrumental to calculus, and therefor physics
As I said earlier calculus is about analyzing functions...
I guess we should start to talk about functions in a very detailed way.
the definition of a function for you will be:
if you have something like
y=f(x)
(where f(x) is some arbitrary polynomial with x's in it)
and for every x you put in you get exactly one number for y back, then
y=f(x) is a function, and y is a function of x
try to soak that in, I'll tell you more about it momentarily
no, just that we get only one value of y for each x
for instance
f(x)=y
I defined that, whether or not it's a function
however if for every x we put in we get only one y back,
then f(x) is a function, and the variable y is a function of the variable x
look at a graph like a circle (I know you haven't studied them much yet this way)|dw:1326043068666:dw|this is not a function because there is more than one y for each value of x...
see you in a bit, I'm gonna have breakfast.
drawing a line from x to its corresponding y value you should know we need a vertical line|dw:1326043155512:dw|see how one x-coordinate corresponds to two different y values y1 and y2?
But what about y=2x+5 ?|dw:1326043286670:dw|this graph has exactly one value of y for each value of x, so it is a function.
the idea above is what's called the 'vertical line test':
if you can draw a vertical line at any point on the graph and intersect it twice, then it is not a function.
when you return look at this
say we have a table of values for some
y=f(x) and we want to know if f(x) is a function....
look at the table
x |y=f(x)
---------
-1 | 2
3 | 7
2 | 0
5 | 2
Is this a function? why or why nmot?
actually look at the x's
each x has one y, so it is a function
we have that f(5)=f(-1)=5 so two different x's give the same y, but that is okay
the only thing against the rules would be to have something like this
x | y
------
1 | 3
2 | 5
1 | 4
because this means that f(1) has two values:
f(1)=3 and 4
so this is not a function...
look at the x to see which gives more than one f(x)...
(I am using the terms f(x) and y here interchangeably)
x | y
-----
2 | 0
3 | 1
5 | 6
3 | 6
1 | 0
function or not ant why?
right
not 1+6=7 of course, it equal the set\[f(3)=\left\{ 1,6 \right\}\]
It doesn't mean it would be a circle but it would fail the vertical line test, watch...
thanks FFM, praise form Caeser :) |dw:1326046327322:dw|now look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?
more than 3?
You only need to hit 2 points to show that this is NOT a function and here it is:|dw:1326046665544:dw|so again we show by this that f(3) has more than one value...
there are infinite points on a line....
a line intersects another line at a point|dw:1326047055394:dw|P is our intersect point
can you draw a vertical line that hits the curve at more than one point?
not so, plug in various x into the function f(x)=0, we get the table
x | y
----
0 | 0
1 | 0
2 | 0
3 | 0
and so on...
and I think you would agree that this chart constitutes a function.
They are all in agreement, the chart, the graph, everything.
we don't need x=0 to get one value of y back; we always get one value of y. The same value of y, namely zero.
do you know what the graph of y=0 looks like?
perfect, what about the line x=0?
(standard cartesian notation here , don't get hung up on the names of variables, we could easily have them to t and p or whatever)
no, this would be
f(x)|x=0
(read f of x such that x is zero)
here f(x) takes on a range of values from negative to positive infinity:
x | f(x)
------
0 | 0
0 | 1
0 | 2
0 | -1
0 | -2
etc.
not a function, right?
no this chart is the same as
x | y
------
0 | 0
0 | 1
0 | 2
0 | -1
0 | -2
I told you that y and f(x) are interchangeable
y is the name of the coordinate which we are using to represent f(x)
so here they are the same
the chart itself is of the same type as you saw before, we have one value of x corresponding to multiple (in this case infinite) values of y, so it is not a function.
exactly, so here x is a function of y
we could have therefor more than one y value for an x value in this case, but not more than one x value for each y, if it is to be a function.
so really the rule is better stated this way:
"a relation is a function if each value of the independent variable corresponds to exactly one value of the dependent variable"
because we can change the names or swap positions of the variables