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sdfgw
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| Joined: 08 Jan 2009 |
| Total Posts: 41681 |
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| 01 Oct 2011 07:56 AM |
is this a genuine question
because i don't know whether to actually explain or just post a whole load of lulzworthy crap |
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| 01 Oct 2011 07:58 AM |
Both at the same time
I want to calculus because I can't take going over quadratic equations for the 5 billionth time in maths |
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sdfgw
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| Joined: 08 Jan 2009 |
| Total Posts: 41681 |
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| 01 Oct 2011 07:58 AM |
ok
lemme through this at you
what is the gradient of y = 3x + 6
if that's easy then
what is the gradient of y = x^2
think about it |
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| 01 Oct 2011 08:02 AM |
Well, the gradient of y = 3x + 6 is going to be 3 unless i am being derp as usual
I know y = x^2 would form a curved line but I have no idea what that would make the gradient |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:06 AM |
yus
of course you're stumped with x^2 because it's curved, and the gradient changes
the closest thing you can get is the gradient between two points
like
draw a line between two points close to each other on the line, then calculate the gradient of that
the first coordinate would be (x, x^2)
the second would be a certain distance leftwards - we'll say 'h' units away so it would be (x+h, (x+h)^2)
you have to work out the gradient of the line between two points
i'll stop here because it's really confusing and i just want to see your reaction so far |
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| 01 Oct 2011 08:07 AM |
omg
So gradient is the european equivalent of the slope?
because here in the US we call it the slope
im going to start calling it the gradient |
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| 01 Oct 2011 08:08 AM |
oh wait
the gradient is a vector isnt it |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:08 AM |
yh I actually found out that it's called 'slope' in america a few weeks ago
before I thought everyone called it gradient but then
slope
I mean
GET SOME CLASS |
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sdfgw
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| Joined: 08 Jan 2009 |
| Total Posts: 41681 |
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| 01 Oct 2011 08:09 AM |
@Hi:
nup, just a number
gradient = change in y / change in x
no length |
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| 01 Oct 2011 08:09 AM |
"I mean
GET SOME CLASS"
i know right
it's stupid :( |
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| 01 Oct 2011 08:11 AM |
Oh god, I've never been amazing at this
So if x = 5 then co-ordinates would be (5, 25) and then ((5+h), (25+h^2))?
This would be so much easier if I could make a graph |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:12 AM |
well when i become the universe king of maths
(note: maths)
i'm going to write a law saying that it must always be referred to as "gradient"
also less survey crap and more binomial expansion in statistics GCSE |
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| 01 Oct 2011 08:13 AM |
| IM IN SEVENTH GRADE AND I DO NOT UNDERSTAND THESE BIG WORDS D: |
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| 01 Oct 2011 08:14 AM |
| Oh yeah, also, Denny burnt all me Chekhov books. Sorry. |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:15 AM |
"So if x = 5 then co-ordinates would be (5, 25) and then ((5+h), (25+h^2))?"
close
first point is right, second one would be (5+h, (5+h)^2)
the formula for gradient of a line between two points is as I said earlier, change in y / change in x
change is x is easy, it's just
5+h - 5 = h
change in y:
(5+h)^2 - 25
= 25 + 10h + h^2 - 25
= 10h + h^2
so the gradient is (10h + h^2) / h = 10 + h
so, at the point (5, 25), the gradient is 10 + however long the line's x is
how much sense did that make because we haven't done this properly algebraically yet |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:16 AM |
"IM IN SEVENTH GRADE AND I DO NOT UNDERSTAND THESE BIG WORDS D:"
welcome
we're doing calculus |
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| 01 Oct 2011 08:19 AM |
how much sense did that make because we haven't done this properly algebraically yet _______
I sort of see how you get from one step to the next
Although I'm not entirely sure why the steps are taken |
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| 01 Oct 2011 08:24 AM |
| How the hell the girl can maths? |
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| 01 Oct 2011 08:24 AM |
How the hell the girl can maths? _______
I can't maths though. That's the point. |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:25 AM |
excellent
the goal is to make an expression for the gradient at any point
before we used 5 - now let's assume we don't know the point, so we use x
point 1: (x, x^2) point 2: (x+h, (x+h)^2)
change in x = always h. convenient, right
change in y = (x+h)^2 - x^2 = x^2 + 2hx + h^2 - x^2 = 2hx + h^2
gradient = 2x + h
now if you can visualise it, the closer the two points, the more accurate the gradient
so if the two points were exactly the same, then the gradient would be exact
and if the two points were the same, h would equal zero
so
the gradient of any point on the line y = x^2 is 2x
fascinating
anyway, that's the basics of the first branch of calculus. 's called differentiation, used to find the gradient of any polynomial
the other branch is called integration, and it's pretty much the exact opposite |
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| 01 Oct 2011 08:30 AM |
and if the two points were the same, h would equal zero _______
Obviously.
And I understand, now. |
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sdfgw
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| Joined: 08 Jan 2009 |
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| 01 Oct 2011 08:33 AM |
that's pretty impressive then
though strictly speaking h isn't 0, it's just 1/infinity, but I skimmed over the concept of limits
the notation for a derivative equation (e.g. y = x^2) is:
dy/dx = 2x
the notation for a derivative function (e.g. f(x) = x^2) is:
f'(x) = 2x
also it'd be exceptionally hard to prove it to you but:
y = x^2 -> dy/dx = 2x
y = x^3 -> dy/dx = 3x^2
y = x^4 -> dy/dx = 4x^3
y = x^5 -> dy/dx = 5x^4
y = x^6 -> dy/dx = 6x^5
PATTURN |
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