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Darkchilde
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Oh dear, I'm sad now. I think my brain must be turning mushy, maths was my strongest subject at school, all the way up to fluid and thermal dynamics, but it just confuses me now.

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trubble76
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So, we need to find the tangent to the parabola at the point (1,1). Then we can find m, the slope from:

f ′ (x)=2×1=2

Where does the "1" come from in this equation? Is it because the "x" co-ordinate is 1?

The Hanging Monkey

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@The Hanging Monkey: Exactly, the coordinate of x is 1. Ah, minor mistake there, should have been a bit clearer. Let me edit that. Post edited and made it a bit clearer.

Darkchilde
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Not a problem, it's probably clear enough, just that I am rusty with some aspects of maths.

I thought I wasn't very good at it when I was at college, but I think I was just lazy and didn't want to work at it

I have a Ph.D. in physical chemistry so obviously I'm not that bad. It will be interesting to learn how to do calculus properly anyway, thanks for making the effort.

The Hanging Monkey

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Oh but why is the gradient 2x ?

You can derive that result as follows.
Imagine a very small increase in x somewhere on the y = x^2 line.
i.e x -> x +dx
y will increase by (x + dx)^2 - x^2 or 2x .dx +dx^2

the gradient at this point is therefore (2x.dx + dx^2) /dx

as dx gets smaller and smaller we can ignore the dx^2 term
so we say the limit of dy/dx (dx->0) = 2x

This approach works for any continous function f(x)

CarlPierce
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I follow that, but I'm probably going to look a bit thick here.....what if dx isn't small?

The Hanging Monkey

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@CarlPierce : the method you describe requires limits, and I said I was not going to go that way. I want to keep it as simple as possible, showing the use of differentiation in maths and physics without getting too complicated in my explanation. For some things I will not explain the provenance, simply will tell as a matter of definition. Otherwise, this would not be an easy primer for differentiation, but a mathematical course.

@The Hanging Monkey : dx is always taken to be small.

Darkchilde
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Well yes, "d" indicates a small change, but I don't see how small changes are relevant to working out a general relationship.

Other than that I followed the mathematics CarlPierce posted fairly easily.

The Hanging Monkey

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The Hanging Monkey wrote:Well yes, "d" indicates a small change, but I don't see how small changes are relevant to working out a general relationship.

Other than that I followed the mathematics CarlPierce posted fairly easily.

Exactly why I have not taken that route. It requires a lot more explanation and requires understanding the concept of a limit.

Darkchilde
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Darkchilde wrote:
The Hanging Monkey wrote:Well yes, "d" indicates a small change, but I don't see how small changes are relevant to working out a general relationship.

Other than that I followed the mathematics CarlPierce posted fairly easily.

Exactly why I have not taken that route. It requires a lot more explanation and requires understanding the concept of a limit.

I would be interested in your explanation of that some time.

The Hanging Monkey

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@The Hanging Monkey : it's a whole subject on itself. Maybe I will go back to the definition of a limit, and see to explain how this all works out later on. With limits you certainly see how and why differentiation works that way.

Darkchilde
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The Hanging Monkey wrote:Well yes, "d" indicates a small change, but I don't see how small changes are relevant to working out a general relationship.

Other than that I followed the mathematics CarlPierce posted fairly easily.

You use the small change idea to find the gradient at a particular point. The idea being that the smaller and smaller the change is the better and better the answer is at a particular point. Darkc doesn't want us to discuss limits so I won't say any more on that.

The general relationship is gotten because I said 'for some X' i.e what I did will work for any value of X on the y = x^2 curve. So gives you the general relationship in terms of x.

In general for a polynominal equation you just reduce the power by 1 and multiply by the original power.
So the derivative of x^2 = 2x and the derivitive of x^3 = 3x^2 etc.

CarlPierce
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CarlPierce wrote:You use the small change idea to find the gradient at a particular point.

And the gradient at any point will give you the general rule?

CarlPierce wrote:The idea being that the smaller and smaller the change is the better and better the answer is

You mean as dx tends towards zero, then ignoring the dx^2 term becomes a better approximation, giving a more accurate answer?

I think I have it

My maths is by no means bad, but it is very piecemeal. I haven't studied maths formally since I was 16 but I've picked up bits here and there throughout my degree and Ph.D. It's very satisfying to have some of the many gaps filled in.

The Hanging Monkey

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The Hanging Monkey wrote:
CarlPierce wrote:You use the small change idea to find the gradient at a particular point.

And the gradient at any point will give you the general rule?

CarlPierce wrote:The idea being that the smaller and smaller the change is the better and better the answer is

You mean as dx tends towards zero, then ignoring the dx^2 term becomes a better approximation, giving a more accurate answer?

I think I have it

My maths is by no means bad, but it is very piecemeal. I haven't studied maths formally since I was 16 but I've picked up bits here and there throughout my degree and Ph.D. It's very satisfying to have some of the many gaps filled in.

Thats right you do 'have it'. In general as dx is very small, mathematically we say as it tends to zero - you can ignore all but lowest power of it.

so another example consider the curve y = x^3

for a small change dx in x, y increases by = (x + dx)^3 - x^3 = x^3 + 3x^2.dx + 3x.dx^2 + dx^3 - x^3 = 3x^2.dx + 3x.dx^2 + dx^3

as dx -> 0 we can disgard the dx^2 and dx^3 terms so dy tends towards = 3x^2.dx
so dy/dx = 3x^2

CarlPierce
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Yup, got that, cheers !

The Hanging Monkey

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Just a thought. When you come to differentiating sines and cosines, would it make sense to introduce the series of these terms to show why we get the result?

e.g. sinx=x+x^3/3!+x^5/5!.... so when differentiated we get 1+x^/2!+x^4/4! etc... (which is cosx).

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campermon
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Thanks for this Vula - great idea!

I'll be back later when I've got more time to see what I can learn from it!
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Spearthrower

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campermon wrote:Just a thought. When you come to differentiating sines and cosines, would it make sense to introduce the series of these terms to show why we get the result?

e.g. sinx=x+x^3/3!+x^5/5!.... so when differentiated we get 1+x^/2!+x^4/4! etc... (which is cosx).

I don't think so, because series is another subject on itself. Taylor series, then Fourier series... I think I am not going to go that deeply; my aim is for people to be able to differentiate most functions, not to teach them a whole course in mathematics for the moment.

Series are very useful for maths and physics, but not for those who just want to know a few basic things in calculus to be able to read the maths in a simple physics book.

Darkchilde
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Scarlett and Ironclad wrote:Campermon,...a middle aged, middle class, Guardian reading, dad of four, knackered hippy, woolly jumper wearing wino and science teacher.

campermon
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