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Differentiation

Can someone explain to me what differentiation is and where to start on it, recommended reading? As i do not do AS or A2 maths i am finding this very difficult. Also can anyone recommend any more maths stuff i will need to learn that won't be covered clearly in physics lessons for A2? Much appreciated thanks.

Reply 1

Dekota

It's where you can differentiate between things. :d

It is hard.

Reply 2

Kyle_S-C

I think the best approach might be to try and get hold of some P1, P2 and P3 books for the Edexcel Maths course, since they have now been replaced by C1-4, so they're actually pretty much obsolete and some schools (including mine) are replacing their stocks and you might be able to get a cheap copy of them. If not, just try any old maths textbooks from a reputable source.

As for internet resources, it tends to be difficult to find anything that's pitched at the right level and is comprehensive enough to be useful (plus accurate and correct is helpful too).

Differentiation is a bit too deep to describe briefly, but it is another way of expression the relationship between one or more variables and in finding various properties of curves and relationships between different variables.

Reply 3

john!!

ok, I dont generally do this

the idea behind all this calculus stuff is that locally everything is linear. if you look at the graph of y=x, of course it is. but if you look at y=x^2, or y=sin x, it looks like a curve. the thing is, you aren't looking close enough. LOCALLY, ie. if you look sufficiently close, you will find that the curve, at a point, has a definite gradient.

problem is, to find the gradient you need TWO points on the curve, and you then find the ratio to the difference in y values to the ratio of the difference in x values and you have your gradient. but when you have two points, it isnt local. so you gradually move those two points closer together, algebraically, and through the use of some new notation, you can make the two points touch to find the actual instantaneous gradient. Of course, if you just put the numbers in, you'll get (0-0)/(0-0) which will make your calculator keel over and die. But by using the concept of limits, you can find the LIMIT, as the change in the x axis TENDS to zero. here's how.

take any function f(x) (thats continuous so things work). chose the two points to be:

point one: (x, f(x)) (these are the coordinates on the curve)
point two: (x+h, f(x+h)) (and the y coordinate should clearly just be f(...) of the x one)

now the gradient is just the change in y over the change in x:

gradient of line connecting the two points = [f(x+h) - f(x)]/[(x+h)-(x)] = [f(x+h) - f(x)]/[h]

naturally you'll want to say, "well, we want to look at it locally, so let's just make h zero... nope, you get 0/0 again. but here's where we use limits. The precise defenition of a limit uses all kinda of dangerous greek symbols, you need to know it is what the function tends to as x tends to something. So the limit of 1/x as x tends to infinity is 0, for example. The gradient of a function f(x) at a point (a,f(a)) on the curve, is defined as:

f'(x) = limit(as h tends to 0) of [f(x+h) - f(x)]/[h]

Take this example. f(x) = x^2

f'(x) = limit(as h tends to 0) of [(x+h)^2 - (x)^2]/[h]
f'(x) = limit(as h tends to 0) of [2hx + h^2]/[h]
f'(x) = limit(as h tends to 0) of 2x + h

and now you can safely plug in h=0 to get f'(x) = 2x. so the gradient of the curve y=x^2 at any point is 2x. neat huh?

physics is more practical than that though...

although I used a ' in f'(x) , generally if you say y = x^2, then you say dy/dx= 2x. "dy" and "dx" are annoying things called differentials (I think), but just remember dy/dx represents the derivative of y. so say you have a boy whose mass varies with time like m = 10sin(t) + 70 kg. freak. but then a doctor asks you "hey, I can only give him his freak pills when his mass is falling at a rate of more than 6kg per second, when's the next time I can feed him?". What do you do? differentiate to find the instantaneous rate of change of mass...

m = 10sin(t) + 70 kg
dm/dt = 10cos(t) > 6
cos(t) > 6/10
t = whatever intervals. (you just learn that sin(t) differentiates to cos(t), convince yourself by looking at a graph).

or say the same rubbish boy manages to take his pills, and now has a mass like m = 50 ln(t) + 30, and he is put in a gravitational field, acceleration g. What is the force on him?

a = g,
a = dv/dt
what differentiates to make g?
v = gt (youll learn rules too, theyre boring and I think the exam board misses the point)

but m = 50 ln(t) + 30
so the momentum, p = mv = 50gt ln(t) + 30gt

And now the force is d(mv)/dt = 50g + 50g ln(t) + 30g = 10g(8 + 5 ln(t)) N.

see, when the problems get harder, it makes life kind of easier when you understand it. Anyway, read a textbook or something. I didnt think a level physics had any differentiation in it (again an example of an exam board missing the point)

--------------

can someone please read that last physics example and convince me that I'm not totally wrong? I re-read it and it looks like utter garbage...

Reply 4

mysteryman

Thanks alot. If anyones wondering i'm on salters horners and our teacher said it would help if we understood differentiation for the simple harmonic motion parts.

Differentiation

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