This is quite tricky to get your head round, I think.
Differentiation is what you do to a curve to obtain its gradient. It works by splitting the curve into infinitesimally tiny chunks and tending towards a limit (ie where you started from). I think I put the proof of it on an earlier post.
Basically it gives you the rate of change in the curve as a formula. Because curves are always changing, the gradient will be different at each point. You plug the x-coordinate into the differentiated gradient formula and there's your gradient at x=whatever.
But if you integrate that curve you get the area under the curve. It works in a similar way to differentiation: it takes the space under the curve and chops it up into tiny rectangles to approximate the area.
It seems to work out that in terms of a formula, integration is the opposite of differentiation (ie raise the power by one and divide by the value of the new power, instead of dropping the power by one and timesing by the new power).
The curve bounding an area with the x-axis is now the value of the area formula differentiated - ie it is the rate of change of the area itself.
Differentiation and integration are inverse processes: that's the theorem of calculus (which means stones used for counting, or something).
If you have a derivative, you can integrate it to get the indefinite integral - which you can then find the full equation for if you know a set of co-ordinates. So by knowing the derivative, you can find the equation of that curve.
If you have a derivative, often you can differentiate again to find a stationary point on the curve - the second derivative test.
I'm still not sure I get this, you know. I still find the connection between the two loose and hard to actually articulate. I can do it with the formulae no problem at all - but understanding the linkage and how rates of change relate to curves is much harder.
Showing posts with label differentiation. Show all posts
Showing posts with label differentiation. Show all posts
Monday, 16 February 2009
Thursday, 22 January 2009
Indices And Differentiation
Yesterday, courtesy of my beautiful and awesomely intelligent maths tutor, I suddenly worked out a way of making differentiation with negative indices slightly easier.
When inputting the x value to find the gradient at that point, to make the final stage of adding it all up much easier, turn your x-2 into 1/x2. It makes calculations a lot, lot easier.
When inputting the x value to find the gradient at that point, to make the final stage of adding it all up much easier, turn your x-2 into 1/x2. It makes calculations a lot, lot easier.
Sunday, 10 August 2008
The dy/dx Post, Or, Differentiation for Dunderheads
Bear in mind that I know nothing at all beyond MPC1: so this post will only cover the basic basics of differentiation because that is all I know about it. Also PLEASE correct the inevitable errors. I'm not going to do the whole equation from first principles because that would just be confusing.
To kick off, a quote from Tom Freeman over at Matt's place:
In the first term, we started on calculus. I looked at the dy/dx stuff the teacher was drawing on the whiteboard, and I said...
"Don't the 'd's cancel?"
The simple answer is no. This is because when you see dy/dx you are not looking at a fraction, but a statement: the change in y with respect to the change in x. Tom knows this of course, but I didn't until a few weeks ago so it's worth pointing out!
Let us assume you have a curve. You want to find its gradient. But you can't because it's constantly changing, and besides, y2-y1/x2-x1 that you used for straight lines doesn't quite seem to cut it, what with all that mucking about going up and down and stuff.
Instead you approximate it by picking point A, which is where you want to know the gradient of the curve. Then you pick another point, B which is some distance off from A on the curve. So the coordinates of A are (x,y) and of B, (x+n, y+p) where n and p are the differences between the two sets of x and y co-ordinates. If you've started at A, and gone on to B, B's coordinates will be a + the increase, and y + the increase.
You draw a line to join A and B, a chord. You can measure the gradient of the chord, obviously (y2-y1/x2-x1). It's a bit ineaxct but it gives an approximation of the gradient of the curve between the two points and so you're nearish to the gradient at A.
But if you think about it, it would be a closer approximation if the line was closer to A...and then closer still and then closer and then closer...
Eventually you'd have such a small line that the gradient of it would be the gradient at point A plus not very much at all, ie a really good approximation of it.
So all along your gradient has been, say, 4 + f, where f is the increase in the gradient, which gets smaller and smaller as you get closer to point A. At point A your gradient is 4 (no increase). This is also the gradient of the tangent at point A.
This is differentiation! Chopping the distance between two points on a curve into smaller and smaller bits until you get invisibly close to point A, to get the gradient at Point A. It's a pain in the neck though to do this all the time, or to draw an infinite amount of tangents with gradients of their own, so we have dy/dx to do it for us.
This is the essence of the formula they give you at A Level. I said I wasn't going to do the proof, because it is a bit confusing; but it's worth looking up because it helps you understand why you are doing this stuff.
What stuff?
Well, say you have the curve y=x2. Dy/dx, the difference in y with respect to the difference in x (ie the gradient, for that is what it means), is 2x.
Reduce the power by one, and put it in front of the x. Piece of the proverbial.
What about the curve y=3x2?
Multiply the power by the number in front of the x, (6), then reduce the power as before. 6x. 6x is the gradient of the curve y=3x2.
Easy!!
But what if there's no power? And you've just got 5x? Well 5x is really 5x1. So you bring the power down as before and multiply, which makes 5x0. Stuff to the 0 is generally 1. So it's 5 times 1, so it's just 5.
If there's no power of the x, get rid of the x!! And you're done!
What if it's a constant? ie 8? Well constants always differentiate to make 0. Think of 8 as 8x0. Bring the 0 down and x it by 8, ie 0, and you get 0x-1. Ie NOWT!
But this is still a gradient formula, not the gradient itself. Well, this changes at every point on the curve, obviously. So you input the x co-ordinate of the point you want. What is the gradient at (2,3) on the curve y=3x2? Well dy/dx gives us 6x, and so put your x co-ordinate into this. 6 x 2 = 12. The gradient at (2,3) is 12.
I'll leave this for now on the question of what to do with a polynomial. The curve y= 3x4 - 2x3 + 6x2 - 3x + 4
you just differentiate bit by bit:
dy/dx = 12x3 - 6x2 + 12x - 3
Does anyone know how to do superscript in blogger? It would really help with these squares!!! (UPDATE: Yay! I've worked it out!)
Here's the proof for anyone who cares. It's come out a bit small but it's here! Click if you're bothered.
To kick off, a quote from Tom Freeman over at Matt's place:
In the first term, we started on calculus. I looked at the dy/dx stuff the teacher was drawing on the whiteboard, and I said...
"Don't the 'd's cancel?"
The simple answer is no. This is because when you see dy/dx you are not looking at a fraction, but a statement: the change in y with respect to the change in x. Tom knows this of course, but I didn't until a few weeks ago so it's worth pointing out!
Let us assume you have a curve. You want to find its gradient. But you can't because it's constantly changing, and besides, y2-y1/x2-x1 that you used for straight lines doesn't quite seem to cut it, what with all that mucking about going up and down and stuff.
Instead you approximate it by picking point A, which is where you want to know the gradient of the curve. Then you pick another point, B which is some distance off from A on the curve. So the coordinates of A are (x,y) and of B, (x+n, y+p) where n and p are the differences between the two sets of x and y co-ordinates. If you've started at A, and gone on to B, B's coordinates will be a + the increase, and y + the increase.
You draw a line to join A and B, a chord. You can measure the gradient of the chord, obviously (y2-y1/x2-x1). It's a bit ineaxct but it gives an approximation of the gradient of the curve between the two points and so you're nearish to the gradient at A.
But if you think about it, it would be a closer approximation if the line was closer to A...and then closer still and then closer and then closer...
Eventually you'd have such a small line that the gradient of it would be the gradient at point A plus not very much at all, ie a really good approximation of it.
So all along your gradient has been, say, 4 + f, where f is the increase in the gradient, which gets smaller and smaller as you get closer to point A. At point A your gradient is 4 (no increase). This is also the gradient of the tangent at point A.
This is differentiation! Chopping the distance between two points on a curve into smaller and smaller bits until you get invisibly close to point A, to get the gradient at Point A. It's a pain in the neck though to do this all the time, or to draw an infinite amount of tangents with gradients of their own, so we have dy/dx to do it for us.
This is the essence of the formula they give you at A Level. I said I wasn't going to do the proof, because it is a bit confusing; but it's worth looking up because it helps you understand why you are doing this stuff.
What stuff?
Well, say you have the curve y=x2. Dy/dx, the difference in y with respect to the difference in x (ie the gradient, for that is what it means), is 2x.
Reduce the power by one, and put it in front of the x. Piece of the proverbial.
What about the curve y=3x2?
Multiply the power by the number in front of the x, (6), then reduce the power as before. 6x. 6x is the gradient of the curve y=3x2.
Easy!!
But what if there's no power? And you've just got 5x? Well 5x is really 5x1. So you bring the power down as before and multiply, which makes 5x0. Stuff to the 0 is generally 1. So it's 5 times 1, so it's just 5.
If there's no power of the x, get rid of the x!! And you're done!
What if it's a constant? ie 8? Well constants always differentiate to make 0. Think of 8 as 8x0. Bring the 0 down and x it by 8, ie 0, and you get 0x-1. Ie NOWT!
But this is still a gradient formula, not the gradient itself. Well, this changes at every point on the curve, obviously. So you input the x co-ordinate of the point you want. What is the gradient at (2,3) on the curve y=3x2? Well dy/dx gives us 6x, and so put your x co-ordinate into this. 6 x 2 = 12. The gradient at (2,3) is 12.
I'll leave this for now on the question of what to do with a polynomial. The curve y= 3x4 - 2x3 + 6x2 - 3x + 4
you just differentiate bit by bit:
dy/dx = 12x3 - 6x2 + 12x - 3
Does anyone know how to do superscript in blogger? It would really help with these squares!!! (UPDATE: Yay! I've worked it out!)
Here's the proof for anyone who cares. It's come out a bit small but it's here! Click if you're bothered.
Wednesday, 6 August 2008
Graphic
Yes I know the clumsy graphic I've chosen is nothing to do with completing the square as such...it is, of course differentiating the curve of y=f(x) (though it'll be a -x2 curve) from first principles, demonstrating how you find the gradient at A by steadily chopping smaller and smaller slices of the distance between A and B until the difference tends to 0 (ie at A). You've therefore used straight line gradients to find a pretty good approximation of the gradient at A (and therefore at any point on the curve).
Oooh, I think I am going to have to be careful here..I don't feel quite so....dogmatic as usual.
UPDATE: actually, looking at it, it seems to be a sort of cubic graph thingy in a positive x stylee. OR it might be something else. This doesn't really affect anything, you can still take dy/dx to find the gradient at any point.
I really do feel on eggshells here. Wow, this is weird.
Oooh, I think I am going to have to be careful here..I don't feel quite so....dogmatic as usual.
UPDATE: actually, looking at it, it seems to be a sort of cubic graph thingy in a positive x stylee. OR it might be something else. This doesn't really affect anything, you can still take dy/dx to find the gradient at any point.
I really do feel on eggshells here. Wow, this is weird.
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