Monday, 16 February 2009

Integration, Differentiation

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.

3 comments:

Anonymous said...

I wish I had taken Calculus...but there wasn't room in my schedule - what with the basket weaving classes and paint by numbers and all that...

I got high marks in Legos class...

:)

Steven Carr said...

Think of a speed-time graph.

The rate of change of the speed is the acceleration ie dv/dt = how rapidly speed is changing.

But the area under a speed-time graph is the distance travelled ie integrate the speed to get the distance.

Bill Haydon said...

Good point. That makes it much more relevant, in a way.