Method of Increments?

[camdenreslink]

I am using this book: http://www.amazon.com/Calculus-Intuitive-Physical-Approach-Second/dp/0486404536/ref=sr_1_1?ie=UTF8&s=books&qid=1275951079&sr=1-1
Calculus: An Intuitive and Physical Approach by: Kline

I've already finished Calc I and II, but I'm brushing up on my skills because I don't feel entirely confident about them.

This is what the method says to do to find a derivative:

s = 10t^2 given t=3

s_{3} + \Delta s = 10(3 + \Delta t)^2

s_{3} + \Delta s = 90 + 60\Delta t + 10\Delta t^2
- (s_{3} = 90)


\frac{\Delta s}{\Delta t} = \frac{60 \Delta t + 10 \Delta t^2}{\Delta t}

\displaystyle{\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t

lim_{\Delta t \rightarrow 0} \displaystyle{\frac{\Delta s}{\Delta t}} = 60

I understand the method and have finished all of the practice problems in the book, but I'm having trouble linking this with the way I was taught to derive in my calculus class with

lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}

[Mark44]

I really hate his abuse of notation, using s3 for s(3). There seem to be a lot of steps missing, which might make it difficult to follow his logic.

For example, \Delta s = s(3 + \Delta t) - s(3) = 10(3 + \Delta t)2 - 90
= 90 + 60 \Delta t + 10 ( \Delta t)2 - 90
= 60 \Delta t + 10 ( \Delta t)2

Dividing by delta t gives you this, the same as above:
\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t

Finally, take the limit to get ds/dt.

Delta t in his exposition is the same as h in what you're used to. Hopefully, my explanation will help you understand the parallels between the two approaches.