Unformatted text preview: deliberately chosen very simple examples to illustrate the concepts. I have
avoided all the rigorous details under which calculations are done. The theory of
Real Analysis and Calculus deals with this. Rigour does not necessarily mean clarity
and ease of understanding.
At the Class XI level this should be the approach. This approach will provide
the confidence to deal with more difficult functions and the motivation to
study the theory.
The notes are intended as thought evoking pointers for students who wish to
study "A Littile More Calculus " by the author. ii Throughout the book we make use of one main example, the projection of a ball, to
illustrate all the concepts and demonstrate all the calculations. Non-science students
may not be familiar with the projectile equation. So a brief explanation follows.
An object may be projected into space with an initial velocity u and an angle of
projection . Let T be the time the object takes to go up and come down. (0, 0) POSITION (2-Dimension) SPEED (2-Dimension) u u u.sin Height y(t) y Range x(t) x (0, 0) u.cos The initial speed u has two components: a ver tical component u . s i n and a
horizontal component u.cos . We know that gravity g acts in the downward or
negative direction. So, with the appropriate units of measure approach in mind,
we may see that the ver tical speed must be u . s i n -- g t [ meters/sec].
+ u.sin VERTICAL SPEED (1-Dimension)
u . s i n -- g t [ meters/sec] (0,0) T T/ 2 u.sin
iii time ertical
Since DISTANCE = SPEED X TIME, the ver tical position or height must be of the
form A1.(u.sin θ ).t -- A2 g t [meters]. The coefficients A1 and A2 may be
determined Analytically (as we shall see in integration) or by physical experiments.
It turns out that A1= 1 and A2 = 1/ 2 . So the function or expression that describes
the ver tical position or height is u . s i n θ . t -- 1 g t 2 .
-- 2 Height y(t) VERTICAL POSITION (1-Dimension) (0, 0) y(t) = u . s i n θ . t -- 1- g t 2
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- Fall '09
- Limit, Δx