Hi,
Students often find it difficult to sketch with
GC curves that are defined parametrically.
The typical issues for students using TI-GC are in
deciding Tmin, Tmax and Tstep.
We'll consider the following example for our
discussion:
The curve C is defined parametrically by
x = (1 + t)^(2/3), y = ln (t^2), t <= -1.
Find the area of the region enclosed by C, the lines x =
0 and x = 1, and the x-axis.
First, we need to sketch the curve and identify the
region visually before we can arrive at a definite integral.
The lines x = 0 and x = 1 give us some indication of what
values to set for Tmin and Tmax.
When x = 0, t = -1 obviously.
When x = 1, (1 + t)^(2/3) = 1 from which t = -2 is
obtained after rejecting the other possibility t = 0, as we are
given that t <= -1 and we know that y = ln (t^2) is not
well-defined when t = 0.
So, we set Tmin = -2 and Tmax = -1.
Never assume that Tmin and Tmax are going to
take positive values, as many students tend to do so for the
sake of simplicity. School exams often expect approaches
contrary to students' thinking :P
What is Tstep, you may ask? It's the interval between two
consecutive T values. Say, we set Tstep to be 0.1, then we'll have
the GC to plot the curve for these values:
-2 (Tmin), -1.9, -1.8, -1.7, -1.6, -1.5, -1.4, -1.3,
-1.2, -1.1, -1 (Tmax).
We have to be careful to select a reasonable Tstep, or
we'll risk having an inaccurate graph. For example, if we were to
let Tstep be 0.5, then the GC will plot the curve inaccurately for
these values:
-2 (Tmin), -1.5, -1 (Tmax).
You may try and compare both graphs on your GC to see
what I mean. The bottomline is more points, better accuracy.
If one were to use a Tstep value of 0.01, he/she risks
precious time because the GC will run for some time to generate the
curve over many values of T:
-2 (Tmin), -1.99, -1.98, ..., -1.03, -1.02, -1.01, -1
(Tmax).
Another bottomline is more points, but more to a
resonable extent.
Thank you.
Cheers,
Wen Shih