Hi PG,
You can use the two point form of a line:
y=(y2-y1)/(x2-x1)*x+b
with x=t.
Since in the first segment we have
x1=0,y1=0, and x2=1, y2=50, we get:
y=(50-0)/(1-0)*x+b
or
y=50/1*x+b
which is
y=50*x+b
then solve for b with either x1 and y1 or with x2 and y2. Using x1 and y1 we get:
b=y-50*x
b=0-50*0
b=0
or with x2 and y2 we get:
b=50-50*1
b=50-50
b=0 again.
For the segment on the interval x=1 and x=3 we have:
x1=1, y1=50
x2=3, y2=-50
so we have:
y=(y2-y1)/(x2-x1)*x+b
which is:
y=((-50)-50)/(3-1)*x+b
or:
y=-100/2*x+b
or
y=-50*x+b
then solving for b:
b=y+50*x
and with x1 and y1 we have:
b=50+50*1
or
b=100
So we have:
y=-50*x+100
for the interval between x=1 and x=3.
The points at the tips are not defined for derivatives because the function
is not continuous there. The function is not continuous there so it gets a
little strange because the same point is defined two different ways if you use
equality signs rather than inequality signs.
Note that we were however able to use those points to find the equations for
the multiple lines. We cant take the derivatives there though.
Yes you are correct
The square wave is not really drawn according to pure math. It's just an approximate representation. The more perfect representation looks like this:
Code:
o-------o o-------o
o-------o o--------o o--------
Note the little circles at the end of each line segment to show the discontinuity there.