Student "edition" found at [csi dot journalspace dot com].

Maybe I shouldn't have started this blog now, not with everything that's been going on.

In my Mathematical Methods One class for the first day of the sixth week of classes, I decided to skip the inequalities part of the third chapter, as well as graphing of equations of more than one degree in two variables.

Instead I borrowed an overhead projector to display a two-dimensional grid on the white board so that I can take up graphing of equations of a line. The projector could not be reserved for the whole week; so it was a good thing I only needed it for the first day of the lecture on the chapter.

We started with the easiest, which is given two points that are on the line. They just have to plot these and draw a line between them.

Then from being given just the slope and a point, they used the denominator of the slope for horizontal displacement and the numerator for vertical displacement to get the second point. From there they were to proceed as in the first case.

Next was the slope-intercept method, where the intercept is actually just another point on either the x or the y-axis. Then this could be treated as point-slope.

The relatively most difficult method, I told them, was when the equation of the line in the form Ax + By + C = 0 was given. Then they had to substitute values for x and y to come up with points that are on the line. The easiest, I recommended, was zero for either, followed by one, then two and so on. Then they would end up with two points again.

For coming up with the equation of the line, the first three methods are now reversed in order of relative ease. Slope-intercept is easiest: just substitute it in the equation y = mx + b, then convert it to standard form. For point-slope, the values of x and y are just substituted in the same equation with the slope to get a value for b. Lastly, with two points, the slope is first computed using the equation (y2-y1)/(x2-x1), then they proceed just like with point-slope.

In DIFEREQ, I gave them an exercise at the start of the class: twenty equations that they had to determine as homogeneous or not, and to what degree. This was because a lot of students were late, probably “sharing knowledge” as to how to answer their problem sets.

The seatwork was to be answered by pairs. There were an odd number of students in the room at that time, so there was one student with no partner. When the others arrived late, they were also in pairs, so he still did not have a partner, until the last student arrived alone.

It took longer than expected though, up to the end of the period, so the next example I wrote on the board I said would be discussed next meeting.

Session 630 concludes here. For now, class dismissed.