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

Last night my cousin asked me to help him with their assignment in differential calculus, given by Maila.

As long as they are problems from a book, and the book is also lent to me, I find the questions easy. I just look up the previous examples in the earlier part of the chapter and "reverse engineer" the process (if it can be called that) from there. Besides, I vaguely recalled answering those types of problems way back in college more than a decade ago.

There were three items in their assignment. The first two were direct applications of the solved problems in the preceding pages.

These were the ones where an applied triangle, like a ladder leaning against a wall, is falling and the top of the ladder is falling to the ground at a certain rate, as well as the bottom of the ladder is receding from the wall. The question is usually at what rate either the top or the bottom of the triangle is moving, particularly when the length of one side is a certain value.

For these problems what is needed is to get the first derivative of the equation (usually Pythagorean) to have a variable or two for the rate, and compute for one of the unknown variables using the original equation so that there is just one unknown, typically one of the rates.

The third problem though did not have one but two triangles. A man with a set height was walking away from a lamppost, also with a set height. The man was walking a certain speed away from the pole. What was being asked for was the rate the tip of the man's shadow was retreating from the pole.

Even though the two triangles had the same angles, there were still two sets of sides involved, only one for each (two total) which are constant. And two additional values given, did not have anything to do with either triangle: the distance of the man from the post, and the rate at which the man walked away from the post.

So the dilemma was to come up with a working equation that used only four two variables: x (the distance of the tip of the shadow from the post) and w (the distance of the man from the post), where x - w = s, the length of the shadow. This was so that dx/dt and dw/dt could be derived. The former is what's asked for; the latter is given.

This meant using not just the Pythagorean equations for both triangles, but also the relationship of certain sides given they are co-angular.

When the differentiated equation was found, x still had to be substituted with an expression using w, because there is no value for x.

My cousin gave me the problem at half past six. It was half past nine when I returned the paper to him.

I guess it helped that the answer (but not the solution) was given at the back of the book, which is usually the case for odd-numbered problems. It made the question more like a verification or proving type of problem than anything.

But I found out today that not even the girl who consistently gets the highest in my mechanics class was able to solve that problem. Hopefully Maila would just make that problem extra credit.

But damn it was exhilarating to be filling papers with equations again, exploring uncertain territories, unlike what I do in class.