Food unlimited

This is not so much about math this time.

Four months ago, I spent some time in Infinite Series which is a common topic in Calculus books. Consider this series:


The first term is 1, the second is one-half, then one-fourth and so on. The next term is always one-half of the previous term. A mathematician can tell us that this never ends. As the series progresses, the terms become smaller and smaller but not exactly zero at the end.

This time, consider the poverty alleviation and food sufficiency program of the Philippines. If Mang Pandoy is given a kilogram of rice, by the logic of this infinite series, he may have unlimited food forever! All he has to do is not to eat all the one kilogram rice at once. Following the infinite series, he has to eat only ½ kilogram rice during the first day. Therefore, at the end of first day, he still has the remaining ½ kilogram rice. On the second day, he has to eat, again only half of what has remained after the first day which is one-half. In doing so, he will save ¼ kilogram of rice at the end of the second day. Remember, he ate only half of ½ kilogram rice. By just eating half of what remains the previous day, he can't run out of rice forever!

Food in every table looks like attainable after all! Even better, make that unlimited food for all.

Isn't this concept worth presenting to the Department of Agriculture and Malacanang?

Ok, back to reality.

Now, you know what can happen to the thoughts of a student who has more of math and less of a meal.

Root of the problem

Find the mass and center of mass of the solid E with the given density function .

This is one of the interesting problems because the conventional first step is to visualize the problem and define the boundaries which will be used later for integration. My instinct is to eliminate any radical sign and express the condition in a more standard form. By squaring equations with radical signs, we can normally see these equations transform into a clearer visual representation of parabola, circle or sphere on some axes.

Although that routine works well most of the time, I found out that it can be misleading also sometimes. The bounding curve for instance in the problem becomes a regular parabola when squared but when left alone, which is best, it shows precisely what it wants: only the curve along the positive y-axis or the positive root of x.

Without heeding this clue, it is a headache to point out the target region in this otherwise simple problem. So, the problem basically hinges on the root of x.

...

Relative work force

A local farm boy (let's call him Bart Tesorero, Jr.) works for his uncle in a corn farm. He was asked one time to bring a 30-lb can of paint up a helical staircase that encircles a silo with a radius of 15 ft. If he is 150 lbs, and he makes exactly three complete revolutions to reach the top of the 60-ft high silo, how much work is done by Bart against gravity in his climb?

This is one problem given in a section for line integrals in Stewart's Calculus. The section teaches one how to calculate, among others work done on a curved path. So, this helical staircase in the problem immediately throws the student into the path of parametric equations to represent that upward winding path. These are typically used to represent the curved path:

The line integral for this work is:

where F is the total force required to push the weight of Bart and the can of paint. Simply put, this must be equal to at least that combined weight.

The line integral is reduced to:

because integration is done vertically or against gravity. So the change in left and right directions with respect to the height is zero because the horizontal and vertical equations have nothing to do with each other. They are independent.

From, an elaborate equation, the work crumbles into an elementary problem.

And an elementary problem it is. That it is served on a calculus book gives it a mystic complexity. But then, here comes an elementary student who doesn't even know basic calculus but good enough to remember that work is simply force times displacement. He gets the total weight as force, and multiply it with the height of the silo as the vertical displacement. He finds the answer faster than a calculus student who has to go through parametric equation first, setting up the line integral, differentiating and integration.

Now that an elementary student can solve this only makes this annoying to calculus students.

It is noteworthy that work here is the same no matter the diameter of the silo is. Or, regardless of how many turns Bart makes towards the top, he will not be “overworked”. Work depends on a directed distance so his movement from one side of the silo onto another side cancels each other. Hence, the work here relative to the horizontal plane becomes irrelevant because the problem asks for the work done upwards.

If the problem asks for power, which now takes the time factor, then the diameter and number of turns around the silo are now important factors. It will surely take Bart longer to climb a wider silo with more winding staircase. Work itself is theoretical and almost meaningless in real world, as far as I'm concerned. Power is more relevant and practical because it is work relative to time.

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Answer: 10,800 ft-lb