Wednesday, January 20, 2010

Calculus Makes It Simpler. Really.

Zach's been giving me crap about using calculus to solve a problem that can be handled with algebra. I'd argue that solving it with algebra requires knowing formulas derived using calculus but he does have a point. Calculus is kind of like the hammer drill of math - it can accomplish tasks other tools can't, but it's a little silly to do drywall with it.

The problem is a proof of an equation to find the acceleration of gravity based on an experiment in which an object is thrown or fired straight up in a tube that's been evacuated of air. That's important because it means that the only force that should act on the object is its own weight, so it should accelerate at exactly the acceleration of gravity. The formula is given below. The problem is to prove it using only the distance between a high and low mark in the tube and the time between crossings of the low mark and crossings of the high mark.


The first thing I did was draw a coordinate system on the original problem.


I made the origin the time at which the object first crosses the lower line. Then the upper line is at height 'H', simplifying the math some. I also figured out pretty quickly that using parts of Δt would get old really fast, so I labeled the points t1 t2, t3 and t4. Then I spent the next couple days banging my head against it and trying to figure out velocity at various points. I eventually realized that I was looking at the wrong side of the curve by starting at the origin and working to the right. I know that the velocity at the point t2 is 0. Since the only force acting on the object is gravity, the velocity v(t) becomes a nice, clean function starting at t2. v(t3) is easily calculated based on the difference in time from t2 to t3 and the acceleration of gravity. v(t4) is easily calculated the same way, and that's when I had my flash of inspiration - I don't need velocity at all!

Because there's no initial velocity to worry about if I ignore everything to the left of t2, which should be a mirror image of everything to the right because this is a constant acceleration problem, if I start at t2 and right down everything I know, I can easily solve the problem. I know that the ball travels the distance H during the time from t3 to t4, and I know the equation describing its vertical position as a function of time. I actually solved the problem using the coordinate system above, but I'm going to give the solution based on the system below because it's even prettier.


If the origin is right in the middle of the graph, v(0)=0, which is great because now I can do a bunch of stuff without having to add a constant or do weirder math than I really need to. Also, because the graph is an even function that's symmetrical about t=0 when it's set up this way, I know that the point at which the ball crosses H, previously labeled t3, is ΔtU/2 and the point at which it crosses the lower mark, at height 0, is ΔtL/2. Now I can set it up as an integral.


You'll notice that I ended up getting the sign backwards at the end of the problem. The reason this happened is that I was looking at up as the positive direction. If the acceleration of gravity is seen as a constant with a positive value, down needs to be the positive direction.

1 comment:

Zach said...

Actually, that is a little more elegant.


Nerd.