## Friday, June 24, 2011

### Khan Academy: "Kinematic equations" exercise is done!

Being sentimental about my school years and following a habit of picking challenges I am certain of being able to handle I vowed to beat Khan Academy's Math exercise course month ago and I'm nearly done with it.

While most of presented problems are supposed to be solved with pen-and-paper toolkit, some explicitly allow use of a calculator. Which I shamelessly abused while solving "Kinematic equations".

8 equations are necessary and sufficient to solve 80% of this exercise - a statement I have formally proven!

If  we examine the problem we can see that there are 5 parameters (or variables) in it: {d, vi, vf, a, t}, 3 of them are known (which 3 of them - varies) and 2 are unknown. By using kinematic formulas we are able to discover unknown parameters from those given to us. Essentially, every formula is a function of 3 arguments: f(x1, x2, x3). That means we need to know certain 3 parameters to discover the new one.

Now, if we adopt a term "state" to denote our progress in solving a problem (i.e. how many parameters we know), we can see that there is 10 different initial states in total - 10 different starting points to play this game.
C(5, 3) = 5! / (3! * (5-3)!) = 10, see wiki:Combination for more details.

Since every formula gives us knowledge of one parameter and there are 2 unknown, we must apply 2 formulae to find out all the unknown. Look at the following diagram I made for you:
On the left side you see 10 initial states (every little circle marks a known parameter), on the right-most of the picture you see the final state, where all 5 parameters are known. Black arrows designate formulas used to transition between states (not all possible paths are shown).

You can also see there are two steps in solving this problem: step1 transitions from one of the initial states to one of the intermediate states, step2 goes further and delivers us to the final state. Each step is an application of appropriate formula (formula is chosen based on particular state, i.e. given set of known parameters).

Now one may ask - "What is the deal with those 2 pink states?". Good question, my observant reader. The problem with initial states {d,vi,a} and {d,vf,a} is that there is no function to discover "t" parameter. You see, in a formula "d = vi*t + a*t^2/2" even when you know {d, vi, a} you need to solve quadratic expression to find out "t" and generally speaking there will be 2 concurrent answer, 2 different values of "t" will fit. Since that duality of an answer there is no function to find t from {d,vi,a}, i.e. you cannot build "t = t(d, vi, a)" function. Same applies to {d,vf,a}. Since the whole approach in this entry is focused around functions, these two pink initial states are exceptions to the solution. We are gonna pretend they are not here, OK? :D

So, now we have 8 initial states.
Important note: solving the problem is equivalent to going from initial state to final state, therefore for a set of functions to be able to solve the problem is necessary and sufficient to be able to perform step1 and step2.

Do you see what i did there? I rephrased a task of "find unknown parameters" into a task of "find a set of functions that goes through step1 and step2".

As mentioned before, every formula is a function of 3 parameters that gives us one unknown parameter. This means every formula can start at precisely one initial state and transitions to intermediate state. And since we have 8 initial states we need at least 8 different functions with different argument sets. Otherwise some initial states would have no 'doorway' to intermediate state.

This is a condition1: "8 functions with different argument sets each corresponding to 8 initial sets". It is necessary and sufficient to make step1.

After step1 is made there is only 1 parameter unknown - therefore there are 5 intermediate states (since there are 5 parameters total). And since any of the parameters can be the last one unknown we need to be sure, that our set of functions have a function to find every parameter there is. I.e. we have functions d=d(x1, x2, x3), vi=vi(x4,x5,x6).... etc. Arguments of these functions don't matter.

This is a condition2: "5 functions of this set must return each of 5 parameters". It will allow step2.

And here we go - I deducted 2 boundary conditions. If a set of functions is within those boundary conditions it is qualified to solve 80% (8 of 10 initial positions, remember) of "Kinematic equations" problem.