Is it cold outside? Why, it's positively polar.

01/31/2012

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It's time for another math blog!  This one shouldn't take too long, but I need to get it written before I can post about other things, so here it goes:  Polar Coordinates!  Up until now we've dealt exclusively with Cartesian coordinates when plotting points on a graph (Cartesian coordinates are the ones you learn in high school geometry), and they look like this:
Pretty simple.  (3,2) accurately describes this point because its value on the x-axis is 3, and its value on the y-axis is 2.  Now, if this were a map and I was telling you how to get to (3,2) from (0,0), I'd tell you to walk 3 units east and 2 units north.  That's essentially what Cartesian coordinates do.  Polar coordinates, on the other hand tell you how to get to a point as the crow flies, to continue the map comparison.  Thus:
So how do we define a set of coordinates that will direct us along our red line to end up at (3,2)?  Do get there from (0,0) as the crow flies, we need two pieces of information:  a heading, and a disatance.  And that's what polar coordinates are:  an angle, expressed in radians (we'll call it θ), and a distance, which is the length of our red line (we'll call this r, for radius), and it's written as (r,θ).

Pretty simple, yes?  By now it should be clear how to find these values given what we already know.  We'll start by finding c using the trusty Pythagorean theorem:

a^2 + b^2 = r^2
2^2 + 3^2 = r^2
4 + 9 = r^2
13 = r^2
√13 = r
So there's one of our two coordinates.  Let's find the value of the angle θ now, shall we?  Seems the easiest way, or at least the way that involves all integers, is to find the arctangent of 2/3.  Plunking tan^-1(2/3) into our trusty calculator in radian mode, we see that the arctangent of 2/3 is approximately 0.59 radians, which coverts nicely to 33.69°, so our polar coordinates would be expressed (√13,33.69°).  Thus:
Now, that was easy enough, but arctan only works in quadrants one and four (in other words, angles whose cosines are positive).  There's a whole slew of functions to allow for that, and to save all those penstrokes, it's usually expressed as atan2 (y,x).  It's a simple way to express ALL THIS:
Which is a complicated way to say "add or subtract 90 or 180 degrees as necessary to figure this out in quadrant one or four". 

Now, converting from polar coordinates to cartesian coordinates is pretty simple too, and I won't go through the process of proving these formulae, but here they are:
x=r cosθ
y=r sinθ
Let's work it out given our same problem from above.x=√13 cos(.59)We rounded that angle to the nearest hundredth, so our answer is a little off, but plugging that in a calculator does give us approximately 3.  Similarly,
y=√13 sin(.59)
Yields a decimal that's pretty close to 2.  So - our calculations were a success!  

That's all for now - soon we begin writing about pre-calc, so stay tuned for that.
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Tan to the negative one, it turns out, doesn't mean you're very pale.

01/12/2012

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I have returned!  Well, in truth, I never left.  I've just been working, studying, and otherwise not writing, and yesterday I realized that I've gotten so far ahead in studying math that it's going to take several posts to catch up!  That's not a bad thing - so I'm going to embark on a whirlwind tour of some of the things I've learned since our last meaningful time together in November (wow - has it been that long?), including some new trig concepts like polar coordinates, some neat problems I've run across, and even some brief forays into the mystical land of calculus!  This first post, though, will just tackle the idea of the Inverse Trig Functions.  Here we go!


Picture
Inverse Functions
So we know about sine, cosine, and tangent... and we even know about cosecant, secant, and cotangent, their multiplicative inverses.  So these inverse functions are the compositional inverse.  I'm talking about arcsin, arccosine, and arctangent - but if you've seen a graphing calculator, you'd probably recognize them by their more common notation: sin^-1, cos^-1, and tan^-1.  Don't let the negative one exponent fool you - this is simply how we denote that we're looking for the inverse function.  So what is that anyway?  Well, look at the triangle here.

Say we know the length of sides a and b but we don't know the value of the angle θ. 

Given what we already know, we can clearly say that tanθ = a / b,  but that doesn't really tell us what the value of θ is.  That's where the inverse functions come in.  See, where the tanθ = a / b, the arctan a / b = θ!  The same is true for the sine, cosine and all the rest.  Simply put, when you're trying to get an angle and you know the values of some sides, you're going to be looking for inverse functions.  Plus - they can be graphed just like our other functions can, but the results may surprise you.  Take a look at what happens when we graph arcsin(x):
The graph if sin(x) is in red - and the graph of arcsin(x) is in blue.  It does something pretty interesting, and if we look at the green vertical line we can see that unlike the sine graph, the arcsine graph is not a function.  

Now - think about this for a second and you may realize why this is the case.  Picture the value of arcsin(x) as the angle at the center of the unit circle (because that's what it is).  That value ranges, in each "quadrant" of the graph, between  π/2 radians and - π/2 radians - or, 90 degrees and -90 degrees.  Now picture those values rolled out as you continue around the circle.  Your angle may be 5π radians, but you're in the same place as if you had only gone  π radians - remember, one trip around the circle is 2π, so reduce in terms of that.  Also, see that the value of sin(π) is the same as the value of sin(0).  They're both zero.

In order to keep this from happening with inverse trig functions, and to actually make them functions, as the name implies, we have to define ranges for them.  If we define the range of arcsin(x) as [- π/x,  π/2] and the domain as [-1,1] we get something much nicer:
Plus, if we hit it with a vertical line test, we can see that yes, it indeed is a function now, and all is right with the world.  This is the part of trig where one has to start thinking a little more aesthetically about the way things work - but after a while, it just becomes intuitive.  There are a lot deeper implications to the inverse functions than I'm going to get into here, but this is a good introduction to the concept, I think.

That's all for now - more to come very soon, promise!
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