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:
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.
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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