WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about polar equations and functions.
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In the previous lesson, we introduced a new way to look at location in the plane, polar coordinates.
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By now, we have seen thousands of graphs (probably literally), where x and y are in relationship with each other.
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As the horizontal changes, the vertical changes somehow.
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Either x and y are both in an equation, and we graph the equation, and we figure out all of the solutions to that equation;
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we graph it; or y is a function of x, and we graph the function--as x changes, how will y respond?
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We can do the exact same thing with polar coordinates: the variables r and θ can be put in some sort of relationship,
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and then we can graph the resulting polar coordinates.
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Before you watch the lesson, make sure you watch the previous lesson (before watching this one).
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You really, really need to understand polar coordinates on their own.
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You have to be able to understand how polar coordinates make locations on the plane before any of this stuff is going to really make sense.
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So, if you are having difficulty with working through polar equations, but you don't have a really good understanding
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of polar coordinates yet, that is the thing--you really want to work on polar coordinates.
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Make sure you have watched the previous lesson before watching this one,
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and that you are comfortable with polar coordinates before you try to work on polar equations and functions.
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And if you can make sense of how polar coordinates work, polar equations and functions probably won't actually be that much harder.
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All right, let's get started.
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We can set up equations and functions using r and θ, exactly the same way as we did for x and y.
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Generally, θ is going to be the independent variable, like x was.
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Our x was allowed to change and vary around, and then r will be our dependent variable, in the same way that y was.
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So, x was allowed to change around, and y responded to x's changes.
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Similarly, here it is going to be θ that will be allowed to change around, and our distance r will respond to the angle that we are at.
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Here are some examples; we could have an equation r = 1 + 2cos(θ) (we plug in a θ, and it tells us an r),
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or a function r(θ) = 3sin(2θ), which is the same thing as...we plug in a θ, and it gives us what the r-value will be.
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So, while r could be the independent, and θ the dependent, such relationships are pretty uncommon in polar equations and functions.
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We normally think in terms of how length changes based on the angle--
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if we go to this angle, what length it will be at--and not the opposite way,
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just like when we are doing a rectangular graph, we normally think in terms of "for this horizontal location,
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what height will I be at? For this horizontal location, what height will I be at?" and not
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"For this height, what horizontal location will I be at?"
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We don't normally think of height and then horizontal; we think of horizontal, and then height,
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because x is the independent, and y is the dependent, just like here θ is the independent and r is the dependent.
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We always, always assume that θ is in radians.
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Whenever we are looking at our θ, it is assumed that θ is going to be in radians.
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That is numbers like π/2, 7π/4, decimal things, etc.
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It could be explicitly put in degrees, but that would be extremely rare.
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Always, always, always use radians, unless you are being told explicitly otherwise, that this thing is in degrees.
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It is extremely uncommon: you almost never see something like that.
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I can't even think of one time I have seen it; so just don't expect that to happen--expect to be using radians when you are working with polar equations and graphs.
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So, always think in terms of radians: you are plugging in radian values, and you are plotting with radian values on your angles.
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All right, how do we graph in polar coordinates?
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We graph polar equations and functions pretty much the same way that we graphed normal stuff.
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You plug in some value for your independent (in this case, θ).
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You plot the point that gets put out; and then you connect the whole thing with curves to make it into a graph.
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So, θ is the independent variable; we plug in some value for it, and then we see what distance r would get out.
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Over here, let's plot some points: if we are looking at the equation r = 1 + 2cos(θ), then r is going to come out once we plug in some θ.
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So, if we plug in 0, well, 1 + 2cos(0)...cos(0) is 1, so 2 times 1 is 2, and 1 + 2 is 3.
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So, we now have the point...not (0,3), but (3,0); that is one little confusing thing--
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the fact that it is not (x,y); it is now (r,θ); so our independent variable is actually the thing that comes second.
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So, don't let that confuse you.
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Our distance out is 3, and our θ is going to be an angle of 0.
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So, we end up getting this point right here; we have an angle of 0; we are at 0 above the starting location.
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We haven't moved at all, and we are out on the third circle out; so we are at (3,0).
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Next, π/4: if we plug in π/4, 1 + 2cos(π/4)...cosine of π/4 is √2/2; 2(√2/2) would be √2;
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so that means that we get 1 + √2 out of this.
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1 + √2...we figure that out with a calculator, so we can actually plot something down.
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That is approximately 2.41; so at this point, we are at angle π/4, so we are on this arc sector line right here.
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Notice, it is broken into eight pieces, so each one of them is going to be π/4, because up here is π/2.
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So, we are at the line of π/4 angle; and then we go out 2.41: (2.41,π/4) is the point that we get out of this.
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So, we are at 2.41 out, somewhere that looks around 2.4, a little more than 2, but even more less than 3.
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So, we get this point right here; OK.
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The same thing if we plug in π/2: if we plug in π/2, 1 + 2cos(π/2)...cosine of π/2 is 0, so we just get 1.
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So, we have the point (1,π/2); and that gets us this point here.
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So now, we think about how these things are going to connect through curves.
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If we are really confused about it, we could just plot down more points.
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We could put down π/6 and π/3, as well.
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And if we want even more points, we could continue to plot down more and more points.
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But we would probably be able to get a pretty good sense with just π/6 and π/3.
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But we can even just figure this out by thinking, "Well, how is it curving--what is happening?"
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So, as the angle goes up, notice that, because it is cosine of θ, as this goes to larger and larger values,
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0 to π/4 to π/2, cosine shrinks down and down.
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So, this portion of our equation, the 2cos(θ), will get smaller and smaller as cos(θ) gets smaller and smaller.
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So, as we get larger and larger, it shrinks down to ultimately this value of 1 right here.
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We go up, and as it gets farther out, it shrinks down; it shrinks down; it cuts through here; it shrinks down; it shrinks down.
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And we get something like that curve.
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We can see it precisely here, now drawn by a computer: r = 1 + 2cos(θ).
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At this point, we just keep plotting more points.
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Here is a new, interesting one to think about: if we plotted 3π/4, well, that is 1 + 2cos(3π/4).
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What does 3π/4 come out to be? That is 1 + 2(-√2/2), so that gets us 1 - √2.
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We approximate that with a calculator; we get -0.41, which means that we have the point (-0.41,3π/4).
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So, here is our 3π/4; it is right here; this is the 3π/4 location.
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We are on this line; but we have -0.41, so we are going in the opposite direction, and we are here for our point.
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We plug in π; 2 times cosine of π; what is cos(π)? It is -1.
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So, 2 times -1 get us -2; 1 - 2 gets us -1; so we have gotten the point (-1,π) out of this.
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So, at the angle of π (here is angle π), we are going to be going opposite the length of 1, and so we are here, as well.
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So, once again, it is continuing to drop down; this part here is continuing to get smaller and smaller, until eventually it becomes negative.
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It actually continues out to this curve here, and then it cuts through.
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It will always have to cut through the origin, because if it gets a 0 in the r, then it has to go through the origin,
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because distant 0 from the origin, distant 0 from the pole, means at the pole; cut through here.
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And then, curve up to here; and that is what we have so far.
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It is computer-drawn there, so it is a little more accurate; but we basically just keep plotting points.
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If we plug in 5π/4, we see that that is at -0.41; so at 5π/4, that would be this angle here.
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So, we are in the opposite now; we are over here at 3π/2; we have length 1, so at 3π/2, we are here; length is 1 in that direction.
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7π/4: we are at a length of approximately 2.41, so at 7π/4...we are at 2.41 from here.
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And we continue this curving; we can also see at this point, perhaps, that is going to end up being symmetric.
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If we looked at the entire table of values, we would see that there is some symmetry going on in the way these distances are coming out,
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and that we are going to end up seeing the top part...this curve so far will happen and just sort of flip over;
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and we will end up getting it out like this.
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And finally, here is a computer-drawn version that is better than my slightly-imperfect drawing.
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And that is what we end up getting out of this.
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You plot points; you work it out; you graph the whole thing.
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Just like graphing with rectangular equations, you don't need to plot a huge number of points.
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It won't hurt; if you plot more points, it is not going to hurt.
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But you really only have to plot enough so that you can sketch the graph.
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Which points should you plot--what are the useful points to plot--what are the interesting points to plot?
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Many polar equations involve trigonometric functions; the interesting points are when the trigonometric function produces a zero,
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when we plug some θ into that trigonometric function and it puts out 0,
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or when it puts out an extreme value (for sine and cosine, that is positive or negative 1).
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Plugging in those values for our θ lets us see what the most extreme values are that our function is going to make,
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which helps us have an understanding of how the whole thing is behaving.
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More points will just make it smoother and easier to make the curves.
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But those are the most important ones of all; so you definitely want to make sure that you are plotting those.
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If we want to try to plot these extreme values and these 0's, we want to figure out where these interesting values will occur.
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How are we going to get to it? Trigonometric functions tend to have a pattern.
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We are used to working with sine and cosine; so we want to think in terms of sine--
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we have to plug in 0 or π/2 or π; if it is sin(θ), we have to plug in 0 or π/2 or π or 3π/2...
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any of these would end up getting an interesting number.
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Sin(0) is 0; sin(π/2) is 1; sin(π) is 0; sin(3π/2) is -1; we are used to working with that from all of our work in trigonometry.
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If we go to something else, though, like cos(5θ), well, it is not just θ anymore; it is 5 times that.
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So, if it were cosine of something, we would want that something to be 0, and then π/2, and then π, and then 3π/2, and then 2π, and so on.
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But in this case, it is 5 times θ; so that has to be taken into account in how our thing works.
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So, the 0 is still just going to be the same; if we plug in a 0, 5 times 0 still gets a 0.
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But if we wanted to try to figure out π/2 = 5θ, what does our θ have to go in to get that interesting first value of π/2?
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We divide both sides by 5, and we would get π/10 = θ.
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π/10 = θ, because if we plug in π/10, 5 times π/10 is π/2; cos(π/2) gets us 0.
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So, that is our next interesting thing.
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Similarly, it is going to continue on this pattern of π/10 being the interval here.
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π/5 (I said that the wrong way...) 5 times π/5 would get us π; cos(π) is -1, an interesting value.
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The cosine of 5 times 3π/10 would get us cos(3π/2), which is 0, an interesting value.
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So, we are thinking in terms of what we have to plug in here, total, to get 0, π/2, π, 3π/2, 2π, the normal, interesting values.
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But then, we have to pay attention to the fact that it is not just θ; it is something interacting with θ.
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So, we have to pay attention to what the numbers are going to be.
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And that helps us figure out what numbers we want to plot in when we are trying to graph.
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Finally, what if it was 2θ + 1? In this case, θ = 0 won't even show up, because we have to figure out what would make this something.
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What would make 2θ + 1 turn into 0?
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Well, if we plug in -1/2 for θ, 2 times -1/2 gets us -1; -1 + 1 gets us 0.
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So, if we plug in -1/2 for our θ, we get out a 0 from it.
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What if we plug in θ = π/4 - 1/2?
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Well, then we have 2θ + 1; so 2(π/4) - 1/2 gets us π/2 - 1 + 1; so we get π/2 out of this.
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Let's work that one out: if we had 2θ + 1 = π/2, we would have to work out how to get to this.
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2θ = π/2 - 1...θ...if we are trying to get to this value of π/2,
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if we want this whole something here to come out to be π/2, the θ will have to end up being π/4 - 1/2.
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And that is where we are getting it.
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It is the same thing if we want this whole thing to be π; we end up needing π/2 - 1/2.
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We want the whole thing to be π/2; we need to plug in 3π/4 - 1/2 for our angle θ.
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Notice that, in each one of these, we are stepping up by π/4 each time.
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Over here, we were stepping up by π/5 each time.
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Once you start to notice the pattern, the pattern will normally continue.
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But it will depend on the specific circumstances of what is inside of your trigonometric function.
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Beyond looking for these interesting points, you might notice repetition in trigonometric functions leading to symmetries in the graph.
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Depending on the way the trigonometric functions are set up, you might notice that all of the same stuff is going to happen here.
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And because of the way that my angle is going to work, there is going to end up being a symmetry occurring in the graph.
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If you notice the symmetry, if you see that a symmetry will certainly occur, just use that to make graphing easier.
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You won't have to plot all of those points, because you will see
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that it is just going to end up doing effectively the same thing, but reversed or flipped in some way.
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If at any time you are unsure how the graph will behave, just plot more points; that is the easiest way to be sure of what is going to happen of all.
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If you are uncertain about what will come next, calculate a bunch of points.
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Just drop them in and connect the curve through those points.
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The more points you have down, the easier it will be to see how the curve works.
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After time, you will develop a sensibility; you will get an intuition for how these curves are going to come out--how they are going to look.
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You won't have to plot as many points.
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But when you are just starting out, if you are uncertain about a graph, plot more points, and you will have a good idea of where to go.
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All right, occasionally you will see an equation that only uses one variable, where there is just one variable there.
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A lot of students get scared; there is no reason to get scared.
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It is totally fine to have one variable; it just means that that variable that you said is fixed, and the other one can change freely.
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For example, if we had r = 2, then that means our distance is always 2.
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But θ--does θ show up in r = 2? θ doesn't show up at all.
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So, θ can be anything over here; in our red one over here, θ can be anything.
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So, if θ can be anything, then that means we set r = 2; so it is going to always be on this length of 2, this distance of 2 from the center.
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But our θ can end up going to any spin at all--it can be anything.
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So, as θ is allowed to spin positive or negative, it is going to always end up being stuck on the circle.
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So, we end up drawing this perfect red circle at a distance of 2 from the origin.
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A similar thing is going on over here with θ = π/3; if θ = π/3, we never mentioned r in there.
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We never mentioned the distance, so that means r can be anything.
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Since r can be anything, we have to be on the angle of π/3.
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But r could be any positive thing, so it could go out forever this way.
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And r could be any negative thing, so it could go out in the opposite direction.
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So, with θ = π/3, with θ at a fixed value, we end up creating a line.
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With r at a fixed value, like r = 2, we end up creating a circle, because either with r (we fix r) we have fixed a distance,
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but we are able to go to any angle (so we are making a circle); or if we fix an angle,
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and we are allowed to go to any distance, we have drawn a line.
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And that is what is going on if we just fix a single variable.
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Converting coordinate types: sometimes, we will want to convert an entire equation or function from polar to rectangular, or vice versa.
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We can do so with the same conversion formulas we figured out and used in the previous lesson.
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So, when we figured out these formulas, they were based off any x, y, r, and θ.
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We didn't say what x, y, r, and θ had to be; so this is going to work for converting the variables in equations.
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They can be used to convert from one variable type to the other variable, to convert from (x,y) to (r,θ) or (r,θ) to (x,y).
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So, previously we had x = rcos(θ), y = rsin(θ) for one set.
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And then, the other set of formulas was r² = x² + y²; tan(θ) = y/x.
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Pretty much any way that you can see these working out to allow you to swap variables around,
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so you can get to the kind of variables you want--go ahead and use it.
00:17:01.100 --> 00:17:05.900
These are the formulas that will allow you to convert from one type of equation to another type of equation,
00:17:05.900 --> 00:17:08.800
to switch from polar to rectangular or rectangular to polar.
00:17:08.800 --> 00:17:15.700
And if you ever forget any of these formulas, if you forget, you can re-derive them by drawing this picture right here,
00:17:15.700 --> 00:17:23.000
because we know that r is always the hypotenuse, and we know that x and y are the horizontal and vertical;
00:17:23.000 --> 00:17:26.300
and we know that θ has to be the angle inside of the triangle.
00:17:26.300 --> 00:17:31.200
So, we can figure out all of these equations: x = rcos(θ); y = rsin(θ);
00:17:31.200 --> 00:17:34.400
r² = x² + y², tan(θ) = y/x,
00:17:34.400 --> 00:17:38.200
just through basic trigonometry, because we know that (right down here) it is a right triangle.
00:17:38.200 --> 00:17:43.400
So, you can just use basic trigonometry and re-derive this if, in the middle of an important exam, they all disappear from your head.
00:17:43.400 --> 00:17:47.800
You can just draw a picture and do basic trigonometry; it is not too hard.
00:17:47.800 --> 00:17:54.600
All right, polar equations allow us to make really interesting graphs--that is to say, crazy, bizarre, cool, strange graphs--
00:17:54.600 --> 00:18:00.300
graphs that look nothing like the kind of graphs that we are used to with rectangular equations.
00:18:00.300 --> 00:18:03.100
They allow us to make really, really interesting stuff very easily.
00:18:03.100 --> 00:18:09.500
For example, look at this red one right here: could you imagine figuring out any way to graph that with rectangular x and y,
00:18:09.500 --> 00:18:13.700
to make an x and y equation that could make this flower-looking thing?
00:18:13.700 --> 00:18:21.100
That looks so unlike what we are used to graphing, but it only takes 2sin(6θ) + 0.5.
00:18:21.100 --> 00:18:25.100
We are able to say that incredible thing, where it has these large petals and these small petals,
00:18:25.100 --> 00:18:30.000
and a bunch of them repeating in this kind-of symmetrical pattern--with very, very little work.
00:18:30.000 --> 00:18:34.300
Very, very little writing is required to be able to make this incredibly detailed picture.
00:18:34.300 --> 00:18:40.500
The same thing over here in the blue picture: we are able to get this weird, squished thing that doesn't really look like anything specific.
00:18:40.500 --> 00:18:44.300
But it is a picture, and it is not very hard to write out.
00:18:44.300 --> 00:18:50.600
Once again, 2cos(θ) - sin(5θ)--we are able to get this very strange-looking picture,
00:18:50.600 --> 00:18:56.500
that there would be no easy way for us to create a graph with x and y coordinates that we would be used to using here.
00:18:56.500 --> 00:18:59.400
But in polar graphs, it is pretty easy to do.
00:18:59.400 --> 00:19:07.800
We can make really interesting things, things that we were really not used to seeing before, with not that long of an equation.
00:19:07.800 --> 00:19:11.500
Polar equations and functions are really a new way of thinking about graphing.
00:19:11.500 --> 00:19:15.000
As such, it is a great time to use a graphing calculator.
00:19:15.000 --> 00:19:19.400
This is the best time for graphing calculators: plot random equations in there.
00:19:19.400 --> 00:19:26.300
Alter equations that you already understand, and just get a sense for how polar stuff works by playing around with graphs on a graphing calculator.
00:19:26.300 --> 00:19:28.400
If you want more information, check out the appendix.
00:19:28.400 --> 00:19:32.600
There is an appendix to the course entirely on graphing calculators--how to use graphing calculators,
00:19:32.600 --> 00:19:36.600
what good graphing calculators are, and even if you don't own one, and you are not going to buy one--
00:19:36.600 --> 00:19:39.400
absolutely, for sure--there are free options out there.
00:19:39.400 --> 00:19:43.100
So, there is lots of cool stuff where you can go on the Web really quickly.
00:19:43.100 --> 00:19:47.900
And in five minutes, you can be graphing polar stuff; probably in one minute, you can be graphing polar stuff.
00:19:47.900 --> 00:19:52.200
And you would be able to get that without having to spend any money on a graphing calculator.
00:19:52.200 --> 00:19:58.700
And just playing around with this stuff is going to help you understand polar graphs massively.
00:19:58.700 --> 00:20:05.600
Trying different things, playing around, putting in things, looking at how changing one number here changed the whole thing--
00:20:05.600 --> 00:20:10.000
just being able to see that incredible speed of responsiveness of a graphing calculator,
00:20:10.000 --> 00:20:12.200
being able to change immediately when you do something...
00:20:12.200 --> 00:20:16.900
You don't have to take all of the time to plot it carefully, because that goes so slowly; it is hard to realize what is going on.
00:20:16.900 --> 00:20:20.300
But if you change one variable in a graphing calculator, and it creates a new graph,
00:20:20.300 --> 00:20:28.400
you will be able to gain this really beautiful intuition of how polar graphs work--how a polar equation creates the graph associated with it.
00:20:28.400 --> 00:20:32.900
I really, really, strongly recommend: if you have a graphing calculator, and even if you don't have a graphing calculator,
00:20:32.900 --> 00:20:36.600
check out the appendix; I will talk about lots of places where you can get free ones,
00:20:36.600 --> 00:20:39.300
where you can just go and play with them right now, immediately.
00:20:39.300 --> 00:20:43.700
And you will be able to get a really good understanding of how polar graphs work with not that much effort.
00:20:43.700 --> 00:20:50.600
Just playing around for 10 or 15 minutes will give you such a better understanding of polar graphs, if you find them difficult, even in the slightest.
00:20:50.600 --> 00:20:52.700
Also, I want to just point out one really important thing.
00:20:52.700 --> 00:20:56.500
When you are using a graphing calculator, pay attention to the interval that θ has.
00:20:56.500 --> 00:21:01.000
Normally, they are going to start your θ being from 0 to 2π.
00:21:01.000 --> 00:21:05.100
And that is going to often be enough to give you the entire graph; but it won't always.
00:21:05.100 --> 00:21:10.000
It might not show you everything; so if it doesn't show you everything you need to see--
00:21:10.000 --> 00:21:13.800
if there is some part missing (you might want to expand it, and you might not even realize that there is something missing)--
00:21:13.800 --> 00:21:20.400
you might want to just try increasing your interval to -6π, +6π, or -10 to +10.
00:21:20.400 --> 00:21:21.800
And just see if that changes the graph.
00:21:21.800 --> 00:21:24.700
If it doesn't change the graph, then you know that a 0 to 2π interval is enough.
00:21:24.700 --> 00:21:29.900
But if it does change the graph, that tells you that you need to think about how big your interval needs to be, to be able to see the whole graph.
00:21:29.900 --> 00:21:35.900
And maybe it won't even be possible to graph the entire interval all at once on one graph, like we are going to see in Example 2.
00:21:35.900 --> 00:21:37.500
All right, we are ready for some examples.
00:21:37.500 --> 00:21:42.200
First, let's graph the function r(θ) = 3sin(2θ).
00:21:42.200 --> 00:21:46.900
We plug in θ's here; it gives some value here, and that tells us what our r is going to be.
00:21:46.900 --> 00:21:51.700
It is just plugging in a number and getting something out of it; and that tells us our r.
00:21:51.700 --> 00:21:54.400
We plug in some value for θ; we get r.
00:21:54.400 --> 00:21:59.100
Let's figure out some values for this; let's figure out how often we need to do this.
00:21:59.100 --> 00:22:04.600
Well, we have 2θ as our thing in here: 2θ...if we were solving for the very first interesting point,
00:22:04.600 --> 00:22:10.000
which would normally occur at π/2 (0 is an interesting point, but we can be certain that 0 is already going to show up,
00:22:10.000 --> 00:22:14.200
and that one doesn't tell us as much), then that is going to be θ = π/4.
00:22:14.200 --> 00:22:19.300
So, we are going to have interesting points occur at an interval of every π/4 that we go out.
00:22:19.300 --> 00:22:22.400
Let's plug in π/4 to figure this out.
00:22:22.400 --> 00:22:30.600
We will plug in θ's; we are going to make a whole bunch of them.
00:22:30.600 --> 00:22:41.700
And here are our r's; and if we plug in 0, 3sin(0)...sin(0) is 0, so we are just going to get 0 in here.
00:22:41.700 --> 00:22:52.100
If we plug in π/4 (we figured out that our first interesting point was π/4), 3sin(2π/4)...2π/4 is π/2; sin(π/2) is 1; 3 times 1 is 3.
00:22:52.100 --> 00:22:54.400
So, we get +3 over here for our r.
00:22:54.400 --> 00:22:57.900
At this point, we can draw a graph; so we can start plotting some things.
00:22:57.900 --> 00:23:01.400
We know that we are going to have to at least get out to a distance of 3.
00:23:01.400 --> 00:23:05.100
And I will tell you that it actually is only going to go out to a total distance of 3, at the maximum.
00:23:05.100 --> 00:23:12.000
So, we will have circles every 3; so here is our first distance circle...
00:23:12.000 --> 00:23:15.800
Pardon me if my circles aren't absolutely perfect; I am but a human.
00:23:15.800 --> 00:23:36.000
The second distance circle; and our third distance circle...OK.
00:23:36.000 --> 00:23:48.000
And then, let's see: since we know that we are going to be based off of π/4, let's cut angles at π/4, as well; cool.
00:23:48.000 --> 00:23:57.900
So now, we can plot some points: at an angle of 0, we go out 0, so our first point is just at the pole, on what we used to call the origin.
00:23:57.900 --> 00:24:05.500
At an angle of π/4, this one right here, we are a distance of 3: 1, 2, 3 distance out.
00:24:05.500 --> 00:24:09.900
Next, let's try π/2, the next π/4 forwards, the next interesting point.
00:24:09.900 --> 00:24:18.600
We plug in sin(2π/2); 2π/2 is π; sin(π) is 0, so we end up getting 3(0) is 0.
00:24:18.600 --> 00:24:23.900
By the time we get back to π/2, this angle here, we are back down to 0; what does that look like?
00:24:23.900 --> 00:24:33.500
As π/4 goes to π/2, it goes down; from 0 up to π/2, we increase to 3, and then we decrease back down to 0.
00:24:33.500 --> 00:24:41.200
We increase to 3; and then we decrease down to 0; from 0 to π/2, we go up to 3, and then down to 0.
00:24:41.200 --> 00:24:51.600
So, as we spin counterclockwise, our thing increases; we touch that, and we spin back down.
00:24:51.600 --> 00:24:55.300
We get smaller and smaller, back down to 0; so that is the first part of our graph.
00:24:55.300 --> 00:25:06.200
Let's see what else is going to happen here: if we plug in 3π/4, sin(2(3π/4))...2(3π/4) is 3π/2;
00:25:06.200 --> 00:25:14.700
sin(3π/2) is -1; 3 times -1 is -3; so at an angle of 3π/4, which is here, we are at -3.
00:25:14.700 --> 00:25:18.000
So, we are going to go in the opposite direction; we are going to go opposite.
00:25:18.000 --> 00:25:21.000
We were going this way, but we are now going to go opposite, because it is -3.
00:25:21.000 --> 00:25:28.400
So, we are out 1, 2, 3 here; and then, at π, the next interesting place, sin(2π) is 0;
00:25:28.400 --> 00:25:32.500
2 times π is 2π, so sine of 2π is 0; so we get 0 once again here.
00:25:32.500 --> 00:25:38.300
And so, it is going to do the same thing, where, as it goes from π/2 to π, it becomes...
00:25:38.300 --> 00:25:44.000
here is 0; here is π/2; here is π; so for the first part, it went up to 3, and then it went back down to 0,
00:25:44.000 --> 00:25:50.400
when it got to π/2; and now it is going to go down to -3, and then it is going to go up to 0.
00:25:50.400 --> 00:25:54.000
So, we are seeing it go up, and then down, and then negative, and then back to 0.
00:25:54.000 --> 00:26:01.200
So, for this part, it is going more and more negative; so we end up seeing it curve out like this as it gets to larger and larger angles.
00:26:01.200 --> 00:26:07.200
We see it spin this way; OK.
00:26:07.200 --> 00:26:20.900
Next, we have 5π/4; at 5π/4, we plug that in; 2 times 5π/4 is going to be 10π/4, which is the same thing as 2π +...
00:26:20.900 --> 00:26:23.300
well, let's just write this out, because that way it is a little less confusing.
00:26:23.300 --> 00:26:39.300
5π/4 times 2 is equal to 10π/4; and look, that is the same thing as 8π/4 + 2π/4.
00:26:39.300 --> 00:26:47.700
So, that is the same thing as 2π + π/2; everything that we have here is just going to end up being the same thing as this.
00:26:47.700 --> 00:26:52.700
The π/4 here will match up to the 5π/4 here; so we are going to end up getting 3.
00:26:52.700 --> 00:27:01.900
With this idea in mind, we could actually realize, really quickly, that π/2 here is going to match to 6π/4, which is 3π/2.
00:27:01.900 --> 00:27:07.400
So, as we go that extra π forward, because we have this 2 here, it is going to end up repeating everything, as well.
00:27:07.400 --> 00:27:11.200
Our angles are going to be new and different and interesting, but the r's will end up repeating.
00:27:11.200 --> 00:27:14.900
We are going to see a repeat of this part here; it is just going to repeat here.
00:27:14.900 --> 00:27:24.400
So, 3π/2 gets us 0; 7π/4 will get us a -3; and at 2π, we will be right back where we started at 0.
00:27:24.400 --> 00:27:26.800
So, it will end up getting back to where we started.
00:27:26.800 --> 00:27:33.400
Let's plot 5π/4; at an angle of 5π/4, we are at this part; we go out a distance of 3, so we are here.
00:27:33.400 --> 00:27:39.000
So, we just got back to the origin when we were at π; the same thing--it curves out and grows larger when it gets out to it.
00:27:39.000 --> 00:27:42.400
And then, it gets smaller and smaller at this point; it drops back down.
00:27:42.400 --> 00:27:44.900
We are starting to see some symmetry; it is like there are petals here.
00:27:44.900 --> 00:27:47.400
"Petals" is a way of talking about this.
00:27:47.400 --> 00:27:55.000
At 7π/4, we are here; but once again, it is -3, so we go in the opposite direction.
00:27:55.000 --> 00:28:04.300
We go out to here, and there we go--they should be perfectly symmetrical, if the graph was absolutely perfect.
00:28:04.300 --> 00:28:07.500
If it was drawn by a computer, they would end up being perfectly symmetrical petals.
00:28:07.500 --> 00:28:12.100
But we can get a pretty good sense of what is going on, even drawing it by hand.
00:28:12.100 --> 00:28:22.300
The second example: Graph the equation r = 1.5^θ/2, where θ is between -2π and 2π; it goes from -2π to 2π.
00:28:22.300 --> 00:28:28.600
At this one, let's start...we will graph...well, we really have no idea of what this is going to end up being.
00:28:28.600 --> 00:28:38.800
We have 1.5r = 1.5^θ/2; we are dividing our θ by 2.
00:28:38.800 --> 00:28:44.000
So, we know that we are going to want 0 in there, because it is right in between -2π and +2π.
00:28:44.000 --> 00:28:46.800
And we want to go all the way down to -2π and all the way up to 2π.
00:28:46.800 --> 00:28:54.600
So, let's do this by π/2; π/2 will be easy to graph, as well, since they are the cross-axis that we normally have in there.
00:28:54.600 --> 00:29:01.100
A distance of r...if we plug in θ at 0 first, well, 1.5 raised to the 0--that is easy.
00:29:01.100 --> 00:29:06.500
We always end up getting 1; if you raise any number to the 0, you end up getting 1 out of it.
00:29:06.500 --> 00:29:12.000
Next, though, this is a little bit more difficult: if we plug in π/2, then what do we get out of this?
00:29:12.000 --> 00:29:14.600
Well, we will have to use a calculator; so here is how we do the first one.
00:29:14.600 --> 00:29:21.800
If we were to try to figure out 1.5, raised to the π/2, over 2...we have π/2, over 2...
00:29:21.800 --> 00:29:26.700
well, that is going to be the same thing as 1.5 to the π/4.
00:29:26.700 --> 00:29:32.300
We might not be able to figure this out by hand, but we could put this into a calculator: 3.14...
00:29:32.300 --> 00:29:39.200
1.57 would be half of that; so it is just going to be plugging in those numbers and getting an approximate value.
00:29:39.200 --> 00:29:44.200
So, this comes out to be approximately...using a calculator, we get 1.37 out of it.
00:29:44.200 --> 00:29:50.500
Sorry, I meant to say 1.57, because 3.14/2 would be equal to 1.57.
00:29:50.500 --> 00:29:58.800
But then, this is 3.14/4, so that is 1.57/2; the point is that we could work it out with a calculator
00:29:58.800 --> 00:30:02.100
and get an approximate decimal value for what that ends up being.
00:30:02.100 --> 00:30:06.100
That comes out to be 1.37; we do the same thing for each one of these.
00:30:06.100 --> 00:30:13.200
We plug in π; so it is 1.5 to the π/2 or 1.5 to the approximately 1.57.
00:30:13.200 --> 00:30:18.200
We raise that...our calculator gives us that that is approximately 1.89.
00:30:18.200 --> 00:30:29.300
3π/2...our calculator gives us that that gives us approximately 2.60...2π...we get approximately 3.57.
00:30:29.300 --> 00:30:36.800
What if we went the other way? -π/2...raising that up there...1.5^-π/2 divided by 2, so 1.5^-π/4...
00:30:36.800 --> 00:30:51.900
we get approximately 0.72; raising it to the -π/2...we get 0.53; raising it to the -3π/2, over 2, gets us 0.38.
00:30:51.900 --> 00:30:56.800
And finally, at -2π, we get 0.28.
00:30:56.800 --> 00:31:03.900
So notice; at 0, if we take 0, and we go up with 0, if we go up, the value gets larger and larger.
00:31:03.900 --> 00:31:08.800
We have it getting larger and larger faster and faster, because remember: this is an exponential function.
00:31:08.800 --> 00:31:13.800
So, it will get faster and faster growing as we go to larger and larger values of θ.
00:31:13.800 --> 00:31:17.600
As we go to negative values of θ, though, it gets smaller and smaller, because it is an exponential function.
00:31:17.600 --> 00:31:22.600
Once again, we are seeing the tail part get really down close to that x-axis.
00:31:22.600 --> 00:31:28.700
So, we can draw this out: the most extreme value that we get out to is 3.57.
00:31:28.700 --> 00:31:32.700
So, I will set our most extreme circle at a distance of 4.
00:31:32.700 --> 00:31:36.000
These single cross bars will be enough, because we are only concerned
00:31:36.000 --> 00:31:41.600
if π/2 is the only real reference angle we have going on here; so that will be OK.
00:31:41.600 --> 00:31:56.600
So, here is a distance of 1 circle; here is a distance of 2 circle; here is a distance of 3 circle; here is a distance of 4 circle.
00:31:56.600 --> 00:32:08.800
Oops, that got a little bit out of hand; here is a distance of 4 circle; that is better.
00:32:08.800 --> 00:32:12.500
OK, that is a pretty reasonable setup for our axes.
00:32:12.500 --> 00:32:20.100
At this point, we can plot some points: at 0, at an angle of 0, we are 1 out; so here is our first point.
00:32:20.100 --> 00:32:29.600
At an angle of π/2, going straight up, we are at 1.37; so we are almost to halfway out.
00:32:29.600 --> 00:32:35.800
At π, this angle here, we are at 1.89, getting pretty close to that distance of 2.
00:32:35.800 --> 00:32:42.300
At 3π/2, we are at 2.60, a little bit over halfway between the 2 and the 3 ring.
00:32:42.300 --> 00:32:48.300
At 2π, we are at 3.57, a little bit over halfway between the 3 and the 4 ring.
00:32:48.300 --> 00:32:53.000
We have this...as we go to a larger and larger angle, the distance out increases.
00:32:53.000 --> 00:32:57.100
We see it spiraling out, the farther out it gets.
00:32:57.100 --> 00:33:05.900
It continues to spiral out this way; so this is what we see as the angle gets larger and larger--it spirals out.
00:33:05.900 --> 00:33:11.000
What if the angle goes negative? Well, at -π/2, remember, this is -π/2,
00:33:11.000 --> 00:33:17.400
because it is talking about going clockwise instead; so at -π/2, we have 0.72.
00:33:17.400 --> 00:33:24.900
0.72 is around here; at -π, that is here, because remember: it is clockwise now;
00:33:24.900 --> 00:33:38.200
we are at 0.53; at -3π/2 (that is this one here), we are at 0.38; at -2π, we are at 0.28.
00:33:38.200 --> 00:33:43.800
So, we end up seeing it continue to spiral in and in and in and in and in.
00:33:43.800 --> 00:33:47.500
-2π to 2π: that is what is stopping this from continuing out forever.
00:33:47.500 --> 00:33:51.800
If this was allowed to keep going forever, we would see it spiral off way out forever.
00:33:51.800 --> 00:33:56.300
If this was allowed to continue forever, we would see it spiral into the center more and more and more and more.
00:33:56.300 --> 00:34:00.700
So, that -2π to 2π--that is why we have to have an interval set--because sometimes,
00:34:00.700 --> 00:34:03.800
if we don't set an interval, we could just keep going forever.
00:34:03.800 --> 00:34:07.400
All right, and that is why we have it set in our graphing calculator, if you are using a graphing calculator.
00:34:07.400 --> 00:34:12.300
You have to pay attention to the interval, because sometimes it will end up cutting off parts of the graph that you want to see.
00:34:12.300 --> 00:34:17.500
You wouldn't see that part where it gets to continue to spin in if you didn't have a larger than 0 as the bottom of your interval.
00:34:17.500 --> 00:34:24.500
You have to go to -2π, -4π, -10π...to really get a sense of just how much that spirals into the center.
00:34:24.500 --> 00:34:30.000
All right, the third example: convert the equation from polar to rectangular, then solve for y in terms of x.
00:34:30.000 --> 00:34:36.100
What were all of our equations? We have x = rcos(θ) as one of our formulas for changing.
00:34:36.100 --> 00:34:46.300
y = rsin(θ) is another formula; r² = x² + y² is another; and tan(θ) = y/x.
00:34:46.300 --> 00:34:51.600
So in this case, we see right away that 6r² times cos²(θ)...
00:34:51.600 --> 00:34:56.100
we are not quite sure about that part, but here is our sin(θ), and here is our sin(θ).
00:34:56.100 --> 00:34:59.600
So, we can swap them out; that is 3y = 7.
00:34:59.600 --> 00:35:03.300
What about this part here--what about r squared, cosine squared, θ?
00:35:03.300 --> 00:35:09.000
Well, we realize that that sounds an awful lot like r cosine θ; so how can we get rcos(θ) to show up there?
00:35:09.000 --> 00:35:17.500
We maybe think, "Oh, well, there is r²; there is cosine squared; we can pull that squared out, and we could write this as 6rcos(θ)."
00:35:17.500 --> 00:35:22.400
And then, that whole thing is squared; and then, minus 3y = 7.
00:35:22.400 --> 00:35:31.100
At this point, we can swap out x = rcos(θ); we have 6 times x² - 3y = 7.
00:35:31.100 --> 00:35:37.600
Add 3y; subtract 7 on both sides; so we have 6x² - 7 = 3y.
00:35:37.600 --> 00:35:45.300
We can divide by 3 on both sides; we could write this as y = 6x² - 7, and that divided by 3.
00:35:45.300 --> 00:35:52.700
We have managed to convert this to a rectangular x and y, and it is now in the form y = stuff involving x.
00:35:52.700 --> 00:35:57.000
What if we are doing the reverse? We are going from rectangular to polar.
00:35:57.000 --> 00:36:00.100
Last time, we went from polar to rectangular; now we are going the other way.
00:36:00.100 --> 00:36:08.800
Solve for r in terms of θ; once again, we have the same x = rcos(θ), y = rsin(θ);
00:36:08.800 --> 00:36:16.200
r² = x² + y²; and tan(θ) = y/x.
00:36:16.200 --> 00:36:22.400
In this case, we see that that is nice; we have rcos(θ), rsin(θ)...we have y; we have x.
00:36:22.400 --> 00:36:26.700
So, we can just swap those out directly; we can swap them out for what we have here.
00:36:26.700 --> 00:36:38.800
We swap them out; y is rsin(θ)...equals 2 times...x is rcos(θ), so 2 times rcos(θ), plus 3.
00:36:38.800 --> 00:36:46.700
At this point, we were told to solve for r in terms of θ, so we need to get our r's on one side, so we can get just r by itself.
00:36:46.700 --> 00:36:54.300
We move the 2rcos(θ) over by subtracting it, so we have rsin(θ), now minus 2rcos(θ), from both sides;
00:36:54.300 --> 00:37:02.300
that equals...+3 is still left over here; now, notice: we have an r here and an r here, so we can use the distributive property in reverse;
00:37:02.300 --> 00:37:11.900
we pull that r out, and we have sin(θ) - 2cos(θ) = 3.
00:37:11.900 --> 00:37:23.000
We now divide that out; so we have just r; so r = 3/(sin(θ) - 2cos(θ)).
00:37:23.000 --> 00:37:28.200
That might be a little surprising; that seems like a fairly complicated thing, if it is just going to give us a line.
00:37:28.200 --> 00:37:34.700
But some things...polar is better at graphing certain kinds of pictures, and rectangular is graphing other kinds of pictures.
00:37:34.700 --> 00:37:39.500
So, it depends on things; rectangular is great for graphing lines; polar is not as great at graphing lines.
00:37:39.500 --> 00:37:42.700
And you might be surprised that that would even end up coming out to be a line.
00:37:42.700 --> 00:37:47.000
Try plugging it into a graphing calculator, and you will see that that ends up giving us y = 2x + 3.
00:37:47.000 --> 00:37:49.000
It is just another way of graphing it.
00:37:49.000 --> 00:37:52.400
You might have a little bit of difficulty seeing why that ends up giving it.
00:37:52.400 --> 00:37:56.600
If you think about it, this bottom part here is going to sort of work as an asymptote,
00:37:56.600 --> 00:38:03.300
as it approaches the same angle that this line is based on, which is why it is going to shoot off infinitely in both the top and the bottom.
00:38:03.300 --> 00:38:05.400
So, think about that for a while; try graphing it.
00:38:05.400 --> 00:38:09.500
Just in general, try to play around with graphing as many polar functions as you can.
00:38:09.500 --> 00:38:14.600
It will really give you such a great sense of how the stuff is working if you just play with it for a while on a graphing calculator.
00:38:14.600 --> 00:38:16.000
All right, we will see you at Educator.com later--goodbye!