Math 121 (Lesieutre); 9.1: Polar coordinates; November 22, 2017

Transcription

1 Math 2 Lesieutre; 9: Polar coordinates; November 22, 207 Plot the point 2, 2 in the plane If you were trying to describe this point to a friend, how could you do it? One option would be coordinates, but another would be to say the distance from the origin and the angle a If your point is on a circle centered at the origin, what does the distance represent? How can you find it given the x and y coordinates? What is it for the point above? Using the Pythagorean theorem, the distance would be For the particular point we re talking about, r = x 2 + y 2 r = = 8 = 2 2 b What about the angle? How can you find it from x and y, and what is it in this example? We can see that tan θ = y x, and so θ = tan y x For this specific point, it s 2 θ = tan = tan = π 2 4 c We have defined polar coordinates To describe a point x, y, we give the two numbers r, θ What are the polar coordinates for this point? The polar coordinates for the point are r, θ = 2 2, π 4 2 On the polar graph paper below, plot and label the three points: A =, 4π, B = 5, 4π, C = 7, π, 2

2 B 4, π C A The ones with a negative radius are weird This means to go to the angle specified, but instead of going distance r in that direction, go the same distance, but in the opposite direction So 5, 9π 2 ends up being the same thing as 5, 7π 2 What are the equations to convert rectangular to polar? Polar to rectangular? We already figured out the first: r = x 2 + y 2 θ = tan y x You have to be a little careful here There are two angles with any given value of tan, and you need to be sure to pick the right one It s not necessarily tan y x See the next problem for an example We have cos θ = x r, so x = r cos θ Similarly, sin θ = y r, and so y = r sin θ Let me set that on its own line to emphasize how important it is: Remember those, both ways x = r cos θ y = r sin θ 4 Consider the point with polar coordinates 4, π a Add the point to the polar graph paper above Done! b What are other ways we could write this point in polar? Find three other possibilities, one with θ > 2π, one with θ < 0, and one with r < 0 We could add 2π to the angle without really changing anything That would be 4, π + 2π = 4, π + 2π 2 = 4, π

3 2 We could also subtract 2π without changing the point 4, π 2π = 4, π 2π = 4, π To get a negative r, just use the angle in the opposite direction, and make the radius negative of what it had been To figure out the opposite angle, either look at the graph paper and find it, or add π 4, 7π Let me stress the upshot of all this: this is three different ways of describing the same point There is more than one way to express a point in polar This is a big difference from rectangular c How would you find the rectangular coordinates of this point? I would use the formula we already came up with You should too x = r cos θ = 4 cos π = 4 2 = 2 y = r sin θ = 4 sin π = 4 2 = 2 So the point is 2, 2 5 Let s try a few where we convert from rectangular to polar A suggestion: make a quick sketch, so you can be sure you put the angle in the right quadant Remember, there are two angles between 0 and 2π with a given tangent, and you have to make sure you pick the right one Find retangular coordinates for the three points 0, 5, 2, 2, and 4, 4 For 0, 5 The radius is = 25 = 5, and the angle is π 2 The polar coordinates are thus 5, π 2 For 2, 2, r = = = = 4, 2 tan θ = 2 θ = 5π π or How do we know whether we want 5π/ or π/? Our point is in the 2nd quadrant, so it has to be the former For 4, 4, r = = 2 = 4 2 tan θ = 4 4 = θ = π 4 or 5π 4 How do we know whether we want π/4 or 5π/4? Our point is in the 4th quadrant, so it has to be the latter r, θ = 4 2, 5π/4

4 You can also express equations for curves using polar coordinates instead of cartesian It s usually pretty straightforward: you just need to plug in r cos θ everywhere you see an x and r sin θ everywhere you see a y, then simplify x 2 + y 2 = 9, 5y 2 = x 2y, y = x 2, x = 4 Let s do them in order This is the formula for a circle of radius x 2 + y 2 = 9 r cos θ 2 + r sin θ 2 = 9 r 2 cos 2 θ + r 2 sin 2 θ = 9 r 2 cos 2 θ + sin 2 θ = 9 r 2 = 9 r = 5y 2 = x 2y 5r sin θ 2 = r cos θ 2r sin θ 5r 2 sin 2 θ = r cos θ 2r sin θ Can t really do anything more with that x = 4 r cos θ = 4 We could call it a day, but often we want to write r = when we can; it s like giving a formula as y = fx This one would be r = 4 sec θ That s the polar equation for a vertical line 7 Now let s try some in the opposite direction: I give you a polar equation, you come up with the rectangular equation Hint: where possible, try to do some algebra to get terms like r sin θ, instead of using the messy formulas r = x 2 + y 2 etc r = 2 csc θ, r = cos θ, r = 7, θ = π 4 Here we go A horizontal line r cos θ =, r = 4 sin θ, r = r = 2 csc θ = 2 sin θ r sin θ = 2 y = 2 sin θ cos θ 4

5 Next, r = cos θ r 2 = r cos θ x 2 + y 2 = x This is a classic trick: multiply both by r to make things easier You can complete the square here, if you want: x 2 x + y 2 = 0 x 2 x y 2 = 9 x 2 + y 2 = 9 This is a circle of radius, centered at, 0 r = 7 r 2 = 49 x 2 + y 2 = 49 This is a circle of radius 7, centered at the origin That s a line θ = π 4 π tan θ = tan 4 y x = y = x Easy one r cos θ = x = r = 4 sin θ r 2 = 4r sin θ x 2 + y 2 = 4y That was the r trick again Completing the square, you get x 2 + y 2 2 = 2 2, so it s a circle of radius 2 centered at 0, 2 That s another line r = sin θ cos θ r sin θ r cos θ = y x = y = x + 5