Math 32, August 25: Polar Cooardinates. Section 1: Polar Coordinates

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1 Math 2, August 2: Polar Cooardinates Section 1: Polar Coordinates For man applications, the Cartesian coordinates and are not the most practical framework for the problem in question. Imagine we are manning a lighthouse late and spot a ship on the horizon: distance direction We can estimate how far the boat is awa in some Cartesian grid (, ) for eample, how far north and how far east the boat is but we cannot be too precise since we are observing the boat from a fied position. We would much rather have a frame of reference which allows us to accuratel communicate the boat s moving without having to move ourselves. The problem, of course, is that Cartesian coordinates divides space into a grid best viewed from the top down. From our vantage point in the lighthouse, however, space is divided according to which direction we are looking and how far things are awa from us. For instance, we could ascertain that the boat is directl northwest of us and use radar to determine it is five miles awa. We could then pass this information on to a third part and, so long as the knew our position, the could find the boat. 1

2 We can change perspective similarl from a mathematical point of view. Rather than imagining points as representing locations on a grid, we can imagine them as being given b a direction (an angle θ) and a distance (a radius r) relative to a fied point (called the pole). This new set of coordinates are called polar coordinates. We will commonl take the pole to be the origin (0, 0), but this need not necessaril be the case. The correspondence between Cartesian points (, ) and polar coordinates (r, θ) is well-defined in the sense that for an point (, ) we can find a representation (r, θ), and for an polar representation (r, θ) we can associate a unique point (, ). We notice a few things: 1. Points (, ) do not have unique polar representation, since θ = θ+2nπ for an n Z so that (r, θ) = (r, θ + 2nπ). The representation ma be made unique b making the restriction π θ π or 0 θ 2π. We ma also allow r to be negative so that we ma have (r, θ) = ( r, θ±π). 2. The point (0, 0) in Cartesian coordinates can be represented b (0, θ) in polar coordinates for an θ R whatsoever. Now consider going from polar to Cartesian coordinates. The equations we need are the following: Polar into Cartesian coordinates: = r cos(θ), = r sin(θ). To see wh this is, we draw our standard right-angle triangle: (,) r θ =r sin( θ) =r cos( θ) and note that we have cos(θ) = r and sin(θ) =. The desired form follows r b rearrangement. Consider the following eamples. 2

3 Eample 1 Find the Cartesian coordinates of the polar points (2, π 2 (0, 7π), and ( 2, π 4 ). ), (, 1π 6 ), Solution: (2, π 2 ): We have = r cos(θ) = 2 cos( π 2 ) = 0 and = r sin(θ) = 2 sin( π 2 ) = 2. Our point is given b (0, 2). (, 1π 6 ): We have = r cos(θ) = cos( π 6 ) = 2 and = r sin(θ) = sin( π 6 ) = 2. Our point is given b ( 2, 2 ). (0, 7π): Notice that if we etend a magnitude of zero from a point, it does not matter which direction we are facing. The point is therefore (0, 0). ( 2, π 4 ): This eample ma appear to be different, since r = 2 is negative, but the evaluation is the same. We have = r cos(θ) = 2 cos( π 4 ) = 2 and = r sin(θ) = 2 sin( π 4 ) = 2 so that our point is ( 2, 2). We can interpret this point as being obtained b rotating π 4 radians and then traveling two units backwards. Now consider having a fied and and determining r and θ. We have the following formulas: Cartesian into Polar coordinates: r = 2 + 2, θ = arctan ( ). To see wh these formulas hold, we can easil see that that the equation for r follows from the Pthagorean Theorem: r 2 = = r = To solve for θ, notice that we can eliminate r from the equations = r cos(θ) and = r sin(θ) b diving. This gives: = sin(θ) ( ) cos(θ) = tan(θ) = θ = arctan.

4 We have, however, lost some information in this process. The ratio / is the same sign if (, ) is in the first or third quadrant (i.e. both positive or both negative) or in the second or fourth quadrant (i.e. opposite signs). Since we have restricted the range of θ = arctan(z) to be π/2 θ π/2, arctan( ) will onl give us the correct value for (, ) in the first and fourth quadrants. We will correct this in the following wa. Quadrant Correction: If (, ) lies in the second or third quadrant, we use the value ( θ = arctan + π. ) Eample 2 Find the polar coordinates of the Cartesian points (1, ), (2, 7), (, 1 ), and ( 1, 1). Solution: (1, ): We plug into the formula to get r = = (1) 2 + ( ) 2 = 2 and θ = arctan( ) = arctan( 1 ) = π. Since (1, ) is in the first quadrant, this has given us the correct θ so we have the polar coordinates (2, π ). (2, 7): Following the formulas, we have r = (2) 2 + ( 7) 2 = and θ = arctan( 7 2 ) Since (2, 7) is in the fourth quadrant, this is the correct values for θ so we have in polar coordinates (approimatel) (, 1.29). (, 1 ): Following the formulas, we have r = ( )2 + ( 1 ) 2 = 4 and θ = arctan( 1 ) = π 6 ; however, the point (, 1 ) is in the third quadrant, so we have to adjust b a factor of π. We have in polar coordinates ( 4, 7π 6 ). ( 1, 1): Following the formulas, we have r = ( 1) 2 + (1) 2 = 2 and 4

5 θ = arctan( 1) = π 4 ; however, the point ( 1, 1) is in the second quadrant, so we need to adjust b a factor of π to get the polar coordinates ( 2, π 4 ). Section 2: Polar Functions We are generall interested in much more than discrete collections of points. In calculus, we are generall interested in sets of points satisfing some algebraic relationship (e.g. functions). Our net logical question, therefore, is to ask how we can represent functions = f() and relations f(, ) = 0 in terms of polar coordinates. We will also be interested in returning from polar coordinates to Cartesian coordinates, just as we were for single points. Eample Convert the following equation into a polar function r = f(θ): = 0. Solution: We know that for a single point that we have the relationships = r cos(θ), and = r sin(θ). If we plug these relations into the equation, we have = r 2 cos 2 (θ) + r 2 sin 2 (θ) 4r sin(θ) = 0 = r 2 = 4r sin(θ) = r = 4 sin(θ). Notice that we can properl sa r is a function of θ that is to sa, for each value of θ we can assign a unique value of r. This was not a propert of the original equations since did not uniquel determine values of (i.e. the graph fails the vertical line test). In that sense, the polar equations are a more natural approach to this equation. Also notice that r can be negative for some values of θ, meaning that as we plot along the ais at angle θ, we plot backwards a magnitude of r. Now consider the following:

6 Eample 4 Convert the following equation into polar coordinates: = Solution: Using our relations above we get r sin(θ) = r 2 cos 2 (θ) + 1 = r 2 cos 2 (θ) r sin(θ) + 1 = 0. Unlike the previous eample, we cannot solve for r uniquel as a function of θ. We are left with the relation f(r, θ) = 0 in the form above. This means that some values of θ ma correspond to multiple values of r. Other values of θ ma not correspond to an value of r at all. These eamples raise an important point about how we should choose to represent functions. For the first eample, we could not represent as a function of, but we could represent r as a function of θ; conversel, for the second eample, we could represent as a function of, but not represent r as a function of θ (at least not choosing (0, 0) as our pole): (,) r (,) t r t We suspect that the first eample is better represented in polar coordinates while the second is better represented in Cartesian coordinates. The net question we can ask is how we transform polar equations into Cartesian equations. We recall that r =

7 so that we can readil replace r readil with s and s. The other relationship, θ = arctan(/), is valid b difficult to appl in practice. For one thing, we had to adjust θ depending upon which quadrant we were in. This was fine for a single point, but for functions we are considering all possible values of on the curve. We can, however, notice that r = implies the following: θ Substitution: cos(θ) = 2 +, sin(θ) = Since most polar equations involve θ onl within trigonometric functions (which can be decomposed into sin(θ) s and cos(θ) s), this is tpicall all the information we need. Eample Convert the following polar equation into Cartesian form: r = cos(θ) 2 sin(θ). Solution: We appl the formulas to get = = = 0 ( = 1 ) 2 + ( + 1) 2 = 2 4. We recognize that this is the equation for the circle with centre (1/2, 1) and a radius of /4. We also notice that, for some values of θ, r has a negative value. We will consider the implications of this in a deeper fashion once we begin graphing polar equations. Now consider the following eample: 7

8 Eample 6 Convert the following polar equation into Cartesian form: r = sin(2θ). Solution: We cannot appl our identities directl since the argument inside the sin( ) is something other than θ (in this case 2θ). We have to appl an identit first to get r = 2 sin(θ) cos(θ). We can now proceed as before: ( ) ( ) = ( 2 + 2) /2 2 = 0. This is the best we can do with this equation. Clearl this was easier to represent in polar coordinates. This suggests that there is something to going to all this trouble after all. 8

9 Suggested Problems 1. Convert the following Cartesian points into polar coordinates (r, θ): (a) (, 2) (b) ( 2, 2) (c) ( 10, 0) 2. Plot the folowing polar points in the (, )-plane and then convert them into Cartesian coordinates (, ): (a) (1, π 4 ) (b) ( 4, π) (c) ( 2, 1π 2 ). Determine the Cartesian form of the follow epressions: (a) r = cos(θ) + 4 sin(θ) (b) r = 1 + sin(θ) (c) r = tan(θ) (d) r = cos(θ) 4. Determine the polar form of the following epressions: (a) = 2 (b) = 4 (c) = 0 (d) 2 2 = 1 9