MATH 118, LECTURES 13 & 1: POLAR EQUATIONS 1 Polar Equations We now know how to equate Cartesian coordinates with polar coordinates, so that we can represents points in either form and understand what we are talking about. Furthermore, we know how to go back and forth between the two representations without losing any information. From a Calculus point of view, however, we are not interested in distinct sets of points we are interested in connected curves of points (i.e. functions). Our net logical question, therefore, is to ask how we can represent functions y = f() and relations f(,y) = 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. Consider the equation + y y = 0. We want to represent this equation in polar coordinates (r,θ), but how can we go about doing this? Well, we know from switching from Cartesian to polar coordinates for a single point that we have the relationship = r cos(θ), and y = r sin(θ). If we plug these relations into the equation, we have + y y = r cos (θ) + r sin (θ) r sin(θ) = 0 = r = r sin(θ) = r = sin(θ). So we have transformed the problem from a relationship between y and into a relationship between θ and r. We notice also that we can properly say r is a function of θ that is to say, for each value of θ we can assign a unique point value of r. This was not a property of the original equations since did not uniquely determine values of y (i.e. the graph fails the vertical line test). In that sense, the polar equations are a more natural approach to this equation. (We also notice that r can be negative for some values of 1
θ, meaning that as we plot along the ais at angle θ, we plot backwards a magnitude of r.) Consider, however, representating the following equation in polar coordinates y = + 1. Using our relations above we get r sin(θ) = r cos (θ) + 1 = r cos (θ) r sin(θ) + 1 = 0. Unlike the previous eample, we cannot solve for r uniquely 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 may correspond to multiple values of r. Other values of θ may not correspond to any 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 y as a function of, but we could represent r as a function of θ; conversely, for the second eample, we could represent y as a function of, but not represent r as a function of θ (at least not choosing (0,0) as our pole) (see Figure 1). We suspect that the first eample is better represented in polar coordinates while the second is better represented in Cartesian coordinates. y y (,y) r (,y) t r t Figure 1: Our two eamples represented in Cartesian coordinates. The first can only be represented as a function by r = f(θ) while the second can only be representated as a function by y = f().
The net question we can ask is how we transform polar equations into Cartesian equations. We recall that r = + y so that we can readily replace r readily with s and y s. But how about θ? We recall that θ = arctan(y/) was, at best, an imperfect transformation since we had to check which quadrant the point was in or risk being off by a factor of π. We can do this for a single point, but it is impractical for functions (we would have to do this for every point lying on the curve!). We can, however, notice that r = + y implies cos(θ) = r = sin(θ) = y r = + y y + y. Since most polar equations involve θ only within trigonometric functions (which can be decomposed into sin(θ) s and cos(θ) s), this is typically all the information we need. Consider transforming the following polar equation into Cartesian variables r = cos(θ) sin(θ). We apply the formulas to get + y = + y + y + y = 0 ( 1 ) + (y + 1) = 5. and y + y We recognize this as the equation for the circle with centre (1/, 1) and a radius of 5/. 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 transform the following polar equation into Cartesian variables r = sin(θ). We cannot apply our identities directly since the argument inside the sin( ) is something other than θ (in this case θ). We have to apply an identity first to get r = sin(θ)cos(θ). 3
We can now proceed as before: ( ) ( ) + y y = + y + y ( + y ) 3/ y = 0. This is the best we can do with this equation. Clearly this was easier to represent in polar coordinates maybe there is something to going to all this trouble after all. 1.1 Graphing Polar Equations Since the Cartesian form of r = sin(θ) was not remotely recognizable to us, our only approach to visualize the graph of the equation is by using polar coordinates; however, we will start by graphing r = cos(θ) sin(θ) using polar equations since we know what the graph looks like based on the Cartesian form. Graph r = cos(θ) sin(θ) using polar coordinates. Our approach will be very similar to what we used when graphing parametric equations. We start by considering the graph of r(θ), or at least considering a chart of important values of the function. Our interpretation, however, will be slightly different than it was for parametrization. We can think of θ as parametrizing an array etending from (0,0), where θ is the angle between the array and the positive -ais. As the array pivots upon the ais (the origin), we assign points on that array at the appropriate distance r. We can set up our table of important points as follows: θ r 0 1 0.63 0 π π 3 1 3 3π 1 5π 6 3 1 π 1.
We could go further, however, we notice that (, π ) and (, 3π ) are the same point, as are (1,0) and ( 1,π). Parametrizing θ through [π,π] would repeat the same points we have already attained. We consider the point where r = 0 as important since it defines a tipping point where r goes from being positive to being negative. We can also graph r = f(θ) to get an idea of important regions of the graph (see Figure 1.1). Putting everything together, we have the circle centred at (1/, 1) of radius 5/, as we epected. r (r 1,t 1 ) y (r 1,t 1 ) t (r,t ) (r,t ) (r 3,t 3 ) (r 3,t 3 ) Figure : Plot of the polar function r = cos(θ) sin(θ) (left) and the plot of the equation in Cartesian coordinates (right). Now consider graphing the function r = sin(θ). When we transformed this equation into Cartesian coordinates, it did not fit the form of any function we recognize polar coordinates is the best we can do. We start by making a chart of critical values and plotting r = sin(θ) (we will need to consider all θ values 0 to π, but notice that sin(θ) repeats after that point): θ r θ r 0 0 π 1 π 0 3π 1 π 0 5π 1 3π 0 7π 1 π 0 Let s consider the region 0 θ π (i.e. the first quadrant). The radius r goes from 0 at θ = 0 (the positive -ais) to 1 at π (a 5 degree angle) 5
and then back to 0 at π (the positive y-ais). Interestingly, this sweeps out what looks like the petal of a flower this is characteristic of polar equations of the form r = asin(b) and r = acos(b). As we sweep out the net quadrant of values π θ π, we see that the same pattern repeats; however, the radius is negative, so that it sweeps out a petal in the fourth quadrant. Repeating the procedure for 0 θ π we see that all four quadrants are filled in to give what appears to be a four-petalled flower (see Figure 1.1). r (r 1,t 1 ) y (r 1,t 1 ) (r 3,t 3 ) t (r 3,t 3 ) (r,t ) (r,t ) Figure 3: Plot of the polar function r = sin(θ) (left) and the plot of the equation in Cartesian coordinates (right). For something different, consider graphing the function r = θ, θ 0. We notice first of all that since θ does not appear on the inside of trigonometric function, we are going to have some difficulty changing this into Cartesian coordinates. We have to approach this in polar coordinates, but we are at a loss for which critical values of θ to put in our chart. Fortunately, things are actually much easier for this eample than in the previous ones. As θ grows, so does the radius. So our curve starts out at the original and slowly rotates counterclockwise, growing at an even rate: this is a spiral (see Figure 1.1)! 1. Derivatives of Polar Equations When we can define r as a function of θ (say, r = f(θ)), we can make use the some of the results from last week regarding parametric equations. This 6
y r=t Figure : Cartesian plot of r = θ, θ 0. is because in the polar form we have (,y) = (r cos(θ),r sin(θ)) = (f(θ)cos(θ),f(θ)sin(θ)) so that and y are completely determined by the parameter θ. In particular, we can now apply the differentiation formula we derived for parametric equations: dy d = y (θ) (θ) = f (θ)sin(θ) + f(θ)cos(θ) f (θ)cos(θ) f(θ)sin(θ). Eample 1: Find the Cartesian coordinates of the points where the tangent line to r = sin(θ) is horizontal. (We will include points with multiple curves running through them so long as at least one curve through the point has a horizontal tangent line.) 7
A point with a horizontal tangent line has a point where the derivative dy d vanishes. We will solve this using polar coordinates. We apply our formula to get dy cos(θ)sin(θ) + sin(θ)cos(θ) = d cos(θ)cos(θ) sin(θ)sin(θ). In order to have a horizontal tangent line, we only need to set the numerator to zero, so that we have 0 = cos(θ)sin(θ) + sin(θ)cos(θ) = (1 sin (θ))sin(θ) + sin(θ)cos (θ) = sin(θ) sin 3 (θ) + sin(θ)(1 sin (θ)) = sin(θ) 6sin 3 (θ) = sin(θ)( 3sin (θ)). This is satisfied if either sin(θ) = 0, or sin(θ) = ± 3. The first is satisfied if θ = nπ, n an integer while the second can be satisfied if either θ 0.955 + nπ or θ 0.955 + nπ, n an integer where arcsin( /3) 0.955 and arcsin( /3) 0.955. We notice that θ = nπ give the same point for all n (r = 0, which is the origin), so we will count it only once. Also, since θ repeats after π, we will reduce our list to just the five angles θ 1 = 0,θ = 0.955,θ 3 = 0.955 + π,θ = 0.955, and θ 5 = 0.955 + π. Since θ parametrizes equation, we can use the formulas to get (,y) = (f(θ)cos(θ),f(θ)sin(θ)) = (sin(θ)cos (θ),sin (θ)cos(θ)). We can solve for the equations (using various trigonometric identities) 8
to get θ 1 = 0 : ( 1,y 1 ) = ( (0,0) ) θ = 0.955 : (,y ) = 3, 3 3 (0.5, 0.770) ( 3 ) θ 3 = 0.955 + π : ( 3,y 3 ) = 3, 3 3 ( 0.5, 0.770) ( ) 3 θ = 0.955 : (,y ) = 3, 3 3 ( 0.5, 0.770) ( 3) θ 5 = 0.955 + π : ( 5,y 5 ) = 3, 3 3 (0.5, 0.770) 3 This matches our intuition, given the graph of r = sin(θ), since we see that along each petal there is a single point where the tangent is horizontal. We have also shown that the curves pass through the origin horizontally as well (see Figure 1.). (,y ) (,y ) ( 1,y 1 ) ( 3,y 3 ) ( 5,y 5 ) Figure 5: Cartesian plot of r = sin(θ) showing five distinct points where the slope of the tangent line is horizontal. 9
1.3 Lengths of Curves in Polar Coordinates In the previous section, we showed that if we could epress r as a function of θ (r = f(θ)), we could think of and y as being parametrized by θ as ((θ),y(θ)) = (f(θ)cos(θ),f(θ)sin(θ)). Last week, we derived the formula for lengths of curves when the curve is given parametrically we can now etend this to polar equations. We have β (d ) ( ) dy L = + dt α dt dt β = [f (θ)cos(θ) f(θ)sin(θ)] + [f (θ)sin(θ) + f(θ)cos(θ)] dθ = = α β α β α [f (θ)] (cos (θ) + sin (θ)) + [f(θ)] (cos (θ) + sin (θ)) dθ [f (θ)] + [f(θ)] dθ. Consider finding the length of the curve given by r = acos(θ), 0 θ π,a > 0. We have f(θ) = acos(θ) and f (θ) = asin(θ). Following the formula, we have π L = a cos (θ) + a sin (θ) dθ = a 0 π 0 1 dθ = a[θ] π 0 = aπ. 10