9.4 POLAR COORDINATES

Transcription

1 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 EXPLORATORY EXERCISES. For the brachistochrone problem, two criteria for the fastest curve are: () steep slope at the origin and () concave down (note in Figure 9.6 that the positive -ais points downward). Eplain wh these criteria make sense and identif other criteria. Then find parametric equations for a curve (different from the ccloid or those of eercises 6) that meet all the criteria. Use the formula of eample. to find out how fast our curve is. You can t beat the ccloid, but get as close as ou can!. The tautochrone problem is another surprising problem that was studied and solved b the same seventeenth-centur mathematicians as the brachistochrone problem. (See Journe Through Genius b William Dunham for a description of this interesting piece of histor, featuring the brilliant et combative Bernoulli brothers.) Recall that the ccloid of eample. runs from (0, 0) to (π, ). It takes the skier k π = π/g seconds to ski the path. How long would it take the skier starting partwa down the path, for instance, at (π/, )? Find the slope of the ccloid at this point and compare it to the slope at (0, 0). Eplain wh the skier would build up less speed starting at this new point. Graph the speed function for the ccloid with 0 u and eplain wh the farther down the slope ou start, the less speed ou ll have. To see how speed and distance balance, use the time formula T = π cos πu du g a cos πa cos πu for the time it takes to ski the ccloid starting at the point (πa sin πa, cos πa), 0 < a <. What is the remarkable propert that the ccloid has? 9. POLAR COORDINATES (, ) You ve probabl heard the cliche about how difficult it is to tr to fit a round peg into a square hole. In some sense, we have faced this problem on several occasions so far in our stud of calculus. For instance, if we were to use an integral to calculate the area of the circle + = 9, we would have A = [ 9 ( )] 9 d = 9 d. (.) 0 FIGURE 9. Rectangular coordinates u r FIGURE 9. Polar coordinates (r, u) Note that ou can evaluate this integral b making the trigonometric substitution = sin θ. (It s a good thing that we alread know a simple formula for the area of a circle!) A better plan might be to use parametric equations, such as = cos t, = sin t, for 0 t π, to describe the circle. In section 9., we saw that the area is given b π 0 (t) (t) dt = = 9 π 0 π 0 ( cos t)( cos t) dt cos tdt. This is certainl better than the integral in (.), but it still requires some effort to evaluate this. The basic problem is that circles do not translate well into the usual - coordinate sstem. We often refer to this sstem as a sstem of rectangular coordinates, because a point is described in terms of the horizontal and vertical distances from the origin (see Figure 9.). An alternative description of a point in the -plane consists of specifing the distance r from the point to the origin and an angle θ (in radians) measured from the positive -ais counterclockwise to the ra connecting the point and the origin (see Figure 9.). We describe the point b the ordered pair (r,θ) and refer to r and θ as polar coordinates for the point.

2 9-9 SECTION 9... Polar Coordinates 7 EXAMPLE. Converting from Polar to Rectangular Coordinates Plot the points with the indicated polar coordinates and determine the corresponding rectangular coordinates (, ) for: (a) (, 0), (b) (, π ), (c) (, π ) and (d) (,π). Solution (a) Notice that the angle θ = 0 locates the point on the positive -ais. At a distance of r = units from the origin, this corresponds to the point (, 0) in rectangular coordinates (see Figure 9.a). (b) The angle θ = π locates points on the positive -ais. At a distance of r = units from the origin, this corresponds to the point (0, ) in rectangular coordinates (see Figure 9.b). (c) The angle is the same as in (b), but a negative value of r indicates that the point is located units in the opposite direction, at the point (0, ) in rectangular coordinates (see Figure 9.b). (d) The angle θ = π corresponds to the negative -ais. The distance of r = units from the origin gives us the point (, 0) in rectangular coordinates (see Figure 9.c). (, q) (, 0) q (, p) p (, q) FIGURE 9.a The point (, 0) in polar coordinates FIGURE 9.b ( The points, π ) ( and, π ) in polar coordinates FIGURE 9.c The point (,π)in polar coordinates FIGURE 9.a Polar coordinates for the point (, ) d EXAMPLE. Converting from Rectangular to Polar Coordinates Find a polar coordinate representation of the rectangular point (, ). Solution From Figure 9.a, notice that the point lies on the line =, which makes an angle of π with the positive -ais. From the distance formula, we get that r = + =. This sas that we can write the point as (, π )inpolar coordinates. Referring to Figure 9.b (on the following page), notice that we can specif the same point b using a negative value of r, r =, with the angle 5π. (Think about this some.) Notice further, that the angle 9π = π + π corresponds to the

3 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-0 h, d p FIGURE 9.b An alternative polar representation of (, ) FIGURE 9.c Another polar representation of the point (, ) REMARK. As we see in eample., each point (, ) inthe plane has infinitel man polar coordinate representations. For a given angle θ, the angles θ ± π, θ ± π and so on, all correspond to the same ra. For convenience, we use the notation θ + nπ (for an integer n) torepresent all of these possible angles. (r, u) same ra shown in Figure 9.a (see Figure 9.c). In fact, all of the polar points (, π + nπ) and (, 5π + nπ) for an integer n correspond to the same point in the -plane. Referring to Figure 9.5, notice that it is a simple matter to find the rectangular coordinates (, ) of a point specified in polar coordinates as (r,θ). From the usual definitions for sin θ and cos θ, weget = r cos θ and = r sin θ. (.) From equations (.), notice that for a point (, ) inthe plane, + = r cos θ + r sin θ = r (cos θ + sin θ) = r and for 0, = r sin θ r cos θ = sin θ = tan θ. cos θ That is, ever polar coordinate representation (r,θ)ofthe point (, ), where 0 must satisf r = + and tan θ =. (.) r r sin u Notice that since there s more than one choice of r and θ, wecannot actuall solve equations (.) to produce formulas for r and θ. Inparticular, while ou might be tempted to write θ = tan ( ) u, this is not the onl possible choice. Remember that for (r,θ)tobe a polar representation of the point (, ),θ can be an angle for which tan θ =, while tan ( ) ( gives ou an angle θ in the interval π, ) r cos u π. Finding polar coordinates for a given point is tpicall a process involving some graphing and some thought. FIGURE 9.5 Converting from polar to rectangular coordinates EXAMPLE. Converting from Rectangular to Polar Coordinates Find all polar coordinate representations for the rectangular points (a) (, ) and (b) (, ). Solution (a) With = and =, we have from (.) that r = + = + =,

4 9- SECTION 9... Polar Coordinates 75 REMARK. Notice that for an point (, ) specified in rectangular coordinates ( 0), we can alwas write the point in polar coordinates using either of the polar angles tan ( ) or tan ( ) + π. You can determine which angle corresponds to r = + and which corresponds to r = + b looking at the quadrant in which the point lies. so that r =±. Also, tan θ = =. One angle is then θ = tan ( ) 0.98 radian. To determine which choice of r corresponds to this angle, note that (, ) is located in the first quadrant (see Figure 9.6a). Since 0.98 radian also puts ou in the first quadrant, this angle corresponds to the positive value of r, sothat (, tan ( )) is one polar representation of the point. The negative choice of r corresponds to an angle one half-circle (i.e., π radians) awa (see Figure 9.6b), so that another representation is (, tan ( ) ) + π.ever other polar representation is found b adding multiples of π to the two angles used above. That is, ever polar representation of the point (, ) must have the form (, tan ( ) + nπ ) or (, tan ( ) + π + nπ ), for some integer choice of n. (, ) (, ) u tan ( ) u tan ( ) p u tan ( ) FIGURE 9.6a The point (, ) FIGURE 9.6b Negative value of r (, ) u tan ( W) p (b) For the point (, ), we have = and =. From (.), we have r = + = ( ) + = 0, so that r =± 0. Further, tan θ = =, FIGURE 9.7 The point (, ) u tan ( W) so that the most obvious choice for the polar angle is θ = tan ( ) 0., which lies in the fourth quadrant. Since the point (, ) is in the second quadrant, this choice of the angle corresponds to the negative value of r (see Figure 9.7). The positive value of r then corresponds to the angle θ = tan ( ) + π. Observe that all polar coordinate representations must then be of the form ( 0, tan ( ) + nπ)or ( 0, tan ( ) + π + nπ), for some integer choice of n. Observe that the conversion from polar coordinates to rectangular coordinates is completel straightforward, as in eample.. EXAMPLE. Converting from Polar to Rectangular Coordinates Find the rectangular coordinates for the polar points (a) (, π 6 ) and (b) (, ). Solution For (a), we have from (.) that = r cos θ = cos π 6 =

5 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- and = r sin θ = sin π 6 =. The rectangular point is then (, ).For (b), we have and = r cos θ = cos.98 = r sin θ = sin 0.8. r The rectangular point is ( cos, sin ), which is located at approimatel (.98, 0.8). The graph of a polar equation r = f (θ) isthe set of all points (, ) for which = r cos θ, = r sin θ and r = f (θ). In other words, the graph of a polar equation is a graph in the -plane of all those points whose polar coordinates satisf the given equation. We begin b sketching two ver simple (and familiar) graphs. The ke to drawing the graph of a polar equation is to alwas keep in mind what the polar coordinates represent. FIGURE 9.8a The circle r = EXAMPLE.5 Some Simple Graphs in Polar Coordinates Sketch the graphs of (a) r = and (b) θ = π/. u Solution For (a), notice that = r = + and so, we want all points whose distance from the origin is (with an polar angle θ). Of course, this is the definition of a circle of radius with center at the origin (see Figure 9.8a). For (b), notice that θ = π/ specifies all points with a polar angle of π/ from the positive -ais (at an distance r from the origin). Including negative values for r, this defines a line with slope tan π/ = (see Figure 9.8b). It turns out that man familiar curves have simple polar equations. 6 FIGURE 9.8b The line θ = π 6 6 EXAMPLE.6 Converting an Equation from Rectangular to Polar Coordinates Find the polar equation(s) corresponding to the hperbola = 9 (see Figure 9.9). Solution From (.), we have 9 = = r cos θ r sin θ = r (cos θ sin θ) = r cos θ. Solving for r,weget r = 9 = 9 sec θ, cos θ so that r =± sec θ. 6 FIGURE 9.9 = 9 Notice that in order to keep sec θ >0, we can restrict θ to lie in the interval π < θ < π,sothat π <θ< π. Observe that with this range of values of θ, the hperbola is drawn eactl once, where r = sec θ corresponds to the right branch of the hperbola and r = sec θ corresponds to the left branch.

6 9- SECTION 9... Polar Coordinates 77 q p w p FIGURE 9.0a = sin plotted in rectangular coordinates FIGURE 9.0b The circle r = sin θ EXAMPLE.7 A Surprisingl Simple Polar Graph Sketch the graph of the polar equation r = sin θ. Solution For reference, we first sketch a graph of the sine function in rectangular coordinates on the interval [0, π] (see Figure 9.0a). Notice that on the interval 0 θ π, sin θ increases from 0 to its maimum value of. This corresponds to a polar arc in the first quadrant from the origin (r = 0) to unit up on the -ais. Then, on the interval π θ π, sin θ decreases from to 0. This corresponds to an arc in the second quadrant, from unit up on the -ais back to the origin. Net, on the interval π θ π, sin θ decreases from 0 to its minimum value of. Since the values of r are negative, remember that this means that the points plotted are in the opposite quadrant (i.e., the first quadrant). Notice that this traces out the same curve in the first quadrant as we ve alread drawn for 0 θ π. Likewise, taking θ in the interval π θ π retraces the portion of the curve in the second quadrant. Since sin θ is periodic of period π, taking further values of θ simpl retraces portions of the curve that we have alread drawn. A sketch of the polar graph is shown in Figure 9.0b. We now verif that this curve is actuall a circle. Notice that if we multipl the equation r = sin θ through b r, weget r = r sin θ. You should immediatel recognize from (.) and (.) that = r sin θ and r = +. This gives us the rectangular equation + = or 0 = +. Completing the square, we get 0 = + ( + ) 0 or, adding to both sides, ( ) ( = +. ) This is the rectangular equation for the circle of radius centered at the point ( 0, ), which is what we see in Figure 9.0b. The graphs of man polar equations are not the graphs of an functions of the form = f (), as in eample.8. EXAMPLE.8 An Archimedian Spiral 0 0 FIGURE 9. The spiral r = θ,θ 0 0 Sketch the graph of the polar equation r = θ, for θ 0. Solution Notice that here, as θ increases, so too does r. That is, as the polar angle increases, the distance from the origin also increases accordingl. This produces the spiral (an eample of an Archimedian spiral) seen in Figure 9.. The graphs shown in eamples.9,.0 and. are all in the general class known as limaçons. This class of graphs is defined b r = a ± b sin θ or r = a ± b cos θ,for positive constants a and b. Ifa = b, the graphs are called cardioids.

7 78 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-5 q p w p FIGURE 9. = + cos in rectangular coordinates 5 EXAMPLE.9 A Limaçon Sketch the graph of the polar equation r = + cos θ. Solution We begin b sketching the graph of = + cos in rectangular coordinates on the interval [0, π], to use as a reference (see Figure 9.). Notice that in this case, we have r = + cos θ>0for all values of θ. Further, the maimum value of r is 5 (corresponding to when cos θ = atθ = 0, π, etc.) and the minimum value of r is (corresponding to when cos θ = atθ = π, π, etc.). In this case, the polar graph is traced out with 0 θ π. Wesummarize the intervals of increase and decrease for r in the following table. Interval cos θ r cos θ [ ] 0, π Decreases from to 0 Decreases from 5 to [ π,π] Decreases from 0 to Decreases from to [ ] π, π Increases from to0 Increases from to [ π, π] Increases from 0 to Increases from to 5 In Figures 9.a 9.d, we show how the sketch progresses through each interval indicated in the table, with the completed figure (called a limaçon) shown in Figure 9.d. FIGURE 9.a 0 θ π FIGURE 9.b 0 θ π FIGURE 9.c 0 θ π FIGURE 9.d 0 θ π EXAMPLE.0 The Graph of a Cardioid Sketch the graph of the polar equation r = sin θ. Solution As we have done several times now, we first sketch a graph of = sin in rectangular coordinates, on the interval [0, π], as in Figure 9.. We summarize the intervals of increase and decrease in the following table.

8 9-5 SECTION 9... Polar Coordinates 79 q p w p FIGURE 9. = sin in rectangular coordinates Interval sin θ r sin θ [ ] 0, π Increases from 0 to Decreases from to 0 [ π,π] Decreases from to 0 Increases from 0 to [ ] π, π Decreases from 0 to Increases from to [ π, π] Increases from to0 Decreases from to Again, we sketch the graph in stages, corresponding to each of the intervals indicated in the table, as seen in Figures 9.5a 9.5d. 5 FIGURE 9.5a 0 θ π 5 FIGURE 9.5b 0 θ π 5 5 FIGURE 9.5c 0 θ π FIGURE 9.5d 0 θ π The completed graph appears in Figure 9.5d and is sketched out for 0 θ π.you can see wh this figure is called a cardioid ( heartlike ). EXAMPLE. ALimaçon with a Loop Sketch the graph of the polar equation r = sin θ. Solution We again begin b sketching a graph of = sin in rectangular coordinates, as in Figure 9.6. We summarize the intervals of increase and decrease in the following table.

9 750 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 q p w p FIGURE 9.6 = sin in rectangular coordinates Interval sin θ r sin θ [ ] 0, π Increases from 0 to Decreases from to [ π,π] Decreases from to 0 Increases from to [ ] π, π Decreases from 0 to Increases from to [ π, π] Increases from to0 Decreases from to Notice that since r assumes both positive and negative values in this case, we need to eercise a bit more caution, as negative values for r cause us to draw that portion of the graph in the opposite quadrant. Observe that r = 0 when sin θ = 0, that is, when sin θ =. This will occur when θ = π 6 and when θ = 5π.For this reason, we epand 6 the above table, to include more intervals and where we also indicate the quadrant where the graph is to be drawn, as follows: Interval sin θ r sin θ Quadrant [ ] 0, π Increases from 0 to Decreases from to 0 First 6 [ π, ] π Increases from to Decreases from 0 to Third 6 [ π, ] 5π Decreases from to Increases from to0 Fourth 6 [ 5π,π] Decreases from to 0 Increases from 0 to Second 6 [ ] π, π Decreases from 0 to Increases from to Third [ π, π] Increases from to0 Decreases from to Fourth We sketch the graph in stages in Figures 9.7a 9.7f, corresponding to each of the intervals indicated in the table. FIGURE 9.7a 0 θ π 6 FIGURE 9.7b 0 θ π FIGURE 9.7c 0 θ 5π 6 The completed graph appears in Figure 9.7f and is sketched out for 0 θ π.you should observe from this the importance of determining where r = 0, as well as where r is increasing and decreasing.

10 9-7 SECTION 9... Polar Coordinates 75 FIGURE 9.7d 0 θ π FIGURE 9.7e 0 θ π FIGURE 9.7f 0 θ π EXAMPLE. AFour-Leaf Rose Sketch the graph of the polar equation r = sin θ. Solution As usual, we will first draw a graph of = sin in rectangular coordinates on the interval [0, π], as seen in Figure 9.8. Notice that the period of sin θ is onl π. We summarize the intervals on which the function is increasing and decreasing in the following table. q p w FIGURE 9.8 = sin in rectangular coordinates p Interval r sin θ Quadrant [ ] 0, π Increases from 0 to First [ π, ] π Decreases from to 0 First [ π, ] π Decreases from 0 to Fourth [ π,π] Increases from to0 Fourth [ ] π, 5π [ 5π, ] π [ π, ] 7π Increases from 0 to Third Decreases from to 0 Third Decreases from 0 to Second [ 7π, π] Increases from to0 Second We sketch the graph in stages in Figures 9.9a 9.9h, each one corresponding to the intervals indicated in the table, where we have also indicated the lines =±,asa guide. This is an interesting curve known as a four-leaf rose. Notice again the significance of the points corresponding to r = 0, or sin θ = 0. Also, notice that r reaches a maimum of when θ = π, 5π,...or θ = π, 5π,...and r reaches a minimum of when θ = π, 7π,...or θ = π, 7π,...Again, ou must keep in mind that when the value of r is negative, this causes us to draw the graph in the opposite quadrant.

11 75 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 FIGURE 9.9a 0 θ π FIGURE 9.9b 0 θ π FIGURE 9.9c 0 θ π FIGURE 9.9d 0 θ π FIGURE 9.9e 0 θ 5π FIGURE 9.9f 0 θ π FIGURE 9.9g 0 θ 7π FIGURE 9.9h 0 θ π Note that in eample., even though the period of the function sin θ is π, ittook θ-values ranging from 0 to π to sketch the entire curve r = sin θ.bcontrast, the period of the function sin θ is π, but the circle r = sin θ was completed with 0 θ π. To determine the range of values of θ that produces a graph, ou need to carefull identif important points as we did in eample.. The Trace feature found on graphing calculators can be ver helpful for getting an idea of the θ-range, but remember that such Trace values are onl approimate. You will eplore a variet of other interesting graphs in the eercises. BEYOND FORMULAS The graphics in Figures 9.5, 9.7 and 9.9 provide a good visual model of how to think of polar graphs. Most polar graphs r = f (θ) can be sketched as a sequence of connected arcs, where the arcs start and stop at places where r = 0or where a new quadrant is entered. B breaking the larger graph into small arcs, ou can use the properties of f to quickl determine where each arc starts and stops.

12 9-9 SECTION 9... Polar Coordinates 75 EXERCISES 9. WRITING EXERCISES. Suppose a point has polar representation (r,θ). Eplain wh another polar representation of the same point is ( r,θ + π).. After working with rectangular coordinates for so long, the idea of polar representations ma seem slightl awkward. However, polar representations are entirel natural in man settings. For instance, if ou were on a ship at sea and another ship was approaching ou, eplain whether ou would use a polar representation (distance and bearing) or a rectangular representation (distance east-west and distance north-south).. In eample.7, the graph (a circle) of r = sin θ is completel traced out with 0 θ π. Eplain wh graphing r = sin θ with π θ π would produce the same full circle.. Two possible advantages of introducing a new coordinate sstem are making previous problems easier to solve and allowing new problems to be solved. Give two eamples of graphs for which the polar equation is simpler than the rectangular equation. Give two eamples of polar graphs for which ou have not seen a rectangular equation. In eercises 6, plot the given polar points (r,θ) and find their rectangular representation.. (, 0). (,π). (,π) ( ) ( )., π 5. (, π) 6. 5, π In eercises 7, find all polar coordinate representations of the given rectangular point. 7. (, ) 8. (, ) 9. (0, ) 0. (, ). (, ). (, 5) In eercises 8, find rectangular coordinates for the given polar point. ( ) ( )., π., π 5. (0, ) ( ) ( ) 6., π 7. 8, π 8. (, ) 0 In eercises 9 6, sketch the graph of the polar equation and find a corresponding - equation. 9. r = 0. r =. θ = π/6. θ = π/. r = cos θ. r = cos θ 5. r = sin θ 6. r = sin θ In eercises 7 50, sketch the graph and identif all values of θ where r 0 and a range of values of θ that produces one cop of the graph. 7. r = cos θ 8. r = cos θ 9. r = sin θ 0. r = sin θ. r = + sin θ. r = cos θ. r = sin θ. r = + cos θ 5. r = + sin θ 6. r = 6 cos θ 7. r = θ 8. r = eθ/ 9. r = cos(θ π/) 0. r = sin(θ π). r = cos θ + sin θ. r = cos θ + sin θ. r = tan θ. r = θ/ θ + 5. r = + cos θ 6. r = sin θ 7. r = + sin θ 9. r = + cos θ 8. r = sin θ 50. r = cos θ 5. Graph r = cos θ sin θ and eplain wh there is no curve to the left of the -ais. 5. Graph r = θ cos θ for π θ π. Eplain wh this is called the Garfield curve. 5. Based on our graphs in eercises and, conjecture the graph of r = a cos θ for an positive constant a. 5. Based on our graphs in eercises 5 and 6, conjecture the graph of r = a sin θ for an positive constant a. 55. Based on the graphs in eercises 7 and 8 and others (tr r = cos θ and r = cos 5θ), conjecture the graph of r = cos nθ for an positive integer n. 56. Based on the graphs in eercises 9 and 0 and others (tr r = sin θ and r = sin 5θ), conjecture the graph of r = sin nθ for an positive integer n. In eercises 57 6, find a polar equation corresponding to the given rectangular equation. 57. = = = = 6. = 6. =

13 75 CHAPTER 9.. Parametric Equations and Polar Coordinates Sketch the graph of r = cos θ first for 0 θ π, then for 0 θ π, then for 0 θ π,..., and finall for 0 θ π. Discuss an patterns that ou find and predict what will happen for larger domains. 6. Sketch the graph of r = cos πθ first for 0 θ, then for 0 θ, then for 0 θ,... and finall for 0 θ 0. Discuss an patterns that ou find and predict what will happen for larger domains. 65. One situation where polar coordinates appl directl to sports is in making a golf putt. The two factors that the golfer tries to control are distance (determined b speed) and direction (usuall called the line ). Suppose a putter is d feet from the hole, which has radius h = 6. Show that the path of the ball will intersect the hole if the angle A in the figure satisfies sin (h/d) < A < sin (h/d). (r, A) (d, 0) (0, 0) A 66. The distance r that the golf ball in eercise 65 travels also needs to be controlled. The ball must reach the front of the hole. In rectangular coordinates, the hole has equation ( d) + = h, so the left side of the hole is = d h. Show that this converts in polar coordinates to r = d cos θ d cos θ (d h ). (Hint: Substitute for and, isolate the square root term, square both sides, combine r terms and use the quadratic formula.) 67. The golf putt in eercises 65 and 66 will not go in the hole if it is hit too hard. Suppose that the putt would go r = d + c feet if it did not go in the hole (c > 0). For a putt hit toward the center of the hole, define b to be the largest value of c such that the putt goes in (i.e., if the ball is hit more than b feet past the hole, it is hit too hard). Eperimental evidence (see Dave Pelz s Putt Like the Pros) shows ( that at other angles A, the distance r [ ] ) A must be less than d + b. The results of sin (h/d) eercises 65 and 66 define limits for the angle A and distance r of a successful putt. Identif the functions r (A) and r (A) such that r (A) < r < r (A) and constants A and A such that A < A < A. 68. Take the general result of eercise 67 and appl it to a putt of d = 5 feet with a value of b = feet. Visualize this b graphing the region 5 cos θ 5 cos θ (5 /6) ( [ ] ) θ < r < 5 + sin (/90) with sin (/90) <θ<sin (/90). A good choice of graphing windows is.8 9 and EXPLORATORY EXERCISES. In this eercise, ou will eplore the roles of the constants a, b and c in the graph of r = af(bθ + c). To start, sketch r = sin θ followed b r = sin θ and r = sin θ. What does the constant a affect? Then sketch r = sin(θ + π/) and r = sin(θ π/). What does the constant c affect? Now for the tough one. Sketch r = sin θ and r = sin θ. What does the constant b seem to affect? Test all of our hpotheses on the base function r = + cos θ and several functions of our choice.. The polar curve r = ae bθ is sometimes called an equiangular curve. To see wh, sketch the curve and then show that dr = br. Asomewhat complicated geometric argument dθ shows that dr = r cot α, where α is the angle between the dθ tangent line and the line connecting the point on the curve to the origin. Comparing equations, conclude that the angle α is constant (hence equiangular ). To illustrate this propert, compute α for the points at θ = 0 and θ = π for r = e θ. This tpe of spiral shows up often in nature, possibl because the equal-angle propert can be easil achieved. Spirals can be found among shellfish (the picture shown here is of an ammonite fossil from about 50 million ears ago) and the florets of the common dais. Other eamples, including the connection to sunflowers, the Fibonacci sequence and the musical scale, can be found in H. E. Huntle s The Divine Proportion.

14 9- SECTION Calculus and Polar Coordinates CALCULUS AND POLAR COORDINATES Having introduced polar coordinates and looked at a variet of polar graphs, our net step is to etend the techniques of calculus to the case of polar coordinates. In this section, we focus on tangent lines, area and arc length. Surface area and other applications will be eamined in the eercises. Notice that ou can think of the graph of the polar equation r = f (θ)asthe graph of the parametric equations = f (t) cos t, (t) = f (t) sin t (where we have used the parameter t = θ), since from (.) = r cos θ = f (θ) cos θ (5.) and = r sin θ = f (θ) sin θ. (5.) In view of this, we can now take an results alread derived for parametric equations and etend these to the special case of polar coordinates. In section 9., we showed that the slope of the tangent line at the point corresponding to θ = a is given [from (.)] to be d d = θ=a From the product rule, (5.) and (5.), we have d dθ = f (θ) sin θ + f (θ) cos θ d dθ (a) d dθ (a). (5.) d and dθ = f (θ) cos θ f (θ) sin θ. Putting these together with (5.), we get d d = f (a) sin a + f (a) cos a θ=a f (a) cos a f (a) sin a. (5.) EXAMPLE 5. Finding the Slope of the Tangent Line to a Three-Leaf Rose Find the slope of the tangent line to the three-leaf rose r = sin θ at θ = 0 and θ = π Solution Asketch of the curve is shown in Figure 9.0a. From (.), we have = r sin θ = sin θ sin θ and = r cos θ = sin θ cos θ. FIGURE 9.0a Three-leaf rose Using (5.), we have d d = d dθ d dθ = ( cos θ) sin θ + sin θ(cos θ) ( cos θ) cos θ sin θ(sin θ). (5.5)

15 756 CHAPTER 9.. Parametric Equations and Polar Coordinates At θ = 0, this gives us d ( cos 0) sin 0 + sin 0(cos 0) d = θ=0 ( cos 0) cos 0 sin 0(sin 0) = 0 = 0. In Figure 9.0b, we sketch r = sin θ for 0. θ 0., in order to isolate the portion of the curve around θ = 0. Notice that from this figure, a slope of 0 seems reasonable. Similarl, at θ = π,wehave from (5.5) that ( cos π ) sin π d + sin π ( cos π ) d = ( θ=π/ cos π ) cos π sin π ( sin π ) = + =. FIGURE 9.0b 0. θ FIGURE 9.0c The tangent line at θ = π d q f FIGURE 9.a = cos sin + sin cos p In Figure 9.0c, we show the section of r = sin θ for 0 θ π, along with the tangent line at θ = π. Recall that for functions = f (), horizontal tangents were especiall significant for locating maimum and minimum points. For polar graphs, the significant points are often places where r has reached a maimum or minimum, which ma or ma not correspond to a horizontal tangent. We eplore this idea further in eample 5.. EXAMPLE 5. Polar Graphs and Horizontal Tangent Lines For the three-leaf rose r = sin θ, find the locations of all horizontal tangent lines and interpret the significance of these points. Further, at the three points where r is a maimum, show that the tangent line is perpendicular to the line segment connecting the point to the origin. Solution From (5.) and (5.), we have d d = d dθ d dθ = f (θ) sin θ + f (θ) cos θ f (θ) cos θ f (θ) sin θ. Here, f (θ) = sin θ and so, to have d = 0, we must have d 0 = d = cos θ sin θ + sin θ cos θ. dθ Solving this equation is not an eas matter. As a start, we graph f () = cos sin + sin cos with 0 π (see Figure 9.a). You should observe that there appear to be five solutions. Three of the solutions can be found eactl: θ = 0,θ = π and θ = π. You can find the remaining two numericall: θ and θ.8. (You can also use trig identities to arrive at sin θ =.) The corresponding 8 points on the curve r = sin θ (specified in rectangular coordinates) are (0, 0), (0.7, 0.56), (0, ), ( 0.7, 0.56) and (0, 0). The point (0, ) lies at the bottom of a leaf. This is the familiar situation of a horizontal tangent line at a local (and in fact, absolute) minimum. The point (0, 0) is a little more trick to interpret. As seen in Figure 9.0b, if we graph a small piece of the curve with θ near 0 (or π), the point (0, 0) is a minimum point. However, this is not true for other values of θ (e.g., π ) where the curve passes through the point (0, 0). The tangent lines at the points (±0.7, 0.56) are shown in Figure 9.b. Note that these points correspond to points where the -coordinate is a maimum. However, referring to the graph, these points do not appear to be of particular

16 9- SECTION Calculus and Polar Coordinates FIGURE 9.b Horizontal tangent lines FIGURE 9.c The tangent line at the tip of a leaf r u FIGURE 9. Circular sector interest. Rather, the tips of the leaves represent the etreme points of most interest. Notice that the tips are where r is a maimum. For r = sin θ, this occurs when sin θ =±, that is, where θ = π, π, 5π,...,or θ = π 6, π, 5π,... From (5.), 6 the slope of the tangent line to the curve at θ = π is given b 6 ( cos π ) sin π d sin π ( cos π ) d = ( θ=π/6 cos π ) cos π 6 6 sin π ( sin π ) = 0 =. 6 6 The rectangular point corresponding to θ = π is given b 6 ( cos π 6, sin π ) ( ) = 6,. The slope of the line segment joining this point to the origin is then. Observe that the line segment from the origin to the point is perpendicular to the tangent line since the product of the slopes ( and )is. This is illustrated in Figure 9.c. Similarl, the slope of the tangent line at θ = 5π 6 is, which again makes the tangent line at that point perpendicular to the line segment from the origin to the point (, ). Finall, we have alread shown that the slope of the tangent line at θ = π is 0 and a horizontal tangent line is perpendicular to the vertical line from the origin to the point (0, ). Net, for polar curves like the three-leaf rose seen in Figure 9.0a, we would like to compute the area enclosed b the curve. Since such a graph is not the graph of a function of the form = f (), we cannot use the usual area formulas developed in Chapter 5. While we can convert our area formulas for parametric equations (from Theorem.) into polar coordinates, a simpler approach uses the following geometric argument. Observe that a sector of a circle( of radius ) r and central angle θ, measured in radians θ (see Figure 9.) contains a fraction of the area of the entire circle. So, the area of π the sector is given b A = πr θ π = r θ. Now, consider the area enclosed b the polar curve defined b the equation r = f (θ) and the ras θ = a and θ = b (see Figure 9.a), where f is continuous and positive on the interval a θ b. Aswedid when we defined the definite integral, we begin b partitioning the

17 758 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- u b u b r f(u) u a A i u u i u u i u a r f(u) FIGURE 9.a Area of a polar region θ-interval into n equal pieces: FIGURE 9.b Approimating the area of a polar region a = θ 0 <θ <θ < <θ n = b. The width of each of these subintervals is then θ = θ i θ i = b a. (Does this look n familiar?) On each subinterval [θ i,θ i ](i =,,...,n), we approimate the curve with the circular arc r = f (θ i ) (see Figure 9.b). The area A i enclosed b the curve on this subinterval is then approimatel the same as the area of the circular sector of radius f (θ i ) and central angle θ: A i r θ = [ f (θ i)] θ. The total area A enclosed b the curve is then approimatel the same as the sum of the areas of all such circular sectors: n n A A i = [ f (θ i)] θ. i= i= As we have done numerous times now, we can improve the approimation b making n larger. Taking the limit as n gives us a definite integral: Area in polar coordinates A = lim n n i= b [ f (θ i)] θ = a [ f (θ)] dθ. (5.6) EXAMPLE 5. The Area of One Leaf of a Three-Leaf Rose Find the area of one leaf of the rose r = sin θ. Solution Notice that one leaf of the rose is traced out with 0 θ π (see Figure 9.). From (5.6), the area is given b π/ A = 0 (sin θ) dθ = π/ sin θ dθ 0 = π/ ( cos 6θ) dθ = (θ 6 ) π/ sin 6θ = π, 0 0 FIGURE 9. One leaf of r = sin θ where we have used the half-angle formula sin α = ( cos α) tosimplif the integrand.

18 9-5 SECTION Calculus and Polar Coordinates 759 Often, the most challenging part of finding the area of a polar region is determining the limits of integration. 5 FIGURE 9.5 r = sin θ EXAMPLE 5. The Area of the Inner Loop of a Limaçon Find the area of the inner loop of the limaçon r = sin θ. Solution Asketch of the limaçon is shown in Figure 9.5. Starting at θ = 0, the curve starts at the point (, 0), passes through the origin, traces out the inner loop, passes back through the origin and finall traces out the outer loop. Thus, the inner loop is formed b θ-values between the first and second occurrences of r = 0 with θ>0. Solving r = 0, we get sin θ =. The two smallest positive solutions are θ = ( sin ) and θ = π sin ( ). Numericall, these are approimatel equal to θ = 0.7 and θ =.. From (5.6), the area is approimatel A = ( sin θ) dθ = 0.7 [ sin θ + 9 ] ( cos θ) dθ 0., ( sin θ + 9 sin θ) dθ u i where we have used the half-angle formula sin θ = ( cos θ) tosimplif the integrand. (Here the area is approimate, owing onl to the approimate limits of integration.) When finding the area ling between two polar graphs, we use the familiar device of subtracting one area from another. Although the calculations in eample 5.5 aren t too mess, finding the points of intersection of two polar curves often provides the greatest challenge. u i u i FIGURE 9.6a r = + cos θ and r = u o 5 FIGURE 9.6b π θ π EXAMPLE 5.5 Finding the Area between Two Polar Graphs Find the area inside the limaçon r = + cos θ and outside the circle r =. Solution We show a sketch of the two curves in Figure 9.6a. Notice that the limits of integration correspond to the values of θ where the two curves intersect. So, we must first solve the equation + cos θ =. Notice that since cos θ is periodic, there are infinitel man solutions of this equation. Consequentl, it is essential to consult the graph to determine which solutions ou are interested in. In this case, we want the least negative and the smallest positive solutions. (Look carefull at Figure 9.6b, where we have shaded the area between the graphs corresponding to θ between π and π, the first two positive solutions. This portion of the graphs corresponds to the area outside the limaçon and inside the circle!) With + cos θ =, we have cos θ =, which occurs at θ = π and θ = π. From (5.6), the area enclosed b the portion of the limaçon on this interval is given b π/ ( + cos θ) dθ = + π. 6 π/ Similarl, the area enclosed b the circle on this interval is given b π/ π/ () dθ = 8π.

19 760 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 The area inside the limaçon and outside the circle is then given b π/ π/ A = π/ ( + cos θ) dθ π/ () dθ = + π 6 8π = + 8π 6.. Here, we have left the (routine) details of the integrations to ou. FIGURE 9.7a r = cos θ and r = sin θ d q f FIGURE 9.7b Rectangular plot: = cos, = sin, 0 π q p w p p FIGURE 9.7c Rectangular plot: = cos, = sin, 0 π In cases where r takes on both positive and negative values, finding the intersection points of two curves is more complicated. EXAMPLE 5.6 Finding Intersections of Polar Curves Where r Can Be Negative Find all intersections of the limaçon r = cos θ and the circle r = sin θ. Solution We show a sketch of the two curves in Figure 9.7a. Notice from the sketch that there are three intersections of the two curves. Since r = sin θ is completel traced with 0 θ π, ou might reasonabl epect to find three solutions of the equation cos θ = sin θ on the interval 0 θ π. However,ifwedrawa rectangular plot of the two curves = cos and = sin, onthe interval 0 π (see Figure 9.7b), we can clearl see that there is onl one solution in this range, at approimatel θ.99. (Use Newton s method or our calculator s solver to obtain an accurate approimation.) The corresponding rectangular point is (r cos θ,r sin θ) ( 0.7,.67). Looking at Figure 9.7a, observe that there is another intersection located below this point. One wa to find this point is to look at a rectangular plot of the two curves corresponding to an epanded range of values of θ (see Figure 9.7c). Notice that there is a second solution of the equation cos θ = sin θ, near θ = 5.86, which corresponds to the point ( 0.7, 0.). Note that this point is on the inner loop of r = cos θ and corresponds to a negative value of r. Finall, there appears to be a third intersection at or near the origin. Notice that this does not arise from an solution of the equation cos θ = sin θ. This is because, while both curves pass through the origin (You should verif this!), the each do so for different values of θ. (Keep in mind that the origin corresponds to the point (0,θ), in polar coordinates, for an angle θ.) Notice that cos θ = 0 for θ = π and sin θ = 0 for θ = 0. So, although the curves intersect at the origin, the each pass through the origin for different values of θ. REMARK 5. To find points of intersection of two polar curves r = f (θ) and r = g(θ), ou must keep in mind that points have more than one representation in polar coordinates. In particular, this sas that points of intersection need not correspond to solutions of f (θ) = g(θ). In eample 5.7, we see an application that is far simpler to set up in polar coordinates than in rectangular coordinates.

20 9-7 SECTION Calculus and Polar Coordinates 76 FIGURE 9.8a Aclindrical oil tank FIGURE 9.8b Cross section of tank.8 EXAMPLE 5.7 Finding the Volume of a Partiall Filled Clinder Aclindrical oil tank with a radius of feet is ling on its side. A measuring stick shows that the oil is.8 feet deep (see Figure 9.8a). What percentage of a full tank is left? Solution Notice that since we wish to find the percentage of oil remaining in the tank, the length of the tank has no bearing on this problem. (Think about this some.) We need onl consider a cross section of the tank, which we represent as a circle of radius centered at the origin. The proportion of oil remaining is given b the area of that portion of the circle ling beneath the line = 0., divided b the total area of the circle. The area of the circle is π,soweneed onl find the area of the shaded region in Figure 9.8b. Computing this area in rectangular coordinates is a mess (tr it!), but it is straightforward in polar coordinates. First, notice that the line = 0. corresponds to r sin θ = 0.orr = 0. csc θ. The area beneath the line and inside the circle is then given b (5.6) as θ θ Area = θ () dθ θ ( 0. csc θ) dθ, where θ and θ are the appropriate intersections of r = and r = 0. csc θ. Using Newton s method, the first two positive solutions of = 0. csc θ are θ. and θ 6.8. The area is then θ θ Area = θ () dθ θ ( 0. csc θ) dθ = (θ cot θ) θ θ The fraction of oil remaining in the tank is then approimatel 5.85/π 0.68 or about.6% of the total capacit of the tank. We close this section with a brief discussion of arc length for polar curves. Recall that from (.), the arc length of a curve defined parametricall b = (t), = (t), for a t b,isgivenb s = b a (d ) + dt ( ) d dt. (5.7) dt Once again thinking of a polar curve as a parametric representation (where the parameter is θ), we have that for the polar curve r = f (θ), This gives us ( ) d + dθ = r cos θ = f (θ) cos θ and = r sin θ = f (θ) sin θ. ( ) d = [ f (θ) cos θ f (θ) sin θ] + [ f (θ) sin θ + f (θ) cos θ] dθ = [ f (θ)] (cos θ + sin θ) + f (θ) f (θ)( cos θ sin θ + sin θ cos θ) + [ f (θ)] (cos θ + sin θ) = [ f (θ)] + [ f (θ)].

21 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 From (5.7), the arc length is then Arc length in polar coordinates s = b a [ f (θ)] + [ f (θ)] dθ. (5.8) EXAMPLE 5.8 Arc Length of a Polar Curve Find the arc length of the cardioid r = cos θ. Solution Asketch of the cardioid is shown in Figure 9.9. First, notice that the curve is traced out with 0 θ π. From (5.8), the arc length is given b 5 FIGURE 9.9 r = cos θ s = = b a π 0 [ f (θ)] + [ f (θ)] dθ = π sin θ + 8 cos θ + cos θ dθ = 0 ( sin θ) + ( cos θ) dθ π cos θ dθ = 6, where we leave the details of the integration as an eercise. (Hint: Use the half-angle formula sin = ( cos ) tosimplif the integrand. Be careful: remember that =!) EXERCISES 9.5 WRITING EXERCISES. Eplain wh the tangent line is perpendicular to the radius line at an point at which r is a local maimum. (See eample 5..) In particular, if the tangent and radius are not perpendicular at (r,θ), eplain wh r is not a local maimum.. In eample 5.5, eplain wh integrating from π to π would give the area shown in Figure 9.6b and not the desired area.. Referring to eample 5.6, eplain wh intersections can occur in each of the cases f (θ) = g(θ), f (θ) = g(θ + π) and f (θ ) = g(θ ) = 0.. In eample 5.7, eplain wh the length of the tank doesn t matter. If the problem were to compute the amount of oil left, would the length matter? In eercises 0, find the slope of the tangent line to the polar curve at the given point.. r = sin θ at θ = π. r = sin θ at θ = π. r = cos θ at θ = 0. r = cos θ at θ = π 5. r = sin θ at θ = 0 6. r = sin θ at θ = π 7. r = sin θ at θ = π 8. r = sin θ at θ = π 6 9. r = cos θ at θ = π 6 0. r = cos θ at θ = π In eercises, find all points at which r is a maimum and show that the tangent line is perpendicular to the radius connecting the point to the origin.. r = sin θ. r = cos θ. r = sin θ. r = + sin θ In eercises 5 0, find the area of the indicated region. 5. One leaf of r = cos θ 6. One leaf of r = sin θ 7. Inner loop of r = sin θ 8. Inner loop of r = cos θ 9. Bounded b r = cos θ 0. Bounded b r = cos θ. Small loop of r = + sin θ. Large loop of r = + sin θ. Inner loop of r = + sin θ. Outer loop of r = + sin θ 5. Inside of r = + sin θ and outside of r = 6. Inside of r = and outside of r = sin θ

22 9-9 SECTION Calculus and Polar Coordinates Inside of r = and outside of both loops of r = + sin θ 8. Inside of r = sin θ and outside r = 9. Inside of both r = + cos θ and r = 0. Inside of both r = + sin θ and r = + cos θ In eercises, find all points at which the two curves intersect.. r = sin θ and r = cos θ. r = + cos θ and r = + 5 sin θ. r = + sin θ and r = + cos θ. r = + sin θ and r = + cos θ In eercises 5 0, find the arc length of the given curve. 5. r = sin θ 6. r = cos θ 7. r = sin θ 8. r = cos θ 9. r = + sin θ 0. r = + sin θ. Repeat eample 5.7 for the case where the oil stick shows a depth of... Repeat eample 5.7 for the case where the oil stick shows a depth of.0.. Repeat eample 5.7 for the case where the oil stick shows a depth of... Repeat eample 5.7 for the case where the oil stick shows a depth of The problem of finding the slope of r = sin θ at the point (0, 0) is not a well-defined problem. To see what we mean, show that the curve passes through the origin at θ = 0,θ = π and θ = π, and find the slopes at these angles. Briefl eplain wh the are different even though the point is the same. 6. For each of the three slopes found in eercise 5, illustrate with asketch of r = sin θ for θ-values near the given values (e.g., π θ π to see the slope at θ = 0) If the polar curve r = f (θ), a θ b, has length L, show that r = cf(θ), a θ b, has length c L for an constant c. 8. If the polar curve r = f (θ), a θ b, encloses area A, show that for an constant c, r = cf(θ), a θ b, encloses area c A. 9. A logarithmic spiral is the graph of r = ae bθ for positive constants a and b. The accompaning figure shows the case where a = and b = with θ. Although the graph never reaches the origin, the limit of the arc length from θ = d to a given point with θ = c,asd decreases to,eists. Show that this total b + arc length equals R, where R is the distance from the b starting point to the origin For the logarithmic spiral of eercise 9, if the starting point P is on the -ais, show that the total arc length to the origin equals the distance from P to the -ais along the tangent line to the curve at P. EXPLORATORY EXERCISES. In this eercise, ou will discover a remarkable propert about the area underneath the graph of =. First, show that a polar representation of this curve is r =.Wewill sin θ cos θ find the area bounded b =, = m and = m for > 0, where m is a positive constant. Sketch graphs for m = (the area bounded b =, = and = ) and m = (the area bounded b =, = and = ). Which area looks larger? To find out, ou should integrate. Eplain wh this would be a ver difficult integration in rectangular coordinates. Then convert all curves to polar coordinates and compute the polar area. You should discover that the area equals ln for an value of m. (Are ou surprised?). In the stud of biological oscillations (e.g., the beating of heart cells), an important mathematical term is limit ccle. A simple eample of a limit ccle is produced b the polar coordinates initial value problem dr dt = ar( r), r(0) = r 0 and dθ = π, θ(0) = θ 0. Here, a is a positive constant. In section dt 7., we showed that the solution of the initial value problem dr dt = ar( r), r(0) = r 0 is r 0 r(t) = r 0 (r 0 )e at and it is not hard to show that the solution of the initial value problem dθ = π, θ(0) = θ 0 is θ(t) = πt + θ 0.Inrectangular coordinates, the solution of the combined initial dt value