9.5 CALCULUS AND POLAR COORDINATES

Transcription

1 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates 9.5 CALCULUS AND POLAR COORDINATES Now that we have introduced ou to 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) as the 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. We start b computing the slope of a tangent line to the polar curve r = f (θ). From our stud of parametric equations in section 9., we know that the slope of the tangent line at the point corresponding to θ = a is given [from (.)] to be d d = θ=a d dθ (a). d dθ (a) (5.) From the product rule, (5.) and (5.), we have d dθ = f (θ) sin θ + f (θ) cos θ and d 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.) Eample 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. 0.5 Solution A sketch of the curve is shown in Figure 9.9a. From (.), we have and = r sin θ = sin θ sin θ = r cos θ = sin θ cos θ. Using (5.), we have Figure 9.9a Three-leaf rose. d d = d dθ d dθ = ( cos θ)sin θ + sin θ(cos θ) ( cos θ)cos θ sin θ(sin θ). (5.5)

2 smi9885_ch09b.qd 5/7/0 :5 PM Page 76 Section 9.5 Calculus 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 = Figure 9.9b 0. θ In Figure 9.9b, 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, we have from (5.5) that ( cos π ) sin π d + sin π ( cos π ) d = ( θ=π/ cos π ) cos π sin π ( sin π ) = + =. In Figure 9.9c, we show the section of r = sin θ for 0 θ π, along with the tangent line at θ = π Figure 9.9c 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 the following eample. Eample 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. d q f Figure 9.0a = cos sin + sin cos. p 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 9.0a). 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. The corresponding 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 9.9b, 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

3 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates Figure 9.0b Horizontal tangent lines Figure 9.0c The tangent line at the tip of a leaf. r u Figure 9. Circular sector. the curve passes through the point (0, 0). The tangent lines at the points (±0.7, 0.56) are shown in Figure 9.0b. 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 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 6,... From (5.), the slope of the tangent line to the curve at θ = π 6 is given b ( 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 6 is given b ( cos π 6, sin π ) ( = 6, ). A quick calculation shows that the slope of the line segment joining this point to the origin is. 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 9.0c. 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, ). For polar curves like the three-leaf rose seen in Figure 9.9a, it is reasonable to epect that we could 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. However, we can convert our area formulas for parametric equations (from Theorem.) into polar coordinates. A simpler approach to get the same result uses the following geometric argument. First, recall that the area of a circle of radius r is πr. Net, consider a sector of a circle of radius r and central angle θ, measured in radians (see Figure 9.). Notice that the sector contains a fraction given b ( θ π ) of the area of the entire circle and so, the area of the sector is 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. As we did when we defined the definite integral, we begin b partitioning the θ-interval into n equal pieces: 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

4 smi9885_ch09b.qd 5/7/0 :5 PM Page 76 Section 9.5 Calculus and Polar Coordinates 76 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. Figure 9.b Approimating the area of a polar region. 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: A n A i = i= n i= [ f (θ i)] θ. As we have seen 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) Eample 5. Find the area of one leaf of the rose r = sin θ. The Area of One Leaf of a Three-Leaf Rose Figure 9. One leaf of r = sin θ. Solution Notice that one leaf of the rose is traced out with 0 θ π (see Figure 9.). Notice that 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 where we have used the half-angle formula sin α = ( cos α) to simplif the integrand. Often, the most challenging part of finding the area of a polar region is determining the limits of integration. 0

5 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates 5 Eample 5. The Area of the Inner Loop of a Limaçon Find the area of the inner loop of the limaçon r = sin θ. Solution A sketch of the limaçon is shown in Figure 9.. 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 smallest two 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 Figure 9. r = sin θ. Figure 9.5a r = + cos θ and r =. 5 Figure 9.5b π θ π. A = ( sin θ) dθ = 0.7 [ sin θ + 9 ] ( cos θ) dθ 0., ( sin θ + 9 sin θ)dθ where we have used the half-angle formula sin θ = ( cos θ) to simplif 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 the following eample aren t too mess, finding the points of intersection of two polar curves often provides the greatest challenge. Eample 5.5 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 9.5a. 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 absolutel necessar 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 9.5b, 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/ Finding the Area between Two Polar Graphs Similarl, the area enclosed b the circle on this interval is given b π/ π/ () dθ = 8π.

6 smi9885_ch09b.qd 5/7/0 :5 PM Page 765 Section 9.5 Calculus and Polar Coordinates 765 Figure 9.6a r = cos θ and r = sin θ. d q f Figure 9.6b Rectangular plot: = cos, = sin, 0 π. q p w p p Figure 9.6c Rectangular plot: = cos, = sin, 0 π. The area inside the limaçon and outside the circle is then given b A = π/ π/ = + π 6 ( + cos θ) dθ π/ π/ 8π = + 8π 6 () dθ Here, we have left the (routine) details of the integrations to ou... In cases where r takes on both positive and negative values, finding the intersection points of two curves is more complicated. 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 9.6a. 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, if we draw a rectangular plot of the two curves: = cos and = sin, on the interval 0 π (see Figure 9.6b), we can clearl see that there is onl one solution in this range, at approimatel θ.99. (Recall that ou can 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 9.6a, 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 9.6c). 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.you need to realize 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. Eample 5.6 Remark 5. Finding Intersections of Polar Curves Where r Can be Negative 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 the following eample, we see an application that is far simpler to set up in polar coordinates than in rectangular coordinates.

7 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates Eample 5.7 Finding the Volume of a Partiall Filled Clinder A clindrical 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 9.7a). What percentage of a full tank is left? Figure 9.7a A clindrical oil tank..8 Figure 9.7b Cross section of tank. Solution Notice that since we wish to find onl 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, so we need onl find the area of the shaded region in Figure 9.7b. 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. or r = 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, is given b 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), = r cos θ = f (θ) cos θ and = r sin θ = f (θ) sin θ. This gives us ( ) d ( ) d + = [ f (θ) cos θ f (θ) sin θ] + [ f (θ) sin θ + f (θ) cos θ] dθ 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)].

8 smi9885_ch09b.qd 5/7/0 :5 PM Page 767 Section 9.5 Calculus and Polar Coordinates 767 From (5.7), we get that the arc length is given b Arc length in polar coordinates s = b a [ f (θ)] + [ f (θ)] dθ. (5.8) Eample 5.8 Arc Length of a Polar Curve Find the arc length of the cardioid r = cos θ. 5 Figure 9.8 r = cos θ. Solution A sketch of the cardioid is shown in Figure 9.8. First, notice that the curve is traced out with 0 θ π. From (5.8), the arc length is given b b π s = [ f (θ)] + [ f (θ)] dθ = ( sin θ) + ( cos θ) dθ = a π 0 sin θ + 8 cos θ + cos θ dθ = 0 π cos θ dθ = 6, where we leave the details of the integration as an eercise. (Hint: Use the half-angle formula: sin = ( cos ) to simplif the integrand. Be careful: remember that =!) EXERCISES 9.5. 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. 9. r = sin θ at θ = 0 0. r = sin θ at θ = π. r = sin θ at θ = π. r = sin θ at θ = π 6. r = cos θ at θ = π 6. r = cos θ at θ = π. In eample 5.5, eplain wh integrating from π to π would give the area shown in Figure 9.5b, 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 5, find the slope of the tangent line to the polar curve at the given point. 5. r = sin θ at θ = π 6. r = sin θ at θ = π 7. r = cos θ at θ = 0 8. r = cos θ at θ = π In eercises 5 8, 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. 5. r = sin θ 6. r = cos θ 7. r = sin θ 8. r = + sin θ In eercises 9, find the area of the indicated region. 9. One leaf of r = cos θ 0. One leaf of r = sin θ. Inner loop of r = sin θ. Inner loop of r = cos θ

9 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates. Bounded b r = cos θ. Bounded b r = cos θ 5. Small loop of r = + sin θ 6. Large loop of r = + sin θ 7. Inner loop of r = + sin θ 8. Outer loop of r = + sin θ 9. Inside of r = + sin θ and outside of r = 0. Inside of r = and outside of r = sin θ. Inside of r = and outside of r = + sin θ. Inside of r = sin θ and outside r =. Inside of both r = + cos θ and r =. Inside of both r = + sin θ and r = + cos θ In eercises 5 8, find all points at which the two curves intersect. 5. r = sin θ and r = cos θ 6. r = + cos θ and r = + 5 sin θ 7. r = + sin θ and r = + cos θ 8. r = + sin θ and r = + cos θ In eercises 9, find the arc length of the given curve. 9. r = sin θ 0. r = cos θ. r = sin θ. r = cos θ. r = + sin θ. r = + sin θ 5. Repeat eample 5.7 for the case where the oil stick shows a depth of.. 6. 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.. 8. 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. 50. For each of the three slopes found in eercise 9, illustrate with a sketch 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. 5. 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. 5. 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 =. We will find the area bounded b =, = m sin θ cos θ 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?) 5. 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 = ar( r), dt r(0) = r 0 and dθ = π, θ(0) = θ 0. Here, a is a positive dt constant. In section 6.5, 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. In rectangular coordinates, the solution of the combined initial value dt problem has parametric equations (t) = r(t) cos θ(t) and (t) = r(t) sin θ(t). Graph the solution in the cases (a) a =, r 0 =,θ 0 = 0; (b) a =, r 0 =,θ 0 = 0; (c) our choice of a > 0, our choice of r 0 with 0 < r 0 <, our choice of θ 0 ; (d) our choice of a > 0, our choice of r 0 with r 0 >, our choice of θ 0. As t increases, what is the limiting behavior of the solution? Eplain what is meant b saing that this sstem has a limit ccle of r =.

10 smi9885_ch09b.qd 5/7/0 :5 PM Page 769 Section 9.6 Conic Sections CONIC SECTIONS So far in this chapter, we have introduced a variet of interesting curves, man of which are not graphs of a function = f () in rectangular coordinates. Among the most important curves are the conic sections, which we eplore here. The conic sections include parabolas, ellipses and hperbolas, which are undoubtedl alread familiar to ou. In this section, we focus on geometric properties that are most easil determined in rectangular coordinates. We visualize each conic section as the intersection of a plane with a right circular cone (see Figures 9.9a 9.9c). Figure 9.9a Parabola. Figure 9.9b Ellipse. Figure 9.9c Hperbola. Depending on the orientation of the plane, the resulting curve can be a parabola, ellipse or hperbola. Focus Parabolas Verte Directri Figure 9.50 Parabola. We define a parabola (see Figure 9.50) to be the set of all points that are equidistant from a fied point (called the focus) and a line (called the directri).aspecial point on the parabola is the verte, the midpoint of the perpendicular segment from the focus to the directri. If the focus of the parabola lies on the -ais and the directri is a horizontal line, it is possible to find a simple rectangular equation of the parabola. Eample 6. Finding the Equation of a Parabola (0, ) Find an equation of the parabola with focus at the point (0, ) whose directri is the line =. (, ) Solution B definition, an point (, ) on the parabola must be equidistant from the focus and the directri (see Figure 9.5). From the distance formula, the distance from (, ) to the focus is given b + ( ) and the distance to the directri is ( ). Since these distances must be equal, the parabola is defined b the equation + ( ) = +. Figure 9.5 The parabola with focus at (0, ) and directri =. Squaring both sides, we get + ( ) = ( + ).

11 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates Epanding this out and simplifing, we get + + = + + or = 8. Solving for gives us = 8. In general, the following relationship holds. Theorem 6. The parabola with verte at the point (b, c), focus at ( b, c + ) a and directri given b the line = c a is described b the equation = a( b) + c. (b, c a ) (b, c) a a c a Figure 9.5 Parabola. Proof Given the focus ( b, c + ) a and directri = c a, the verte is the midpoint (b, c) (see Figure 9.5). For an point (, ) on the parabola, its distance to the focus is given b ( b) + ( c ), a while its distance to the directri is given b c + a. Setting these equal and squaring as in eample 6., we have ( ( b) + c ) ( = c + ). a a Epanding this out and simplifing, we get the more familiar form of the equation: = a( b) + c, as desired. Notice that the roles of and can be reversed. We leave the proof of the following result as an eercise. Theorem 6. The parabola with verte at the point (c, b), focus at ( c + a, b) and directri given b the line = c a is described b the equation = a( b) + c. We illustrate Theorem 6. in the following eample. 5 Eample 6. A Parabola Opening to the Left For the parabola = 0, find the verte, focus and directri. Figure 9.5 Parabola with focus at ( 5, 0) and directri =. Solution To put this into the form of the equation given in Theorem 6., we must first solve for. We have =. The verte is then at (, 0). The focus and directri are shifted left and right, respectivel from the verte b a =. This puts the focus at (, 0) = ( 5, 0) and the directri at = ( ) =. We show a sketch of the parabola in Figure 9.5.

12 smi9885_ch09b.qd 5/7/0 :5 PM Page 77 Section 9.6 Conic Sections 77 Eample 6. Finding the Equation of a Parabola 8 6 (, ) (, ) Find an equation relating all points that are equidistant from the point (, ) and the line = 6. Solution Referring to Figure 9.5, notice that the verte must be at the point (, ) (i.e., the midpoint of the perpendicular line segment connecting the focus to the directri) and the parabola opens down. From the verte, the focus is shifted verticall b a = units, so a = ( ) = 8. An equation is then = 8 ( ) +. Figure 9.5 Parabola with focus at (, ) and directri at = (, ) 8 (, ) Figure 9.55 Parabola with focus at (, ) and directri at =. Find an equation relating all points that are equidistant from the point (, ) and the line =. Solution Referring to Figure 9.55, notice that the verte must be halfwa between the focus (, ) and the directri =, that is, at the point (, ) and the parabola opens to the right. From the verte, the focus is shifted horizontall b = a units, so that a = 8. An equation is then Eample 6. A Parabola Opening to the Right = 8 ( + ) +. You see parabolas nearl ever da. As we discussed in section 5.5, the motion of man projectiles is approimatel parabolic. In addition, parabolas have a reflective propert that is etremel useful in man important applications. This can be seen as follows. For the parabola = a indicated in Figure 9.56a, draw a horizontal line that intersects the parabola at the point A. Then, one can show that the acute angle α between the horizontal line and the tangent line at A is the same as the acute angle β between the tangent line and the line segment joining A to the focus. You ma alread have recognized that light ras are reflected from a surface in eactl the same fashion. (You ma recall that the angle of incidence must equal the angle of reflection.) In Figure 9.56b, we indicate a b Focus Focus A a Figure 9.56a Reflection of ras. Figure 9.56b The reflective propert.

13 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates number of ras (ou can think of them as light ras, although the could represent other forms of energ) traveling horizontall until the strike the parabola. As indicated, all ras striking the parabola are reflected through the focus of the parabola. Due to this reflective propert, satellite dishes are usuall built with a parabolic shape and have a microphone located at the focus to receive all signals. This reflective propert works in both directions. That is, energ emitted from the focus will reflect off the parabola and travel in parallel ras. For this reason, flashlights utilize parabolic reflectors to direct their light in a beam of parallel ras. Eample 6.5 Design of a Flashlight A parabolic reflector for a flashlight has the shape =. Where should the lightbulb be located? Solution Based on the reflective propert of parabolas, the lightbulb should be located at the focus of the parabola. The verte is at (0, 0) and the focus is shifted to the right from the verte a = 8 units, so the lightbulb should be located at the point ( 8, 0). Focus Focus Figure 9.57a Definition of ellipse. Ellipses The second conic section we stud is the ellipse. We define an ellipse to be the set of all points for which the sum of the distances to two fied points (called foci, the plural of focus) is constant. This definition is illustrated in Figure 9.57a. We define the center of an ellipse to be the midpoint of the line segment connecting the foci. The familiar equation of an ellipse can be derived from this definition. For convenience, we assume that the foci lie at the points (c, 0) and ( c, 0) for some positive constant c (i.e., the lie on the -ais, at the same distance from the origin). For an point (, ) on the ellipse, the distance from (, ) to the focus (c, 0) is ( c) + and the distance to the focus ( c, 0) is ( + c) +. The sum of these distances must equal a constant that we ll call k. We then have ( c) + + ( + c) + = k. Subtracting the first square root from both sides and then squaring, we get ( ( + c) + ) ( = k ( c) + ). Epanding this out gives us + c + c + = k k ( c) + + c + c +. Now, solving for the remaining term with the radical gives us k ( c) + = k c. In order to get rid of the radical, we square both sides and epand, to get k 8k c + k c + k = k 8k c + 6c or (k 6c ) + k = k k c. To simplif this epression, we set k = a, to obtain (6a 6c ) + 6a = 6a 6a c. Notice that since a is the sum of the distances from (, ) to (c, 0) and from (, ) to

14 smi9885_ch09b.qd 5/7/0 :5 PM Page 77 Section 9.6 Conic Sections 77 (0, b) ( c, 0) (c, 0) ( a, 0) (0, b) (a, 0) Figure 9.57b Ellipse with foci at (c, 0) and ( c, 0). ( c, 0) and the distance from (c, 0) to ( c, 0) is c, we must have a > c, so that a > c > 0. Dividing both sides of the equation b 6 and defining b = a c, we get b + a = a b. Finall, dividing b a b leaves us with the familiar equation a + b =. In this equation, notice that can assume values from a to a and can assume values from b to b. The points (a, 0) and ( a, 0) are called the vertices of the ellipse (see Figure 9.57b). Since a > b, we call the line segment joining the vertices the major ais and we call the line segment joining the points (0, b) and (0, b) the minor ais. Notice that the length of the major ais is a and the length of the minor ais is b. More generall, we have the following. Theorem 6. The equation ( 0 ) + ( 0) = (6.) a b with a > b > 0 describes an ellipse with foci at ( 0 c, 0 ) and ( 0 + c, 0 ), where c = a b. The center of the ellipse is at the point ( 0, 0 ) and the vertices are located at ( 0 ± a, 0 ) on the major ais. The endpoints of the minor ais are located at ( 0, 0 ± b). The equation ( 0 ) + ( 0) = (6.) b a with a > b > 0 describes an ellipse with foci at ( 0, 0 c) and ( 0, 0 + c) where c = a b. The center of the ellipse is at the point ( 0, 0 ) and the vertices are located at ( 0, 0 ± a) on the major ais. The endpoints of the minor ais are located at ( 0 ± b, 0 ). In the following eample, we use Theorem 6. to identif the features of an ellipse whose major ais lies along the -ais. Eample 6.6 Identifing the Features of an Ellipse Identif the center, foci and vertices of the ellipse =. Figure =. Solution From (6.), the equation describes an ellipse with center at the origin (0, 0). The values of a and b are 6 and 9, respectivel, and so, c = a b = 7. Since the major ais is parallel to the -ais, the foci are shifted c units to the left and right of the center. That is, the foci are located at ( 7, 0) and ( 7, 0). The vertices here are the -intercepts (i.e., the intersections of the ellipse with the major ais). With = 0, we have = 6 and so, the vertices are at (±, 0). Taking = 0, we get = 9 so that =±. The -intercepts are then (0, ) and (0, ). The ellipse is sketched in Figure 9.58.

15 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates 6 6 Figure 9.59 ( ) ( + ) + =. 5 Theorem 6. can also be used to identif the features of an ellipse whose major ais runs parallel to the -ais. Eample 6.7 Identif the center, foci and vertices of the ellipse An Ellipse with Major Ais Parallel to the -ais ( ) + ( + ) 5 =. Solution From (6.), the center is at (, ). The values of a and b are 5 and, respectivel, so that c =. Since the major ais is parallel to the -ais, the foci are shifted c units above and below the center, at (, ) and (, + ). Notice that in this case, the vertices are the intersections of the ellipse with the line =. With =, we have ( + ) = 5, so that = ± 5 and the vertices are (, 6) and (, ). Finall, the endpoints of the minor ais are found b setting =. We have ( ) =, so that = ± and these endpoints are (0, ) and (, ). The ellipse is sketched in Figure Find an equation of the ellipse with foci at (, ) and (, 5) and vertices (, ) and (, 6). Solution Recall that the center is the midpoint of the foci, in this case (, ). You can now see that the foci are shifted c = unit from the center. The vertices are shifted a = units from the center. From c = a b, we get b = =. Notice that the major ais is parallel to the -ais, so that a = is the divisor of the -term. From (6.), the ellipse has the equation ( ) ( ) + =. Eample 6.8 Finding the Equation of an Ellipse Focus a A a Focus Figure 9.60 The reflective propert of ellipses. Much like parabolas, ellipses have some useful reflective properties. As illustrated in Figure 9.60, a line segment joining one focus to a point A on the ellipse makes the same acute angle with the tangent line at A as does the line segment joining the other focus to A. Again, this is the same wa in which light and sound reflect off a surface, so that a ra originating at one focus will alwas reflect off the ellipse toward the other focus. A surprising application of this principle is found in the so-called whispering room of the U.S. Senate. The ceiling of this room is elliptical, so that b standing at one focus ou can hear everthing said on the other side of the room at the other focus. (You probabl never imagined how much of a role mathematics could pla in political intrigue.) Eample 6.9 A Medical Application of the Reflective Propert of Ellipses A medical procedure called shockwave lithotrips is used to break up kidne stones that are too large or irregular to be passed. In this procedure, shockwaves emanating from a transducer located at one focus are bounced off of an elliptical reflector to the kidne stone located at the other focus. Suppose that the reflector is described b the equation + = (in units of inches). Where should the transducer be placed? 8

16 smi9885_ch09b.qd 5/7/0 :5 PM Page 775 Section 9.6 Conic Sections 775 Focus (, ) Focus Figure 9.6 Definition of hperbola. b a c a a c a b Solution In this case, c = a b = 8 = 8, so that the foci are 6 inches apart. Since the transducer must be located at one focus, it should be placed 6 inches awa from the kidne stone and aligned so that the line segment from the kidne stone to the transducer lies along the major ais of the elliptical reflector. Hperbolas The third tpe of conic section is the hperbola. We define a hperbola to be the set of all points such that the difference of the distances between two fied points (called the foci) is a constant. This definition is illustrated in Figure 9.6. Notice that it is nearl identical to the definition of the ellipse, ecept that we subtract the distances instead of add them. The familiar equation of the hperbola can be derived from the definition. The derivation is almost identical to that of the ellipse, ecept that the quantit a c is now negative. We leave the details of the derivation of this as an eercise. An equation of the hperbola with foci at (±c, 0) and parameter a (equal to the difference of the distances) is a b =, where b = c a. An important feature of hperbolas that is not shared b ellipses is the presence of asmptotes. For the hperbola a =, we have b b = a or = b a b. Notice that lim ± = ( b ) lim ± a b = b a. That is, as ±, b a, so that ±b a and so, =±b are the (slant) a asmptotes as shown in Figure 9.6. More generall, we have the following. Figure 9.6 Hperbola, shown with its asmptotes. Theorem 6. The equation ( 0 ) ( 0) = (6.) a b describes a hperbola with foci at the points ( 0 c, 0 ) and ( 0 + c, 0 ),where c = a + b. The center of the hperbola is at the point ( 0, 0 ) and the vertices are located at ( 0 ± a, 0 ). The asmptotes are =± b a ( 0) + 0. The equation ( 0 ) ( 0) = (6.) a b describes a hperbola with foci at the points ( 0, 0 c) and ( 0, 0 + c), where c = a + b. The center of the hperbola is at the point ( 0, 0 ) and the vertices are located at ( 0, 0 ± a). The asmptotes are =± a b ( 0) + 0.

17 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates 6 In the following eample, we show how to use Theorem 6. to identif the features of a hperbola. Eample 6.0 Identifing the Features of a Hperbola 6 6 Figure =. 6 For the hperbola =, find the center, vertices, foci and asmptotes. 9 Solution From (6.), we can see that the center is at (0, 0). Further, the vertices lie on the -ais, where = (set = 0), so that =±. The vertices are then located at (, 0) and (, 0). The foci are shifted b c = a + b = + 9 = units from the center, to (±, 0). Finall, the asmptotes are =±. A sketch of the hperbola is shown in Figure 9.6. Eample 6. Identifing the Features of a Hperbola 8 For the hperbola asmptotes. ( ) 9 ( + ) 6 =, find the center, vertices, foci and 8 8 Solution Notice that from (6.), the center is at (, ). Setting =, we find that the vertices are shifted verticall b a = units from the center, to (, ) and (, ). The foci are shifted verticall b c = a + b = 5 = 5 units from the center, to (, ) and (, 6). The asmptotes are =± ( + ) +. A sketch of the hperbola is shown in Figure Figure 9.6 ( ) ( + ) 9 6 =. Eample 6. Finding the Equation of a Hperbola Find an equation of the hperbola with center at (, 0), vertices at (, 0) and (0, 0) and foci at ( 5, 0) and (, 0). 6 Solution Notice that since the center, vertices and foci all lie on the -ais, the hperbola must have an equation of the form of (6.). Here, the vertices are shifted a = units from the center, and the foci are shifted c = units from the center. Then, we have b = c a = 5. Following (6.), we have the equation ( + ) 5 =. 6 Much like parabolas and ellipses, hperbolas have a reflective propert that is useful in applications. It can be shown that a ra directed toward one focus will reflect off the hperbola toward the other focus. We illustrate this in Figure Figure 9.65 The reflective propert of hperbolas. Eample 6. An Application to Hperbolic Mirrors A hperbolic mirror is constructed in the shape of the top half of the hperbola ( + ) =. Toward what point will light ras following the paths = k reflect (where k is a constant)?

18 smi9885_ch09b.qd 5/7/0 :5 PM Page 777 Section 9.6 Conic Sections 777 E Parabola F Hperbola Figure 9.66 A combination of parabolic and hperbolic mirrors. Solution For the given hperbola, we have c = a + b = + =. Notice that the center is at (0, ), and the foci are at (0, 0) and (0, ). Since ras of the form = k will pass through the focus at (0, 0), the will be reflected toward the focus at (0, ). As a final note on the reflective properties of the conic sections, we briefl discuss a clever use of parabolic and hperbolic mirrors in telescope design. In Figure 9.66, a parabolic mirror to the left and a hperbolic mirror to the right are arranged so that the have a common focus at the point F. The verte of the parabola is located at the other focus of the hperbola, at the point E, where there is an opening for the ee or a camera. Notice that light entering the telescope from the right (and passing around the hperbolic mirror) will reflect off the parabola directl toward its focus at F. Since F is also a focus of the hperbola, the light will reflect off the hperbola toward its other focus at E. In combination, the mirrors cause all incoming light to be focused at the point E. EXERCISES 9.6. Each fied point referred to in the definitions of the conic sections is called a focus. Use the reflective properties of the conic sections to eplain wh this is an appropriate name.. A hperbola looks somewhat like a pair of parabolas facing opposite directions. Discuss the differences between a parabola and one half of a hperbola (recall that hperbolas have asmptotes).. Carefull eplain wh in eample 6.8 (or for an other ellipse) the sum of the distances from a point on the ellipse to the two foci equals a.. Imagine plaing a game of pool on an elliptical pool table with a single hole located at one focus. If a ball rests near the other focus, which is clearl marked, describe an eas wa to hit the ball into the hole. In eercises 5 6, find an equation for the indicated conic section. 5. Parabola with focus (0, ) and directri = 6. Parabola with focus (, ) and directri = 7. Parabola with focus (, 0) and directri = 8. Parabola with focus (, 0) and directri = 9. Ellipse with foci (0, ) and (0, 5) and vertices (0, ) and (0, 7) 0. Ellipse with foci (, ) and (, ) and vertices (, ) and (, 5). Ellipse with foci (, ) and (6, ) and vertices (0, ) and (8, ). Ellipse with foci (, ) and (5, ) and vertices (, ) and (6, ). Hperbola with foci (0, 0) and (, 0) and vertices (, 0) and (, 0). Hperbola with foci (, ) and (6, ) and vertices (0, ) and (, ) 5. Hperbola with foci (, ) and (, 6) and vertices (, ) and (, 5) 6. Hperbola with foci (0, ) and (0, ) and vertices (0, 0) and (0, ) In eercises 7 8, identif the conic section and find each verte, focus and directri. 7. = ( + ) 8. = ( + ) 9... ( ) + ( ) 9 ( + ) 6 ( ) 9 = 0. ( + ) 6 + = =. ( + ) ( ) = ( + ) =. ( + ) 9 = 5. ( ) + 9 = ( + ) = 6 7. ( + ) ( ) = 6 8. ( + ) ( ) =

19 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates In eercises 9, graph the conic section and find an equation. 9. All points equidistant from the point (, ) and the line = 0. All points equidistant from the point (, 0) and the line =. All points such that the sum of the distances to the points (0, ) and (, ) equals 8. All points such that the sum of the distances to the points (, ) and (, ) equals 6. All points such that the difference of the distances to the points (0, ) and (0, ) equals. All points such that the difference of the distances to the points (, ) and (6, ) equals 5. A parabolic flashlight reflector has the shape =. Where should the lightbulb be placed? 6. A parabolic flashlight reflector has the shape =. Where should the lightbulb be placed? 7. A parabolic satellite dish has the shape =. Where should the microphone be placed? 8. A parabolic satellite dish has the shape =. Where should the microphone be placed? 9. In eample 6.9, if the shape of the reflector is + =, how far from the kidne stone should the transducer be placed? 0. In eample 6.9, if the shape of the reflector is + 5 =, how far from the kidne stone should the transducer be placed?. If a hperbolic mirror is in the shape of the top half of ( + ) =, to which point will light ras following the 5 path = c ( < 0) reflect?. If a hperbolic mirror is in the shape of the bottom half of ( ) =, to which point will light ras following the 8 path = c ( > 0) reflect?. If a hperbolic mirror is in the shape of the right half of =, to which point will light ras following the path = c( ) reflect?. If a hperbolic mirror is in the shape of the left half of 8 =, to which point will light ras following the path = c( + ) reflect? 5. If the ceiling of a room has the shape 00 + =, where 00 should ou place the desks so that ou can sit at one desk and hear everthing said at the other desk? 6. If the ceiling of a room has the shape =, where 00 should ou place the desks so that ou can sit at one desk and hear everthing said at the other desk? 7. A spectator at the 000 Summer Olmpics Games throws an object. After seconds, the object is 8 meters from the spectator. After seconds, the object is 8 meters from the spectator. If the object s distance from the spectator is a quadratic function of time, find an equation for the position of the object. Sketch a graph of the path. What is the object? 8. Halle s comet follows an elliptical path with a = 7.79 (Au) and b =.5 (Au). Compute the distance the comet travels in one orbit. Given that Halle s comet completes an orbit in approimatel 76 ears, what is the average speed of the comet? 9. All of the equations of conic sections that we have seen so far have been of the form A + C + D + E + F = 0. In this eercise, ou will classif the conic sections for different values of the constants. First, assume that A > 0 and C > 0. Which conic section will ou get? Net, tr A > 0 and C < 0. Which conic section is it this time? How about A < 0 and C > 0? A < 0 and C < 0? Finall, suppose that either A or C (not both) equals 0; which conic section is it? In all cases, the values of the constants D, E and F do not affect the classification. Eplain what effect these constants have. 50. In this eercise, ou will generalize the results of eercise 9 b eploring the equation A + B + C + D + E + F = 0. (In eercise 9, the coefficient of was 0.) You will need to have software that will graph such equations. Make up several eamples with B AC = 0 (e.g., B =, A = and C = ). Which conic section results? Now, make up several eamples with B AC < 0 (e.g., B =, A = and C = ). Which conic section do ou get? Finall, make up several eamples with B AC > 0 (e.g., B =, A = and C = ). Which conic section is this?

20 smi9885_ch09b.qd 5/7/0 :5 PM Page 779 Section 9.7 Conic Sections in Polar Coordinates CONIC SECTIONS IN POLAR COORDINATES The conic sections are among the most studied curves in mathematics. As we saw in section 9.6, their unique properties enable us to make use of them in a variet of important applications. In addition to the definitions given in section 9.6, there are a variet of alternative definitions of the conic sections. One such alternative, utilizing an important quantit called eccentricit, is especiall convenient for studing conic sections in polar coordinates. We introduce this definition in this section, and review some options for parametric representations of conic sections. For a fied point P (the focus) and a fied line l not containing P (the directri), consider the set of all points whose distance to the focus is a constant multiple of their distance to the directri. The constant multiple e > 0 is called the eccentricit. Note that if e =, this is the usual definition of a parabola. As shown below, for other values of e we get the other conic sections. Theorem 7. The set of all points whose distance to the focus is the product of the eccentricit e and the distance to the directri is (i) an ellipse if 0 < e <, (ii) a parabola if e = or (iii) a hperbola if e >. (, ) Proof We can simplif the algebra greatl b assuming that the focus is located at the origin and the directri is the line = d > 0. (We illustrate this in Figure 9.67 for the case of a parabola.) For an point (, ) on the curve, observe that the distance to the focus is given b + and the distance to the directri is d. We then have + = e(d ). (7.) d Figure 9.67 Focus and directri. Squaring both sides gives us + = e (d d + ). Finall, if we gather together the like terms, we get ( e ) + de + = e d. (7.) Note that (7.) has the form of the equation of a conic section. In particular, if e =, (7.) becomes d + = d, which is the equation of a parabola. If 0 < e <, notice that ( e )>0 and so, (7.) is the equation of an ellipse (with center shifted to the left b the -term). Finall, if e >, then ( e )<0 and so, (7.) is the equation of a hperbola. Notice that the original form of the defining equation (7.) of these conic sections includes the term +. You should think of polar coordinates antime ou see this

21 smi9885_ch09b.qd 5/7/0 :5 PM Page Chapter 9 Parametric Equations and Polar Coordinates epression. Recall that in polar coordinates, r = + and = r cos θ. Equation (7.) now becomes r = e(d r cos θ) or r(e cos θ + ) = ed. Finall, solving for r, we have r = ed e cos θ +. This is the polar form of an equation for the conic sections with focus and directri oriented as in Figure As ou will show in the eercises, different orientations of the focus and directri can produce different forms of the polar equation. The possibilities are summarized in the following result. Theorem 7. The conic section with eccentricit e > 0, focus at (0, 0) and the indicated directri has the polar equation ed (i) r =, if the directri is the line = d > 0, e cos θ + ed (ii) r =, if the directri is the line = d < 0, e cos θ ed (iii) r =, if the directri is the line = d > 0 or e sin θ + ed (iv) r =, if the directri is the line = d < 0. e sin θ Notice that we proved part (i) above. The remaining parts are derived in similar fashion and are left as eercises. In the following eample, we illustrate how the eccentricit effects the graph of a conic section. Eample 7. The Effect of Various Eccentricities Find polar equations of the conic sections with focus (0, 0), directri = and eccentricities (a) e = 0., (b) e = 0.8, (c) e =, (d) e =. and (e) e =. Solution B Theorem 7., observe that (a) and (b) are ellipses, (c) is a parabola and (d) and (e) are hperbolas. B Theorem 7., all have polar equations of the form r = e e cos θ +. The graphs of the ellipses r =.6 0. cos θ + and r =. 0.8 cos θ + are shown in Figure 9.68a. Note that the ellipse with the smaller eccentricit is much more nearl circular than the ellipse with the larger eccentricit. Further, the ellipse with e = 0.8 opens up much farther to the left. In fact, as the value of e approaches, the ellipse will open up farther to the left, approaching the parabola with e =, r =, also shown in Figure 9.68a. For values of e >, the graph is a hperbola, opening up to the right and left. For instance, with e =. and e =, we have the cos θ + hperbolas r =.8. cos θ + and r = 8 (shown in Figure 9.68b), where we cos θ +