Solutions to the Eercises of Chapter 4 4A. Basic Analtic Geometr. The distance between (, ) and (4, 5) is ( 4) +( 5) = 9+6 = 5 and that from (, 6) to (, ) is ( ( )) +( 6 ( )) = ( + )=.. i. AB = (6 ) +( 7 ( )) = 5 + 6 = 4. AC = (6 ) +( 7 ( )) = 6 + 5 = 4. BC = ( ) +( ( )) = 8 + = 8. So AB + AC = BC.SobPthagoras, ABC is a right triangle. ( )( ) ii. With the side AB as base, the height is AC. Sothe area is 4 4 = 4.. AB = ( ) +( ) = 6 + 64 = 80. BC = ( 5) + ( 5) = 4+6= 0 AC = ( 5) +( 5) = 6 + 44 = 80 So AB + BC = 6 5+ 4 5=4 5 + 5=6 5= 6 5=AC. 4. i. = ii. = 5. i. The and aes taken together.
ii. =means that either =or =. So the graph consists of the two lines = = 6. This is the set of all (, ) with either positive and negative, or negative and positive. So it is the shaded region (without the aes): 7. This is the strip (including the indicated boundaries): = 4 = = 0 8. Because < isequivalent to << and < isequivalent to <<, this consists of all points within, but not on, the rectangle below. = = = =
9. To show that the midpoint of the line segment from P (, )top (, )is ( +, + ) proceed as follows: Drop perpendiculars and form the point (, ). Then use the fact that the midpoint of the segment from to on the ais is + and that from to (, ) + ( +, ) + (, ) (, ) + (, ) (, ) on the ais is +. Consider the points (, + ) ( and ) +,. It remains to notice (b considering similar triangles) that ( +, ) + it is the midpoint of the segment from P (, )top (, ). 0. Use of the formula developed in Eercise 9 shows that the midpoint of the line segment joining the points (, ) and (7, 5) is (4, 9) and that of the segment joining the points (, 6) and (8, ) is ( 7, ). 4B. Circles, Parabolas, and Ellipses. This is a parabola opening downward. To get a precise idea of the graph, complete the squares = ( 4) = ( +( ) ( ) 4) = ( ) +( 9 4 + 6 4 )=( ) + 5 4. Note that the highest point on the parabola is (, 5 4 ). It crosses the -ais when = ± 5, - 4 so at = and 4.
. Dividing through b 6, we get 4 + =. This is an ellipse with semimajor ais a =4and semiminor ais b =. For the graph see Figure 4.8.. This is a parabola opening to the right starting at the origin. (, ) (, -) 4. Alook atthe standard equation of the circle shows that this is a circle of radius 7 centered at (, 5). ( 4 5. Note that + =and hence that ) + 9 ( =. This is an ellipse with semimajor 6) ais 4 6 and semiminor ais. Notice that the major ais is on the -ais and the minor ais is on the -ais. The general shape of the graph is obtained b rotating the ellipse of Figure 4.8 b 90. 6. = + = 7. The points of intersection of the line and the parabola are obtained b appling the quadratic formula to the equation +6 =0. Doing so, shows that the coordinates of the points of intersection are ±. Since the parabola opens upward, the situation is as pictured. _ + = 8 _ 4
For P =(, ) tolie in the parabolic section, both + and +6 +7 8must hold. Wh is the first of these two conditions superfluous? 8. Completing the square transforms = +4 +7to =( +4 + ) +7and hence to =( +) +. The smallest value is and it occurs when =. So (, ) is the lowest point on the graph. The points of intersection of the line =7and the parabola are obtained b setting +4 +7=7and solving for. Since ( +4)=0, this shows that =0and = 4. So the points of intersection are S =( 4, 7) and S =(0, 7). The verte of the parabolic section is V =(, ). The area of the triangle S VS is (4)(4) = 8. Therefore b Archimedes s theorem, the area of the parabolic section S VS is 4 8=0. 9. Consider the equation = +5together with the general equation ( ) ( ) ( ) a a = + b c +. (b c) b c (b c) In this case, =, a =,and a +b c =5. Sob c =,a =(b c) =, and (b c) b c (b c) 6 a +b c = 0(b c) = 5. Since (b+c)(b c) =b c = 5 = 4,b+c = 4 6 = 8 = 56. 9 9 9 6 Using b c = 57 55,weget b = and c =. Refer to the tet and conclude that the focus is 6 (a, b) = (, ) 57 and that the directri is the line = 55. 0. This is ( ) +( +) =5.. Completing the square with both variables, we get 0 = + 4 +0 += 4 + +0 + = 4 +( )+ +0 +(5 5 )+ = ( ) +( +5) 5 +. So, ( ) +( +5) = +4+5=6=4.Sothe graph is a circle. Its center is (, 5) and its radius is 4.. Proceed as above and complete the square with + + a + b + c =0: 0 = + + a + b + c = + a + + b + c ( a ) ( a ) ( ) ( ) b b = + a + + + b + + c ( = + a ) ( + + b ( a ) ( ) b + c. ) Therefore, ( + ) a ( + + b ( ) = a ( ) + b ) c. Since the left side cannot be negative, ( a ) ( + b ) c must be greater than or equal to 0 if there are to be an points on the graph of this equation. So the condition is ( a ( ) + b ( ) c. Let r = a ( ) + b ) c. Since ( ) + a ( + + b ) = r,weare dealing with a circle with center ( a, b ) and radius r. 5
. Since the equation is 5 + =,the semimajor ais is a =5and the semiminor is b =. The linear eccentricit is e = a b = 5 =, and the astronomical eccentricit is ε = e a = 5. 4. Since the string is stretched it will alwas form a triangle with base the segment F F. So the base has length e = a b. This means that the sum of the lengths of the remaining two sides of the triangle is equal to a. Hence the sum of the distances from the tip of the pencil to the points F and F is equal to a. Therefore what is being traced out is an ellipse with focal points F and F, constant k =a, and linear eccentricit e. The rest follows from the discussion in Section 4.5, especiall Figure 4.8. 5. Take a line segment of fied length and let P be a fied point on it. Let a and b be the lengths of the segments on the two sides of P as shown. Let the segment be in tpical position in the first quadrant and put P =(, ). B similar triangles, = b. Square both sides to get a b a P = (, ) b b - = b. Therefore, + =. Check that this holds regardless of the quadrant in a b a b which the segment is placed. So the points P produced in this wa coincide with the ellipse a + b =. 6. Review the basics about hperbolas from Section.. Let k be a positive constant and let e be one half the distance between F and F. So the focal points are ( e, 0) and (e, 0). Let F =(e, 0) and F =( e, 0). The hperbola determined b F,F and k, consists of all points P =(, ) such that PF PF = k. A look at the diagram below shows that e + PF >PF.Soe>PF PF and e > PF PF = k. Note that P =(, ) ison P F e F 6
the hperbola precisel if PF PF = ±k. This translates to ( e) + ( + e) + = ±k. So ( e) + = ±k + ( + e) +. After squaring both sides, etc., this equation can be transformed in successive steps to ( e) + = k ± k ( + e) + +( + e) + ( e) = k ± k ( + e) + +( + e) e + e = k ± k ( + e) + + +e + e ±k ( + e) + = k +4e 4k (( + e) + )=k 4 +8k e +6e 4k ( +e + e + )=k 4 +8k e +6e 4k +8k e +4k e +4k = k 4 +8k e +6e 4k 6e +4k = k 4 4k e 4(k 4e ) +4k = k (k 4e ). Dividing through b k (k 4e ), gives 4 + 4 =. Hence =. k k 4e k 4e k 4 4 Recall that e >k. So 4e >k, and 4e k k > 0. With a = = k and b = 4 4e k,wenowhave The graph is given b a b =. 4e k 4 = = - _b a = _b a b a 4C. Some Geometr and Trigonometr 7. The radian measure of θ is s =0.69 and P =(cos θ, sin θ) (0.769, 0.69). 8. Since θ is equal to both 5 and s,weget that s = 0. 9. Consider θ =7.5. Because 7.7 =.8, we get that 7.7 =(.8)(π). Since θ is positive, π it follows that P θ is obtained b going around the unit circle.8 revolutions in the clockwise 7
direction starting from the point (, 0). Two complete revolutions return us to the starting point (, 0). Since (0.8)(π) =(.8) ( ) π,itremains to proceed another three quarters of a revolutions in the clockwise direction to the point (0, ), and then another 0.8 of a 4 quarter revolution to locate P θ. It follows that P θ is in the fourth quadrant. More precision is obtained b recalling that P θ =(cos θ, sin θ) =(cos 7.5, sin 7.5) (0.4, 0.97). For θ =.8, do a similar thing in the clockwise direction. Because P θ = (cos θ, sin θ) = (cos.8, sin.8) ( 0.99, 0.6). This time P θ is in the third quadrant, close to the point (, 0). 0. Start with the portion of the graph of the cosine shown below on the left. First consider 0 θ π.ascos θ goes from cos 0 = to cos π =0,notice that sec θ = will move cos θ sec θ π 0 π cos θ π π 0 π π from to a larger and larger positive number. At θ = π, note that sec π = = is cos π 0 not defined. The graph on the right captures this information. Then do the same thing for π θ 0 and π θ π The rest of the graph follows the pattern alread established.. The first three identities follow from similar identities for the cosine. To verif the fourth, start with sin θ + cos =and divide through b cos θ. 4D. Computing Orbital Information. From Table 4., a =5.08 AU and ε =0.0484. So the linear eccentricit is e = εa =0.58 AU and the semiminor ais is b = a e = 7.069 0.064 = 7.0057 = 5.967 AU. After consulting Figure 4.8 observe that the greatest distance from Jupiter to the Sun is a + e =5.4546 AU and that the least distance is a e =4.95 AU. To convert to miles use the estimate AU = 9 0 6 miles. For eample, the greatest distance from Jupiter to the Sun is 507 0 6 miles.. For Mars, a =.57 =.575 T.8809.578 a =0.9999. For Jupiter, = 5.08 = 40.85 =.0009. For T.86 40.78 =.000. The fact that all these ratios are equal to one is a Saturn, = 9.588 = 867.9 T 9.4577 867.756 directl related to the definition of the units used. B definition, AU is the semimajor ais of the Earth and ear is equal to the period of the Earth s orbit. So in these units, the ratio a is equal to for the Earth. B Kepler s law all the ratios a must be equal to (or more T T 8
accuratel, close to ) in these units. 4. i. The formula for the area of a circular sector of radius r and angle θ is r θ. The radian measure of the angle PSP arc PP is. Itfollows that the area of the circular sector a e PSP is arc PP (a e) = (a e)(arc a e PP ). The same computation verifies the other formula. ii. Let t be this common time. B Kepler s second law and part (i), (a e)(arc PP )= (a + e)(arc QQ ). Since v P t = arc PP and v A t = arc QQ,itfollows that (a e)v P t = (a + e)v At. The formula v P va = a+e follows. Dividing the numerator and denominator a e b a, shows that v P va = +ε. A look at Table 4. tells us that its value for the Earth is ε +0.067 =.067 +0.0557 =.040. For Saturn the value is =.0557 =.80. 0.067 0.98 0.0557 0.944 5. i. Counting das, hours, and minutes, shows that t ve =75.6 das for 995. Adding the length of spring, i.e., 9.769 das to t ve =75.6, gives us t ss = 68.95 das for 995. ii. For t ve, the strateg of Section 4.8 gives β =.0 and then β = β = β 4 =.7. iii. Making the indicated substitutions gives r = 0.9958 AU and α =.5 radians. iv. For t ss, the strateg of Section 4.8 gives β =.8969 and then β = β = β 4 =.9009. It follows that r =.06 AU and α =.9048 radians. 6. Since e =0, we find that a = b. Sothe ellipse + =and the circle + = a coincide. a b In reference to Figure 4.4, this means that the focus S of the ellipse is the center O of the circle and that the points P and P 0 coincide. So the segment SP coincides with the segment OP 0. It follows that α = β, r = a, and b Kepler s formula (or a direct argument), that α = β = πt. The method of successive approimations of Section 4.8 is not needed because T r = a and α = β has an eplicit epression in terms of t. 4E. The Orbit of Halle s Comet 7. B Kepler s third law a =in the units AU and ears because this ratio is for the Earth. T Since T =76ears for Halle, we get a = T =76. It follows that a =76 / =7.94 AU. Since the minimum distance between Halle and the Sun is a e = d =0.59 AU, where e is the linear eccentricit, we see that e = a d =7.94 0.59 =7.5 AU. Halle s semiminor ais is b = a e =4.56 AU and its astronomical eccentricit is ε = e = 7.5 a =0.967. 7.94 for Halle s greatest distance from the Sun is a + e =7.94+7.5 =5.9 AU. The ratio v P va Halle is v P va = 7.94+7.5 7.94 7.5 = 5.9 0.59 60. 8. B assumption and Figure 4.9, the Earth s orbit is a circle with center (e, 0) and radius. So ( e) + =is an equation of the orbit. To find the -coordinates of the points H and H of Figure 4.40, we need to solve the equations ( e) + =and + =for a b 9
. B substituting and taking common denominators, we get = ( e) + = b + a a +a e a e = a e +a e a e a b a b a b So a b = a e +a e a e and e a e + a e a + a b =0. Because a e a + a b simplifies to a (e +b )=a (a ), we get b using the quadratic formula that = a e ± 4a 4 e 4e a (a ) e = a e ± ae a (a ) e = a ± a e. Because a +a = a + a > a = a, itisnot possible for = a +a. (Since = a is the e e e a e -intercept of the ellipse in Figure 4.40.) So we can conclude that = a a = a(a ). So e e = 7.94(6.94) =7.5 AU. Inserting this value of into = ( e), gives us = 7.5 (7.5 7.5) = (0.7) = 0.0 =0.97. It follows that the -coordinates of the points H and H are 0.98 and 0.98 respectivel. To sketch a more accurate version of Figure 4.40, place H and H in such a wa that the vertical segment H H is 0.7 units to the right of the Sun S. Notice that the trajector of Halle (within the Earth s orbit) is much steeper than suggested in the figure. 9. The value of r is equal to the length of the segment SP = SH. Since H lies on the circle with center S and radius AU (in other words on the Earth s orbit), it follows that r =. Because 0.98 is the -coordinate of H, notice that sin α = 0.98 = 0.98. Using the inverse sine button of our calculator, will give ou α =.7 (in radians). Now use Gauss s formula tan α = +ε tan β to compute β. BEercise 7,ε= e = 7.5 =0.967. So ε a 7.94 tan β = ε +ε tan α β =(0.)(0.8) =0.. B taking an inverse tan, =0.. So β =0.. Inserting what we alread know into Kepler s formula β ε sin β = πt πt, tells us that 0. (0.967)(0.) =. Solving for t, we T 76 get t (0.0)(76) =0. ears. We have shown that Halle requires about 0. ears or 44 das π to move from perihelion to H. A repetition of this discussion (or an appeal to smmetr) will show that Halle requires the same number of das to move from H to perihelion. It follows that Halle remains inside the Earth s orbit for about 88 das. 40. Recall that in general β β i ε i. To insure that β β i 0.000, we need to achieve ε i 0.000. For Halle, ε =0.967. To be on the safe side we will take ε =0.968. (There is no information about the fourth decimal place.) B squaring again and again, we see that ε 56 < 0.0004. So this gets us close. Multipling b 0.968 si more times shows that ε 6 < 0.000. 4. This task is left to the student. 0
4F. More Trigonometr and Gauss s Formula 4. Enough hints have been supplied. 4. The first part is obvious. For the second, use sin ϕ + cos ϕ =. 44. The equalit tan α = cos α +cos α sin ϕ = cos ϕ with ϕ = α That tan β with ϕ = β. So is verified b using the equalities cos ϕ = +cos ϕ and cos α. That = +ε cos β cos β ε follows b use of cos α =. +cos α ε +cos β ε cos β = cos β +cos β, uses cos ϕ =+cos ϕ and sin ϕ = cos ϕ again, this time tan α = +ε β ε tan. Now suppose that 0 β<π. Refer to the basic diagram from Kepler s discussion and notice that 0 α<π.so 0 β < π and 0 α < π. Refer to the graph of the tangent and notice that tan α and tan β are both positive. So b taking square roots, tan α = +ε tan β in this ε case. If π<β<π, then π<α<π. Now both tan α and tan β are negative. So again, tan α = +ε tan β.ifβ = π, then b the basic diagram from Kepler s discussion α = π. So ε neither tan α and tan β are defined in this case. (Have a look at the graph of the tangent.) This presents no problem since the basic point is to determine α in terms of β. 4G. A stud of Kepler s Formulas Correction: In Eercise 45 the formula tan α = b sin β a(cos β ε) is incorrectl written as tan α = b sin β a(cos β e). 45. In the new figure, P and P 0 both lie below the -ais, X is on the left of O, and is negative. Check that SX is equal to e in this situation also. Let α = XSP and β = XOP 0 and notice that α = α + π and β = β + π. Buse of the Eamples 4. and 4., the verifications A X = (, 0) O β α S N P = (, ) P = (, ) 0 0
b sin β a(cos β ε) of the equations r = a( ε cos β) and tan α = go through virtuall unchanged. Onl Kepler s equation remains. The sectors referred to in the argument below, as well as the related sections, will be those determined b the angles π α and π β rather than α and β. Inreference to the circle, Area section P 0 XN = Area sector P 0 ON + Area P 0 XO. Because the sector P 0 ON is determined b the angle π β, its area is equal to a (π β). The area of the triangle P 0 XO is ( )( 0)= ( )( a sin β). So Area section P 0 XN = a (π β)+ a sin β. Now turn to the ellipse. B Cavalieri s principle, Area section PXN = b ( a a (π β)+ ) a sin β = ba(π β)+ b sin β. Note net that A t is equal to the area of the full ellipse, minus the area of the section just computed, plus the area of PXS.So A t = abπ ba(π β) b sin β + (e )( ) = abβ b sin β (e )b sin β = abβ eb sin β = ab(β ε sin β). The rest of the argument is identical to the one in the tet. Alternativel, the argument in the tet can be retained with the following understanding: Let the sector P 0 ON be that determined b β, the section P 0 XN to be that with perimeter the circular arc from N to A to P 0 and the segments P 0 X and XN, and let the elliptical section PXN be that with perimeter the elliptical arc from N to A to P, and the segments PX and XN. 46. Let α, β, and t be the parameters for the same position in the first orbit. In going from the first orbit to the second, π is added to both α and β and T is added to t. Observe that a, b, ε, and r are the same for both orbits. So the question is as to whether the formulas are valid with α = α +π in place of α, β = β +π in place of β, and t = t + T in place of t. This follows quickl from the identities in Eamples 4. and 4.. For eample, β ε sin β = β +π ε sin β = πt T πt +πt +π = T = πt T.