Math 5335 Section 2 Fall 2005 Solutions to December 8 homework problems

Transcription

1 Math 5335 Section 2 Fall 2005 Solutions to December 8 homework problems PROBLEM 9.7 To find an intersection points, we have to solve the following sstem of equations: = 6, ( ) =. We epand (and slightl simplif) the second equation to = 0. Substituting into this from the first equation, we have = 0, or = 3. If we substitute this into the first equation, we obtain 2 = 3. Since this has no real solutions, we conclude that there are no intersection points. Note: If we plot the two circles, we see that the smaller one [with center (,0) and radius = ] is completel contained inside the larger one [with center (0,0) and radius = 6 ]. Thus the circles do not intersect, which certainl is consistent with the solution presented here. But a complete solution definitel involves doing the algebra, or somehow proving that the circles don t intersect. PROBLEM The figure and the equations of the lines are eactl as the were given in the back of the tet. To describe the Poincaré segments, we also need inequalities to specif the range of -coordinates. Thus: Segment Equation of Poincaré line -range of segment ("2,4)(2,4) = 20, > ("2,4)(0,4) (+) = 7, > (0,4)(2,4) ( ) = 7, > PROBLEM 0.5. The circle that gives the Poincaré line has center (- 4, 0) and radius 2. Therefore it intersects the -ais at the points (- 6, 0) and (- 2, 0). These two points are therefore the direction indicators of the Poincaré line ( + 4) = 4, > 0. The one on the left belongs to the ra given b the inequalit - 3, and the one on the right belongs to the ra given b the inequalit - 3. Thus: Ra Direction indicator ( + 4) = 4, > 3 ( 6, 0) ( + 4) = 4, > 3 ( 2, 0) Note: Literall correct descriptions of the Poincaré ras could be understood to include the inequalit > 0 as well as the inequalit that specifies the range of -values. But that is often omitted because it is understood that we re working in the upper half plane.

2 December 8 HW solutions page 2 PROBLEM 0.2. The Poincaré line ( 2) =, > 0 has direction indicators (, 0) and (3,0), while the Poincaré line ( + 2) =, > 0 has direction indicators (-, 0) and (-3, 0). An line which is asmptoticall parallel to both of these must have one direction indicator in common with each. Here are the possibilities: Direction indicators (ω, 0) ρ Poincaré line (, 0) and (, 0) (0, 0) =, > 0 (3, 0) and ( 3, 0) (0, 0) = 3, > 0 (3, 0) and (, 0) (, 0) 2 ( ) = 4, > 0 (, 0) and ( 3, 0) (, 0) 2 ( + ) = 4, > 0 PROBLEM The epression ma( 2, ) refers to the larger of the two numbers 2 and. Therefore, the inequalit ma( 2, ) describes the set of all points that are above the parabola = 2 and also above the line =. It is shown in the following figure: Generall, the hperbolic area of a region Ω is given as the integral " dd. If we want to do the integral in a single piece, then we use the backwards(?) order of integration, i.e., we integrate first with respect to, treating the right side of the parabola as being given b the equation = and the left side of the parabola as being given b the equation = ". In this wa, we get the following integral: 2 * d ' % 2 ) d = " ( = d = 2 "3 2 d = "4" 2 = = = 4. 2 ="

3 December 8 HW solutions page 3 Alternativel, if we wish to do the more traditional(?) order of integration, where we integrate first with respect to (and therefore last with respect to ), then we have to divide the region into three pieces, as shown in the following figure: B smmetr, the region on the right has the same area as the region on the left. Therefore, the area is given b the following sum of integrals: " d ' " " d ' ) d + 2 ) d = % 2 * ( % 2 ( = d + 2 d = " " 2 2 = = = " d + 2 " " = = = 4. = = 2 d {Please see the net page for Problem 0.3.}

4 December 8 HW solutions page 4 PROBLEM 0.3. We calculate the hperbolic area as π α β, where α and β are the angular measures at the non-asmptotic vertices. B smmetr (see the figures below), we have α = β, so that the hperbolic area is π 2α. Method. We find cos α b calculating the inner product of the two tangent vectors at A. Clearl, (,0) is the unit tangent vector of the horizontal ra. The radius of the circular arc is = 2 5, so that it is given parametricall as X(t) = 2 5(cost,sint). Hence, we find a tangent vector at A = X(θ) b calculating X "() = 2 5(sin,cos). Therefore, the unit tangent vector is ( sinθ, cosθ), as indicated in the figure. And then we calculate: ( sin!, cos!) (-2,4) = B (!,0) "! (0,) A= (2,4) cosα = ( sinθ,cosθ),(0,) = cosθ. Referring to the figure and noting that the circle has radius = 2 5, we see that cos" = 2 =. Therefore, cos" =, so that " = arccos. Hence the hperbolic area is " 2arccos. Alternativel, we can write it as " 2arcsin When finding the numerical value, remember to set our calculator for radians, rather than degrees. Thus, the numerical value is Method 2. Again, we have smmetr, so that the hperbolic area is π 2α. The Poincaré ras that form the sides of the angle are shown in dark blue in the figure. We use the formula from Theorem 7 in 0.3, with (g,h) = (2,4), the direction indicators of the two ras being ("2 5,0) and (,0). Therefore, we appl version (0.4) of the formula, as follows: " = arccos (2 5 2)2 4 2 ' ) %(2 5 2) ( = arccos ( 5 +)2 2 2 ' ) %( 5 +) ( (-2,4) = B (-2"5,0) (!,0)! A= (2,4)

5 December 8 HW solutions page 5 We calculate ( 5 +) 2 = = Therefore, the formula works out as follows: " = arccos % ( = arccos + 5 % ( ' '. While this looks different from the previous answer, we can show (optionall) that it s reall the same, as follows: = " = = = 5 5 = 5. We can describe the beginning of this process b saing that we multiplied numerator and denominator b the conjugate of the denominator. While it resembles the comple conjugate, it s a slightl different entit, namel the conjugate in the quadratic number field Q( 5). {This is the field of rational numbers with 5 adjoined; elements of Q( 5) are of the form a + b 5, where a and b are rational numbers. Obviousl, sums and differences eist in this algebraic sstem. A bit of calculation is needed in order to check that products and quotients also eist inside the sstem. Our method of simplification above does, however, ehibit the method needed for proving that quotients eist.} Back to the class homepage.