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UT M 408D - Homework 2 - Solutions

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jones (bwj276) – Homework 2 – pascaleff – (55960) 1This print-out should have 17 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsFind an equation for the tangent line to thegraph o fr = 4e−θ− 5at the point P corresponding to θ = 0.1. y = 4x + 12. y = 4x + 43. y + 4x = −44. 4y = x + 1 correct5. 4y = x + 46. 4y + x = −1Explanation:The usual point-slope formula can be usedto find an equation for the tangent line to thegraph of a polar curve r = f (θ) at a pointP once we know t he Cartesian coordinatesP (x0, y0) of P and the slope of the tangentline at P .Now, w henr = 4e−θ− 5 ,thenx(θ) = ( 4 e−θ− 5) cos θ ,whiley(θ) = (4e−θ− 5) sin θ .Thus in Cartesian coordinates, the point Pcorresponding to θ = 0 is (−1, 0). On theother hand,x′(θ) = −4e−θcos θ − sin θ(4e−θ− 5) ,whiley′(θ) = −4e−θsin θ + cos θ(4e−θ− 5) ,so the slope at P is gi ven bydydxθ=0=y′(0)x′(0)=14.Consequently, by t he point sl ope formula, thetangent line at P has equationy =14x + 1which after simplification becomes4y = x + 1 .002 10.0 pointsFind the slope of the ta ngent line to thegraph o fr = 4 + sin θat θ = π/6.1. slope =4√3 − 14 +√32. slope = −14√33. slope = −35√34. slope = −4√3 + 14 +√35. slope = −53√3 correct6. slope =4√3 − 1√3 − 4Explanation:The graph of a polar curve r = f(θ) canexpressed by the parametric equationsx = f (θ) cos θ , y = f(θ) sin θ .In this form the slope of the tangent line tothe curve is given bydydx=y′(θ)x′(θ).jones (bwj276) – H omework 2 – pascaleff – (55960) 2But whenr = 4 + sin θ ,we see thaty′(θ) = cos θ sin θ + (4 + sin θ) cos θ ,whilex′(θ) = cos θ cos θ − (4 + sin θ) sin θ .But theny′π6=4√3 + 1√32,whilex′π6=−42.Consequently, at θ = π/6,slope =dydxθ=π/6= −53√3 .keywords: polar function, slope, parametricequations, tangent line,003 10.0 pointsFind the slope of the ta ngent line to thegraph o fr = 3eθ− 1at θ = π/4.1. slope = 6eπ/4+ 12. slope = 6e−π/43. slope =13eπ/4− 14. slope = 3eπ/45. slope =13eπ/4+ 16. slope = 6eπ/4− 1 correctExplanation:The graph of a polar curve r = f(θ) canexpressed by the parametric equationsx = f (θ) cos θ , y = f(θ) sin θ .In this form t he slope of the pol ar curve isgiven bydydx=y′(θ)x′(θ).Now, w henr = 3eθ− 1 ,we see thaty′(θ) = 3eθsin θ + (3eθ− 1) cos θ ,whilex′(θ) = 3eθcos θ − (3eθ− 1) sin θ .But theny′π4=1√2(6eπ/4− 1) , x′π4=1√2.Consequently, at θ = π/4,slope =dydxθ=π/4= 6eπ/4− 1.004 10.0 pointsFind the area of the region bounded by thepolar curver =√2 ln θ + 6as well as the rays θ = 1 and θ = e.1. area =12(3e − 2)2. area =12(3e + 4)3. area = 3e + 44. area = 2(3e − 2)5. area = 2(3e + 4)6. area = 3e − 2 correctExplanation:jones (bwj276) – H omework 2 – pascaleff – (55960) 3The area of the region bo unded by thegraph of the polar function r = f (θ) as wellas the rays θ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .Whenf(θ) =√2 ln θ + 6 , θ0= 1 , θ1= e ,therefore, the area of the enclosed region isthus gi ven by the integralA =12Ze1(√2 ln θ + 6)2dθ=12Ze1(2 ln θ + 6) dθ .To evaluate this last integral we use Integra-tion by Parts, for thenA =12h2θ ln θ + 6 θie1−12Ze12 dθ=12h2θ ln θ + 4 θie1.Consequently,area = A = 3e − 2 .005 10.0 pointsThe shaded region inlies inside the polar curve r = 4 sin θ andoutside the polar curve r = 2 sin θ. Determinethe a r ea of this region.1. area = 3π correct2. area = 12π3. area = 6π4. area = 35. area = 66. area = 12Explanation:As the graphs show, the pola r curves i nter-sect when4 sin θ = 2 sin θ ,i.e. at θ = 0, π. T hus the area o f the shadedregion is gi ven byI =12Zπ0n(4 sin θ)2−(2 sin θ)2odθ .Consequently,I =12Zπ012 sin2θ dθ .To evaluate this last i ntegral we use thedouble angl e formulasin2θ =12(1 − cos 2θ) .For thenZπ012 sin2θ dθ = 6Zπ0(1 − cos 2θ) dθ= 6hθ −12sin 2θiπ0= 6π .Consequently, the shaded region hasarea = 3π .Are there any other ways of calculating thisarea wit hout using i ntegration?jones (bwj276) – H omework 2 – pascaleff – (55960) 4keywords: area, polar cooor di nates, definiteintegral, circle,006 10.0 pointsFind the area of the shaded regioninside the graph ofr = 1 + 2 cos θ .1. area = π −3√342. area =32π3. area =12π −3√34correct4. area = π +3√345. area = π6. area =π27. area =12π +3√34Explanation:The area of a region bounded by the graphof the polar function r = f(θ) and the raysθ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .In this question the function isr = 1 + 2 cos θ ;on the other hand, to determine the rays θ =θ0and θ = θ1bounding the shaded region,note that as θ ranges from 2π/3 to π thegraph(i) passes though the origin when r = 0,i.e., when θ = 2π/3,(ii) and then crosses the x -axis when θ = πat r = −1.Thus the area of t he shaded region is given bythe integralA =12Zπ2π/3(1 + 2 cos θ)2dθ .But(1 + 2 cos θ)2= 1 + 4 cos θ + 4 cos2θ= 3 + 4 cos θ + 2 cos 2θ ,sincecos2θ =121 + cos 2θ.HenceA =12Zπ2π/33 + 4 cos θ + 2 cos 2θdθ=12h3θ + 4 sin θ + sin 2θiπ2π/3.Consequently,area = A =12π −3√34.keywords: polar graph, area, cardioid, polarintegral007 10.0 pointsFind the area of the region enclosed by thegraph o f the polar functionr = 3 + cos θ .jones (bwj276) – H omework 2 – pascaleff – (55960) 51. area = 9π2. area =192π correct3. area = 20π4. area = 10π5. area =172πExplanation:The area of the region bo unded by thegraph of the polar function r = f (θ) andthe rays θ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .On the other hand, the graph ofr = 3 + cos θis the cardioid similar to the one shown inso in this case we can take θ0= 0 and θ1= 2π.Thus the area of the region enclosed by thegraph is given by the integralA =12Z2π0(3 + cos θ)2dθ .Now(3 + cos θ)2= 9 + 6 cos θ + cos2θ=192+ 6 cos θ +12cos 2θ ,sincecos2θ =121 + cos 2θ.But then,A =12Z2π0192+ 6 cos θ +12cos 2θdθ=12h192θ + 6 sin θ +14sin 2θi2π0.Consequently,area = A =192π.keywords: p olar gra ph, area, cardioi d008 10.0 pointsFind the area of one loop of the graph ofthe polar functionr = 4 cos 2θ .1. area =2916π2. area =3116π3. area =74π4. area =158π5. area =


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