New version page

# UT M 408D - Homework 2 - Solutions

Pages: 12
Documents in this Course

## This preview shows page 1-2-3-4 out of 12 pages.

View Full Document
Do you want full access? Go Premium and unlock all 12 pages.
Do you want full access? Go Premium and unlock all 12 pages.
Do you want full access? Go Premium and unlock all 12 pages.
Do you want full access? Go Premium and unlock all 12 pages.

Unformatted text preview:

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 =

View Full Document Unlocking...