jones bwj276 Homework 2 pascaleff 55960 This print out should have 17 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Find an equation for the tangent line to the graph of r 4e 5 at the point P corresponding to 0 so the slope at P is given by dy dx 4y x 1 002 3 y 4x 4 10 0 points Find the slope of the tangent line to the graph of r 4 sin 4 4y x 1 correct 5 4y x 4 at 6 6 4y x 1 Explanation The usual point slope formula can be used to find an equation for the tangent line to the graph of a polar curve r f at a point P once we know the Cartesian coordinates P x0 y0 of P and the slope of the tangent line at P Now when 1 slope 2 slope 3 slope 4 slope r 4e 5 5 slope x 4e 5 cos 6 slope then while y 4e 5 sin Thus in Cartesian coordinates the point P corresponding to 0 is 1 0 On the other hand x 4e cos sin 4e 5 while y 4e y 0 1 x 0 4 which after simplification becomes 2 y 4x 4 0 Consequently by the point slope formula the tangent line at P has equation 1 x 1 y 4 1 y 4x 1 1 sin cos 4e 5 4 3 1 4 3 1 3 4 3 3 5 4 3 1 4 3 5 3 correct 3 4 3 1 3 4 Explanation The graph of a polar curve r f can expressed by the parametric equations x f cos y f sin In this form the slope of the tangent line to the curve is given by y dy dx x jones bwj276 Homework 2 pascaleff 55960 But when r 4 sin 2 The graph of a polar curve r f can expressed by the parametric equations x f cos y f sin we see that y cos sin 4 sin cos while x cos cos 4 sin sin In this form the slope of the polar curve is given by dy y dx x Now when r 3e 1 But then y 6 while x 4 3 1 3 2 6 Consequently at 6 dy slope dx 4 2 6 5 3 3 keywords polar function slope parametric equations tangent line 003 at 4 1 slope 6e 4 1 2 slope 6e 4 1 3e 4 1 4 slope 3e 4 5 slope y 3e sin 3e 1 cos while x 3e cos 3e 1 sin But then 1 y 6e 4 1 4 2 slope dy dx 004 4 6e 4 1 Find the area of the region bounded by the polar curve r 2 ln 6 as well as the rays 1 and e 1 3e 2 2 1 2 area 3e 4 2 1 area 3 area 3e 4 4 area 2 3e 2 3e 4 1 5 area 2 3e 4 Explanation 1 4 2 10 0 points 1 6 slope 6e 4 1 correct x Consequently at 4 10 0 points Find the slope of the tangent line to the graph of r 3e 1 3 slope we see that 6 area 3e 2 correct Explanation jones bwj276 Homework 2 pascaleff 55960 The area of the region bounded by the graph of the polar function r f as well as the rays 0 1 is given by the integral Z 1 1 A f 2 d 2 0 When f 2 ln 6 0 1 1 e therefore the area of the enclosed region is thus given by the integral Z 1 e A 2 ln 6 2 d 2 1 Z 1 e 2 ln 6 d 2 1 To evaluate this last integral we use Integration by Parts for then ie 1 Z e 1h 2 ln 6 2 d A 2 1 2 1 ie 1h 2 ln 4 1 2 2 area 12 3 area 6 4 area 3 5 area 6 6 area 12 Explanation As the graphs show the polar curves intersect when 4 sin 2 sin i e at 0 Thus the area of the shaded region is given by Z o 1 n 2 2 I 4 sin 2 sin d 2 0 Consequently 1 I 2 Consequently area A 3e 2 005 10 0 points 3 Z 12 sin2 d 0 To evaluate this last integral we use the double angle formula sin2 The shaded region in 1 1 cos 2 2 For then Z Z 2 12 sin d 6 0 0 1 cos 2 d h i 1 6 sin 2 6 2 0 Consequently the shaded region has lies inside the polar curve r 4 sin and outside the polar curve r 2 sin Determine the area of this region 1 area 3 correct area 3 Are there any other ways of calculating this area without using integration jones bwj276 Homework 2 pascaleff 55960 keywords area polar cooordinates definite integral circle 006 10 0 points Find the area of the shaded region 4 on the other hand to determine the rays 0 and 1 bounding the shaded region note that as ranges from 2 3 to the graph i passes though the origin when r 0 i e when 2 3 ii and then crosses the x axis when at r 1 Thus the area of the shaded region is given by the integral 1 A 2 Z 1 2 cos 2 d 2 3 But 1 2 cos 2 1 4 cos 4 cos2 3 4 cos 2 cos 2 inside the graph of r 1 2 cos since 1 cos 1 cos 2 2 2 1 area 2 area 3 area 4 area 3 3 4 3 2 1 3 3 correct 2 4 3 3 4 5 area 6 area 2 3 3 1 7 area 2 4 Explanation The area of a region bounded by the graph of the polar function r f and the rays 0 1 is given by the integral Z 1 1 A f 2 d 2 0 In this question the function is r 1 2 cos Hence 1 A 2 Z 2 3 3 4 cos 2 cos 2 d i 1h 3 4 sin sin 2 2 2 3 Consequently 3 3 1 area A 2 4 keywords polar graph area cardioid polar integral 007 10 0 points Find the area of the region enclosed by the graph of the polar function r 3 cos jones bwj276 Homework 2 pascaleff 55960 since 1 area 9 2 area cos2 5 1 1 cos 2 2 But then Z 1 1 2 19 6 cos cos 2 d A 2 0 2 2 i2 1 1 h 19 6 sin sin 2 2 2 4 0 19 correct 2 3 area 20 4 area 10 Consequently 17 5 area 2 area A Explanation The area of the region bounded by the graph of the polar function r f and the …