hw2 - * HW2 * segerman * (58615) 1 This print—out should...

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Unformatted text preview: * HW2 * segerman * (58615) 1 This print—out should have 15 questions. Multiple—choice questions may continue on the next column or page — find all choices before answering. 001 10.0 points Locate the points given in polar coordinates 5. P:A Q1. R:O 6. P2. Q20 R:A correct Explanation: To convert from polar coordinates to Carte— sian coordinates we use :23:7°cos6, y:rsin6. For then the points P(—3, lit) Q(2, in), R(2, E71“), 2 6 4 correspond to P : O Q 2 O R : A in Cartesian coordinates. keywords: polar coordinates, Cartesian coor— dinates, change of coordinates, 002 10.0 points Which, if any, of A. (2, l7Tf/6), B. (4, 27r/3), C. (—2, —57r/6), are polar coordinates for the point given in Cartesian coordinates by P(7\/§, l)? 1. C only 2. B only 3. all of them 4. A only 5. B and C only 6. A and C only correct 7. A and B only 8. none of them Explanation: To convert from Cartesian coordinates to polar coordinates we use the relations: 3::7‘cosél, y:rsin6, sothat r2 = 332+y2, tant9 = :1: For the point P(7\/§, l) in Cartesian co— ordinates, therefore, one choice of r and 6 is * HW2 * segernian * (58615) 2 7‘ : 2 and 6 : Eur/6, but there are equivalent solutions for r < 0 as well as values of (9 dif— 1. 2?" 2 sec 6tant9 fering by integer multiples of 7r. For the given choices we thus see that 2. 2r : sec 6 cot 6 A. TRUE: difiers frorn Eur/6 by 27?. B. FALSE: 7‘ and 6 incorrect. C. TRUE; 4. r = 2sect9tant9 3. r : 2csc6tan6 —2005(57T/6) : _\@: —251n(57T/6) : 1- 5. r 20sct900tt9 correct . 6. 2r : csc6cot6 003 10.0 p01nts Explanation: A ' t P ' ' ' C t ' d' t po1n is given in ar es1an coor 1na es We have to substitute for SE: y in by 13(71, 1). Find polar coordinates (r, ('9) of this point With r < 0 and 0 3 6 < 27?. 2 i y 7 2:1: \/— 77? 1- (* 33?) using the relations 2 mzrcosfi, yzrsinél. . , 4 7 ln this case the Cartesian equation becomes 3. (—x/Z —) correct 4 7‘2 sin26 : 27° cosG. f 37? 4' (_ 3: Consequently, the polar form of the equation is 5 5. (ix/g, r = 2csct900tt9 37? 6. (ix/5, I) 005 10.0 points Explanation: Find a Cartesian equation for the curve Since the relationship between Cartesian , , given by the polar equation coordinates and polar coordinates is 3::rcosél, y:rsin6, 7’ : 45mg- the point 13(71, 1) in Cartesian coordinates can be given in polar coordinates as 1, (35 + 2)2 W y? = 4 7 2 2 p(_\/§7 I”) a 2. (:1: 2) y 4 : 0 004 10.0 points Find a polar equation for the curve given by the Cartesian equation 39:23:. 6.12:2+(y+2)2:4 * HW2 * segerman * (58615) 3 7. 3:2 W (y i 2)2 = 4 correct 8. (x——2)2+y2+4 : 0 Explanation: We have to replace r and 6 in the polar equation r = 4sint9 using the relations :23:7°cos6, y:rsin6. As a first simplification, notice that r2 = 4rsint9. But then 2 2324—112 : r : 4rsin6 : 43;, in which case 3:2 + y2 7 4y 2 0 . Consequently, by completing the square we get the Cartesian equation m2+(yi2)2 = 4 006 10.0 points Which one of the following polar functions has graph 1. r = 2sect9 2. r : 20056 3. r = 2csct9 correct 4. 7‘ : 2sin6 5. 7‘ : 2 6. t9 = 2 Explanation: When the graph of a polar function cannot be determined directly, it is sometimes more convenient to use the relations :1: = rcosél, y = rsmt9, to convert the polar form to Cartesian form and then use standard knowledge of Cartesian graphs. This is often the case with special lines and circles, so let’s look at the six polar functions listed above. 1.In Cartesian form 2 cos (9 7‘ : 2sec€ : becomes :2: : 2 and its graph is a vertical line to the right of the origin. 2. In Cartesian form 2 T 050 Sin 9 becomes 3; : 2 and its graph is a horizontal line lying above the sic—axis. 3. After multiplication, r = 2 cosél can be written as 7‘2 —2rcos6 : 0, which in Cartesian form becomes 3324—92—22: : 0, is, (33—1)2—|—y2 :1. Its graph is a circle centered on the :IJ—aXis to the right of the origin and passing through the origin. 4. After multiplication, r = 2sint9 can be written as 7‘2 —2rsin6 : 0, 7 HW2 * segerman * (58615) 4 Which in Cartesian form becomes m2+y272y = 0, i.e., 5132+(y7 1)2 =1. Its graph is a circle centered on the y—aXis above the origin and passing through the ori— gin. 5. Converting r : 2 to Cartesian form does not help at all, Indeed, in terms of polar coordinates, its graph consists of all points distance 2 from the origin, ta, its graph is a circle centered at the origin having radius 2. 6. Converting 6 : 2 to Cartesian form does not help much. Indeed, in terms of polar coordinates its graph consists of all points on the line through the origin making an angle 6 = 2 (in radians) With the the m—aXis. This line Will have negative slope because 7r/ 2 < theta < 7? When 6 = 2. Consequently, the given graph is that of the polar function r : 2csc€ keywords: polar graph, polar function, line, Cartesian graph, circle 007 10.0 points Find the slope of the tangent line to the graph of r :2—sin6 att9=7r/6. 1 1. slope: —§\/§ 2 sloei 2fi—1 ' p 2+\/§ 3 sloei —2fi+1 ' p 2+\/§ 2 3—1 4. slope: [— £72 5. slope : —3\/§ 1 6. slope 2 fig 3c0rrect Explanation: The graph of a polar curve 7‘ : f(6) can expressed by the parametric equations y : f(6) sin6l. In this form the slope of the tangent line to the curve is given by :2: : f0?) c056, d_y : W9) d3: 513’(6) . But When r = 2 * sint9, we see that y’(t9) = * cost9sint9 + (2 * sinél) cost9, While 56(9) 2 * cosgél * (2 * sinél) sint9. But then ’(W) * “g0 m y 6 i 2 7 While we — L i 1 6 _ 2 ’ Consequently, at 6 : 7T/6, d 1 4” *——\/§. 1 : i S ope da: ear/6 3 008 10.0 points Find an equation for the tangent line to the graph of r = 3679 i 4 at the point P corresponding to 6 : 0. 1. y = 33: + 1 2. y + 33: = *3 3. y : 393+?) * HW2 * segerman * (58615) 5 4. 3y 2 3:773 5. 3y : :23——1correct 6. 3y—l—zz: : —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(6) at a point P once we know the Cartesian coordinates P(:2:0, yo) of P and the slope of the tangent line at P. Now, when then while y(6) = (38ft? * 4) sint9. Thus in Cartesian coordinates, the point P corresponding to (9 = 0 is (*1, 0). On the other hand, 33’09) = *3€_9 cost? * sint9(3e_9 * 4), while y’(t9) = *38i6 sint9 + cos (9(38i9 * 4), so the slope at P is given by d_y : MO) d3: 9:0 :23’(0) 1 3 . Consequently, by the point slope formula, the tangent line at P has equation 1 y : §(x—l—1) which after simplification becomes 3y ::I:+1 009 10.0 points Find the area of the region bounded by the polar curve r : V 6 sin 6 as well as the rays 6 : 0 and 6 : 7T/3. 1 9 . area : — 8 3 2. area : 5 correct 11 3. area = 7 8 5 4. area = — 4 5. area : 1 Explanation: The area of the region bounded by the graph of the polar function r = f(t9) and the rays 6 = (90, ('91 is given by the integral a. A :1/ f(6)2d6. 290 When 1‘09) 2 V6sin6, Tl' (9020,6123, therefore, the area of the enclosed region is thus given by the integral 7r/3 A = 1/ (x/6sint9)2dt9 0 2 7r/3 : 3/ sinGdQ. 0 Consequently, 7?/3 3 area = 3[icos9] = i 0 2 010 10.0 points Find the area of the shaded—region inside the polar curve 7‘ : 2cos6 shown in A W When gbl : 7r/6 and gbg : 7r/3. 1 1. : — — area 27? 1 2. area = 77v 2 1 3. area : 77r— 2 1 4. area : —7r— 6 1 5. area = —7r7 6 1 6. area = 77v 6 Explanation: The area of the region bounded by the f(6) and graph of the pol $242?) 7 Elam/g) _; — i(2+fi) ~ :mfi) ar function r * HW2 * segerrnan * (58615) 6 Thus 7r/3 A = / (1+cos26)d6 far/6 1 . GT/3 2 [6 + 7 Sin 26] . 2 —7r/6 Consequently, area : 1 1 i i 3 27r+2f keywords: polar graph, polar integral, double angle, 011 10.0 points Find the area of the shaded region shown in the rays 61 and 62 is given by the integral 192 between the graphs of the spiral r : 46 and A : _ f(6)2 d9. the circler = 2sin6. 2 91 1 i 1 4 3 1 When . area 7 — ) 2 260893 91 2 i951: 62 Z 9623 2. area : :(3W3+ therefore, 1 4 1 7r/3 3. area : —(—7r3— 1) A = 400526669. 4 3 far/6 4 1 . area = —7r —7r On the other hand, 4 3 2 1 1 2 2 cos 6 : §(1—|—cos26). 5. area : §7T(§7T — 1) correct 6 1 (2 2+1) . area = *7? 7w 2 3 Explanation: The area of a region bounded by the graphs of polar function r = 1‘09) and r = 9(6) be— tween the rays 6 : (90, (91 is given by the integral 1 31 2 2 A : 5/90 m6) —g<e) Me. Now in the given example, 60 : 0 while ('91 = 7r/2. Thus the shaded region has 1 717/2 area = (16192 *4sin2 ('9) d9. 0 On the other hand, 1 sin2t9 = §(l * cos2t9). Thus 1 717/2 2 area = (166 72+2c0s26)d6 0 /2 = 1[166326+sin26]fi 2 3 0 12 = {4377:}. 2 3 Consequently, the shaded region has 1 2 2 area = 77r(77r * l) 2 3 keywords: definite integral, polar integral, area between curves, spiral, circle, 012 10.0 points Which one of the following conic sections has graph 1. m2—23:+y—1 : 0 2. y2 W 2yimil = 0correct 3. 2227722279'71 = 0 4. x2—2x—y—1 : 0 5. y2——2y—l—:E—1 : 0 6. $2772m+yil = 0 7. $721,751,171 = 0 8. y2—2y+:c—1 : 0 Explanation: The graph is that of a parabola opening to the right and having vertex at (*2, *1). Thus it is the graph of (y+1)2 : A(:z:+2), for some A > 0. But the graph also passes through (—1, —2), so A : 1. Consequently, the graph is that of the conic section (y+1)2 : m+23 which after expansion and simplification can be rewritten as y2+2y—:z:—1 : 0 013 10.0 points Which one of the following conic sections has graph HW2 * segerrnan * (58615) 8 1. 4x27322+9y2713y : 23 2. 9132—1823—4112—81/ : 41 23 correct 3. 42:2 7 33: 4 9y? +134; 4. 92:2 +134: 7 4y2 7 By 41 5. 9$2—18m—4y2+8y : 41 6. 4x2+8x+9y2+18y : 23 Explanation: The graph is a shifted ellipse centered at (1, —1) whose major axis has length 6 and minor axis has length 4. Thus one equation for this ellipse is _ 2 2 <2: 1)+(y+1) :1, 9 4 which after expansion and simplification can be rewritten as 43:2 732493912 +18y = 23 014 10.0 points Which one of the following conic sections has graph 8y 2 7 correct 3. m2—2m+4y2——8y+1 : 0 4. 2227242743127314 : 7 5. 4y273y72272x =1 6. 44127327422722 :1 7. 4y2+8$—2:2——2y :1 8. x2+2m—4y2—8y : 7 Explanation: The graph is a shifted hyperbola having asymptotes which intersect at (*1, l) as well as vertices at (13 l) and (*3, 1). Thus one equation for this hyperbola is (4+1)? (471)? 11 4 1 i ’ which after expansion and simplification can be rewritten as 1232+2123—4y2+8y : 7 015 10.0 points Find an equation for the parabola having vertex (0D 0) and focus (0, *2). 1. 51:2 : 3y 7 HW2 * segerman * (58615) 4. 51:2 —— 8y : Ocorrect 5. 51:2 2 2y 6. m2+3y = 0 Explanation: The parabola With equation 2 :1: = 4ay has focus (0, a) and vertex (0D 0). Conse— quently, When a : —2, the parabola has equa— tion m2+8y = 0 ...
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This note was uploaded on 09/28/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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hw2 - * HW2 * segerman * (58615) 1 This print—out should...

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