Homework 3 - This print—out should have 16 Questions....

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: This print—out should have 16 Questions. Multiple-choice questions may continue on the next column or page w find all choices a before answering. Locate the points given in polar coordinates by P( among 1 1, "2'7? 001 W( Vincent (jmv777) — HWO3 m Gilbert W (57495) »- 1 10.0 points 1 6 2, we) , 3(2, 3w), 1. P: Q- 2. P19 Q 3. P : 43 Q 4. P : Q. 5. Pz® Q 6. P-:0 Q1 Expianation: To convert from polar coordinates to Carte~ 81811 COOIdlflatBS we use a: 2: r0059, correct 3; : rsinél. For then the points P(l, ‘37?) (2(2, $71”) , R(2, 271'), correSpond to A Q: in Cartesian coordinates. 13:0 R: keywords: polar coordinates, Cartesian coor! dinates, change of coordinates, 002 10.0 points A point P is given in Cartesian coordinates by P(—~1,l). Find polar coordinates (139) of this point with r < 0 and 0 g 6 < 29¢. 3.. (we, 3;) correct 2. we 3. (5/1 3341 4- war) 5- we a we) Explanation: Since the relationship between Cartesian coordinates and polar coordinates is mmrcosfi, y =rsin6, the point P(—1, 1) in Cartesian coordinates can be given in polar coordinates as 3:), P(w~\/§, 4 003 10.0 points Vincent Univ???) —— HW03 m Gilbert — (57495) 3 1. r = 2(2ch 2. r = 2sec8 3. r :: 2 4. r = 2sint9 5. r m 2cos9 6. 6 z 2 correct Explanation: When the graph of a polar function cannot be determined directiy, it is sometimes more convenient to use the relations a : roost}, y = rsind, 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 loci: at the six polar functions listed above. ' 1.In Cartesian form 2 m 2 9 = w r sec €086 becomes x m 2 and its graph is a vertical line to the right of the origin. 2. In Cartesian form r 2 203m? : 51nd becomes y = 2 and its graph is a horizontal line lying above the rwaxis. 3. After multiplication, r = Boost} can be written as r2 — 2rcosd = 0, which in Cartesian form becomes m2+y2-293 = 0, is, (m—i)2+y2 = 1. Its graph is a circle centered on the naxis to the right of the origin and passing through the origin. 4. After multiplication, r = 2sin6 can be written as r2 m Ersint) m e, which in Cartesian forrn becomes $2+y2m2y m U, i.e.,m2+{y~— l)2 21' Its graph is a circle centered on the ymaxis above the origin and passing through the ori— gin. 5. Convertingr =: 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, i.e., 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 9 m 2 (in radians) with the the $~aXiS. This line will have negative slope because 7r/2 < theta < it when 9 = 2. Consequently, the given graph is that of the polar function 9:2 keywords: polar graph, polar function, line, Cartesian graph, circle 006 10.0 points Use the graph in Cartesian coordinates Vincent (jaw???) w HW03 — Gilbert m (57495) 5 “ -Which One of the following could be the graph of the polar function r m 511136? correct Explanation: The graphs of 7" = sin6, r = cost? are Circles passing through the origin. This question shows how the graphs change as the angle variable changes from 6 to muitipies 26, 39, etc. To determine the polar graph of r m sin 39 look first at its Cartesian graph Then, by tracing in polar coordinates the value of r as 6 varies, looking especiaily at Vincent (jinx/r777) — ewes - Gilbert M (57495) 7 5. ymintercept :2 g Explanation: _ The usual point—slope formula can be used to find an equation for the tangent line to the graph of a polar curve 7" 2 f0?) at a point P once we know the Cartesian coordinates 13(330, yo) of P and the slope of the tangent line at P. Now, when then while y(0) m (356 -~ 4)sin6. r:i‘hns in Cartesian coordinates, the point P corresponding to 6 2: 0 is (~1, 9). On the other hand, ac’(9) 2 ~36” cost? w sint9(86“6 —- 4), y’(6)_ = "—366 51nd + cost'5’(3e"‘9 — 4), so the slope at P is given by gig M0) m 1 aimless: w’(0) m 3' Consequently, by the point slepe formuia, the tangent line at P has equation 1 andsohas 1 y-intercept m ~57 010 10.0 points Find an equation for the tangent line to the graph of ' r m 2eos9 m4sin6 ate 2 Tr/li. 1 . 2. y = Bat—1 l 3 =w~ 2 y 2mi- 1 4 2w w-l y 253 5. y m 22: + 1 correct 6. y m 233~i~2 Explanation: The usuai point—siope formula can be used . to find an equation for the tangent line to the graph of a polar curve 7* = f(6) at a point P once we know the Cartesian coordinates P(:to, ya) of P and the slope of the tangent Eine at P. Now the graph of r m f (6) can expressed by the parametric equations y = f(6)sin9, and in this forrn the slope of the tangent line is given by a: = f(8)cosd, dy y’(9) _ f’(9)sin6+f(9)cos9 "6i; "" we) ‘ f’(6)cos6— name In the given case of r = 2oosf9 —— 431119, therefore, y’(6) = 2(eos2 6 — sin? 8) - Seine} c086, whiie 33(0) = W4: sin 69 cos 9 —- éi(eos2 6 — sin 9). Consequently, the tangent line has axe/4) m 2 m’(7r/4) ' siope = Vincent (jrnv777) — HWOB — Gilbert m (57495) 9 Explanation: The area of the region bounded by the graph of the polar function r = f(0) and the rays 6 x 80, 81 is given by the integral 1 91 A = w f(9)2d9. 2 60 When ' ma) = «60059, 90 = 0, 61 a 7‘" '6‘ 1 therefore, the area of the enclosed region is thus given by the integral 1 7T/6 A = —/ (V60086)2d6l {3 2. 11/6 2: 3/ 0039658. 0 Consequently, area: 3[sin9]:/6 m 3 ml 013 10.0 points Find the area of the region bounded by the polar curve r m x/7ln9-1—4 as well as the rays 6 = 1 and 6 m 5. El 1 1. area #28 ~i- 3) 2. area m Lie—r3 l 3. area = §(£le+3) correct 1 4. area = g(4e+8) I 5. area = g(2e+3) 6. area m 2e+3 Expkanation: The area of the region bounded by the graph of the polar function r we fit?) as well as the rays 6 = 60, 61 is given by the integral 1 61 A m - flame. 2 (,0 When fit?) m v7ln9+4, 90=1,91=€, therefore, the area of the enclosed region is thus given by the integral 6 A a (x/71n6+4)3d9 1 1 8 2 w/ (71n6+4)d6. 2 1 To evaluate this last integrai we use Integra» tion by Parts, for then 1 e 1 e A _ §l791n9+4allmjfl we 1 e m ilremewasll. Censequently, area 2 A m %(46+3) . 014 10.0 points Find the area of the region bounded by the polar curve r x 48mg . as well as the rays (9 = O and 0 = 2. 1. area m 4(1-— 8—4) correct 2. area m 8(1—-e"4) 3. area m 16(1—m 6'25 4. area m 8(1— 8—2) 5. area. m 16(1me_4‘) 6. area m Min—64) Vincent (jmv’???) —~ HWOB w Gilbert w (57495) 11 Find the area of the shaded region shown in between the graphs of the spiral r = 26 and. the circle 7‘ as sin 9. 1% (1:722 + 1) ' ' 2 2. area §w(§w2 — 1) correct 1. area m 3. area m éw<§n2 +1) 4. area = %(§7r3w 1) 5. area m 1—16C—g7r34n1) 6. area = E£6~<§7r3- 1) Explanation: The area of a region bounded by the graphs of polar function r =: f0?) and r m 9(6) be— tween the rays 9 = 90, 91 is given by the integral " 1 91 A = 5 lflgl2"9(9l2|d9- New in the given example, 09 = 0 while 91 : 7r / 2. Thus the shaded region has 1 TF/2 area = E f (492 — sin2 (9)019. 0 On the other hand, 1 sin26 2 5(1 m 008249). Thus area m é/GWB (492 m m; + “12"— cos 28) all? a $5393 —~ «2316 + isinZQE/g = -:-<—:-«3 — a - Consequently, the shaded region has area = —;—7T(i:—7r2m l) . keywords: definite integral, polar integral, area between curves, spiral, circle, ...
View Full Document

Page1 / 6

Homework 3 - This print—out should have 16 Questions....

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online