Homework 3

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

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

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

View Full Document

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

View Full Document

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

View Full Document
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

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online