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Unformatted text preview: This print—out should have 16 Questions.
Multiplechoice 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+y2293 = 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
yintercept 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 wl
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/7ln91—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(9l2d9 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, ...
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 Fall '09
 Gilbert

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