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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.
 Fall '07
 Gilbert

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