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Unformatted text preview: −1 and that x = ± 5/2 if y = 5. The curves intersect y (– √ 5 , 5 )
2 ( √ 5 , 5)
2 at ( 5/2, 5) and (− 5/2, 5), and thus the area is
√
5/2
√
A=2
( 5 + 8x2 − 2x) dx x 0 √
√
5√
5 2√
( 5 − 1) +
2 ln(2 + 5)
=
2
4
(b) Eliminate x to get y 2 = 1, y = ±1. Use either equation
to ﬁnd that x = ±2 if y = 1 or if y = −1. The curves
intersect at (2, 1), (2, −1), (−2, 1), and (−2, −1),
and thus the area is
√
5/3
1
A=4
1 + 2x2 dx
3
0
2 +4 √
= 5/3 1
3 1
1 + 2x2 − √
7 3x2 − 5 dx √
√
√
10 √
5
1√
ln 5
2 ln(2 2 + 3) +
21 ln(2 3 + 7) −
3
21
21 y (–2, 1) (2, 1)
x (–2, –1) (2, –1) Exercise Set 12.4 430 (c) Add both equations to get x2 = 4, x = ±2.
√
Use either equation to ﬁnd that y = ± 3 if x = 2
or if x = −2. The curves intersect at
√
√
√
√
(2, 3), (2, − 3), (−2, 3), (−2, 3) and thus
1 (–2, √ 3) (2, √ 3)
x 2 7 − x2 dx + 4 A=4 y 0 7 − x2 − x2 − 1 dx 1 √
= 4 3 + 14 sin−1 (–2, – √ 3) √
√
2√
7 − 4 3 + 2 ln(2 + 3)
7 (2, – √ 3) 45. (a) P : (b cos t, b sin t); Q : (a cos t, a sin t); R : (a cos t, b sin t)
(b) For a circle, t measures the angle between the positive xaxis and the line segment joining
the origin to the point. For an ellipse, t measures the angle between the xaxis and OP Q,
not OR.
46. (a) For any point (x, y ), the equation
y = b sinh t has a unique solution t, 3 (b) −∞ < t < +∞. On the hyperbola,
3 y2
x2
= 1 + 2 = 1 + sinh2 t
2
a
b
= cosh2 t, so x = ±a cosh t. 3 3 47. (a) For any point (x, y ), the equation y = b tan t has a unique solution t where −π/2 < t < π/2.
y2
x2
On the hyperbola, 2 = 1 + 2 = 1 + tan2 t = sec2 t, so x = ±a sec t.
a
b
3 (b) 3 3 3 48. By Deﬁnition 12.4.1, (x + 1)2 + (y − 4)2 = (y − 1)2 , (x + 1)2 = 6y − 15, (x + 1)2 = 6(y − 5/2)
49. (4, 1) and (4, 5) are the foci so the center is at (4, 3) thus c = 2, a = 12/2 = 6, b2 = 36 − 4 = 32;
(x − 4)2 /32 + (y − 3)2 /36 = 1
50. From the deﬁnition of a hyperbola,
(x − 1)2 + (y − 1)2 − (x − 1)2 + (y − 1)2 − x2 + y 2 = 1, x2 + y 2 = ±1, transpose the second radical to the right hand side of the equation and square and simplify to get ±2 x2 + y 2 = −2x − 2y + 1, square and simplify again
to get 8xy − 4x − 4y + 1 = 0.
51. Let the ellipse have equation
9/2 16 1 − V =2
0 4x2
81 4x2
4 2 y2
x+
= 1, then A(x) = (2y )2 = 16 1 −
81
4
81 dx = 96 , 431 Chapter 12 52. See Exercise 51, A(y ) = 53. Assume √ y2
x2
+ 2 = 1, A = 4
a2
b 54. (a) Assume √ 81
3
4 3x2 = 1− y2
4 ,V = √ 81
3
2 2 1−
0 √
dy = 54 3 y2
4 a 1 − x2 /a2 dx = πab b
0 y2
x2
+ 2 = 1, V = 2
a2
b a πb2 1 − x2 /a2 dx =
0 4
πab2
3 (b) In Part (a) interchange a and b to obtain the result.
55. Assume bx
y2
dy
x2
=− √
+ 2 = 1,
,1 +
a2
b
dx
a a2 − x2
a S=2
0 2πb
a 2πa 1 − y 2 /b2
0 2 dx
dy b S=2 2 = a4 − (a2 − b2 )x2
dx = 2πab
a2 − x2 1 − x2 /a2 56. As in Exercise 55, 1 + dy
dx = a4 − (a2 − b2 )x2
,
a2 (a2 − x2 )
a
c
b
+ sin−1
a
c
a ,c = a2 − b2 b4 + (a2 − b2 )y 2
,
b2 (b2 − y 2 ) b4 + (a2 − b2 )y 2
dy = 2πab
b2 (b2 − y 2 ) a b a+c
+ ln
b
c
b ,c = a2 − b2 57. Open the compass to the length of half the major axis, place the point of the compass at an end
of the minor axis and draw arcs that cross the major axis to both sides of the center of the ellipse.
Place the tacks where the arcs intersect the major axis.
58. Let P denote the pencil tip, and let R(x, 0) be the point below Q and P which lies on the line L.
Then QP + P F is the length of the string and QR = QP + P R is the length of the side of the
triangle. These two are equal, so P F = P R. But this is the deﬁnition of a parabola according to
Deﬁnition 12.4.1.
59. Let P denote the pencil tip, and let k be the diﬀerence between the length of the ruler and that
of the string. Then QP + P F2 + k = QF1 , and hence P F2 + k = P F1 , P F1 − P F2 = k . But this
is the deﬁnition of a hyperbola according to Deﬁnition 12.4.3.
60. In the x y plane an equation of the circle is x 2 + y 2 = r2 where r is the radius of the cylinder. Let
P (x, y ) be a point on the curve in the xy plane, then x = x cos θ and y = y so x2 cos2 θ + y 2 = r2
which is an equation of an ellipse in the xy plane.
61. L = 2a =
T = c= D 2 + p2 D 2 = D
a2 − b2 = 1 + p2 (see ﬁgure), so a = 1
D
2 1 + p2 , but b = 1
D,
2 1
1
12
D (1 + p2 ) − D2 = pD.
4
4
2
pD D 62. y = 12
1
x , dy/dx =
x, dy/dx
4p
2p =
x=x0 1
x0 , the tangent line at (x0 , y0 ) has the formula
2p x2
x2
x0
x0
12
(x − x0 ) =
x − 0 , but 0 = 2y0 because (x0 , y0 ) is on the parabola y =
x.
2p
2p
2p
2p
4p
x0
x0
Thus the tangent line is y − y0 =
x − 2y0 , y =
x − y0 .
2p
2p
y − y0 = Exercise Set 12.4 432 dy
dx 63. By implicit diﬀerentiation, =−
(x0 ,y0 ) b2 x0
if y0 = 0, the tangent line is
a2 y0 b2 x0
2
2
y y0 = − 2 (x − x0 ), a2 y0 y − a2 y0 = −b2 x0 x + b2 x2 , b2 x0 x + a2 y0 y = b2 x2 + a2 y0 ,
0
0
a y0
2
but (x0 , y0 ) is on the ellipse so b2 x2 + a2 y0 = a2 b2 ; thus the tangent line is b2 x0 x + a2 y0 y = a2 b2 ,
0
2
2
x0 x/a + y0 y/b = 1. If y0 = 0 then x0 = ±a an...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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