7.
x = cosh t, y = sinh2 t.
Parabola x 2 − y = 1, or y = x 2 − 1, traversed left to right.
10.
x = cos t + sin t, y = cos t − sin t, (0 ≤ t ≤ 2π )
The circle x 2 + y 2 = 2, traversed clockwise, starting and ending at (1, 1).
11.
4
y = t 3 − 3t
2
1+t
dy
dx
8t
= 3(t 2 − 1)
=−
2
2
dt
dt
(1 + t )
Horizontal tangent at t = ±1, i.e., at (2, ±2).
Vertical tangent at t = 0, i.e.,
√ at (4, 0).
Self-intersection at t = ± 3, i.e., at (1, 0).
y
x=
t=−1
√
t=± 3
t=0
x
t=1
Fig. R-11
14.
y = t 3 − 12t
x = t 3 − 3t
dy
dx
= 3(t 2 − 1)
= 3(t 2 − 4)
dt
dt
Horizontal tangent at t = ±2, i.e., at (2, −16) and (−2, 16).
Vertical tangent at t = ±1, i.e., at (2, 11) and (−2, −11).
dy
t2 − 4
> 0 if |t| > 2 or |t| < 1
Slope
= 2
<
0 if 1 < |t| < 2
dx
t −1
Slope → 1 as t → ±∞.
y
(−2,16)
(2,11)
(−2,−11)
x=t 33 −3t
y=t −12t
(2,−16)
Fig. R-14
x
17.
x = et − t, y = 4et/2 , (0 ≤ t ≤ 2). Length is
Z
2p
(et − 1)2 + 4et dt
Z 2p
Z 2
t
2
(e + 1) dt =
=
(et + 1) dt
0
0
2
= (et + t) = e2 + 1 units.
L=
0
0
23. r = 1 + 2 cos(2θ )
y
r=1+2 cos 2θ
x
Fig. R-23
26. Area of a small loop:
Z
1 π/2
(1 + 2 cos(2θ ))2 dθ
A =2×
2 π/3
Z π/2
=
[1 + 4 cos(2θ ) + 2(1 + cos(4θ ))] dθ
π/3
π/2
1
= 3θ + 2 sin(2θ ) + sin(4θ ) 2
π/3
√
3 3
π
sq. units.
= −
2
4
28. r cos θ = x = 1/4 and r = 1 + cos θ intersect where
1
4 cos θ
2
4 cos θ + 4 cos θ − 1 = 0
√
√
−4 ± 16 + 16
± 2−1
cos θ =
=
.
8
2
1 + cos θ =
√
Only ( 2−1)/2 is between −1 and 1, so is a possible value of cos θ . Let θ0 = cos−1
Then
v
p
!2
u
√
√
u
2
−
1
1+2 2
t
.
=
sin θ0 = 1 −
2
2
By symmetry, the area inside r = 1 + cos θ to the left of the line x = 1/4 is
Z 1 π
1 + cos(2θ )
A =2×
dθ + cos θ0 sin θ0
1 + 2 cos θ +
2 θ0
2
π
3
1
= (π − θ0 ) + 2 sin θ + sin(2θ ) 2
4
θ0
p
√
√
( 2 − 1) 1 + 2 2
+
4
! q
!
√
√
√
2
−
1
2
−
9
3
=
π − cos−1
+ 1+2 2
sq. units.
2
2
8
y
x=1/4
r=1+cos θ
θ0
x
Fig. R-28
√
2−1
.
2
2. Let S1 and S2 be two spheres inscribed in the cylinder, one on each side of the plane that
intersects the cylinder in the curve C that we are trying to show is an ellipse. Let the spheres
be tangent to the cylinder around the circles C1 and C2 , and suppose they are also tangent
to the plane at the points F1 and F2 , respectively, as shown in the figure.
C1
A1
S1
C
P
F2
F1
C2
A2
S2
Fig. C-2
Let P be any point on C. Let A1 A2 be the line through P that lies on the cylinder, with A1
on C1 and A2 on C2 . Then P F1 = P A1 because both lengths are of tangents drawn to the
sphere S1 from the same exterior point P. Similarly, P F2 = P A2 . Hence
P F1 + P F2 = P A1 + P A2 = A1 A2 ,
which is constant, the distance between the centres of the two spheres. Thus C must be an
ellipse, with foci at F1 and F2 .
18. The two curves r 2 = 2 sin 2θ and r = 2 cos θ intersect where
2 sin 2θ = 4 cos2 θ
⇔
4 sin θ cos θ = 4 cos2 θ
(sin θ − cos θ ) cos θ = 0
sin θ = cos θ or cos θ = 0,
h√ π i
2,
and P2 = (0, 0).
4
√
dr
= 4 cos 2θ . At P1 we have r = 2 and dr/dθ = 0. Thus
For r 2 = 2 sin 2θ we have 2r
dθ
the angle ψ between the curve and the radial line θ = π/4 is ψ = π/2.
For r = 2 cos θ we have dr/dθ = −2 sin θ , so the angle between this curve and the radial
r = −1, and ψ = 3π/4. The two curves
line θ = π/4 satisfies tan ψ =
dr/dθ θ =π/4
3π
π
π
intersect at P1 at angle
− = .
4
2
4
The Figure shows that at the origin, P2 , the circle meets the lemniscate twice, at angles 0
and π/2.
i.e., at P1 =
y
x
r 2 =2 sin 2θ
r=2 cos θ
Fig. 6-18
11.
Z
dy
3y
dx
dy
=
⇒
=3
dx
x −1
y
x −1
⇒ ln |y| = ln |x − 1|3 + ln |C|
⇒ y = C(x − 1)3 .
Since y = 4 when x = 2, we have 4 = C(2 − 1)3 = C, so the equation of the curve is
y = 4(x − 1)3 .
8.
y
Q
y = f (x)
L
P(x, y)
x
(L , 0)
Fig. C-8
If Q = (0, Y ), then the slope of P Q is
dy
y−Y
= f ′ (x) =
.
x −0
dx
Since |P Q| √
= L, we have (y − Y )2 = L 2 − x 2 . Since the slope dy/d x is negative at P,
dy/d x = − L 2 − x 2 /x. Thus
y=−
Z √
L2 − x2
d x = L ln
x
L+
√
L2 − x2
x
!
−
p
L 2 − x 2 + C.
Since y = 0 when x = L, we have C = 0 and the equation of the tractrix is
L+
y = L ln
√
L2 − x2
x
!
−
p
L 2 − x 2.
Note that the first term can be written in an alternate way:
y = L ln
x
L−
√
L2
−
x2
−
p
L 2 − x 2.
18. Since f (x) =
2
> 0 on [0, ∞) and
π(1 + x 2 )
2
π
Z
0
∞
2
dx
2 π −1
=
lim
tan
(R)
=
= 1,
R→∞ π
1 + x2
π 2
therefore f (x) is a probability density function on [0, ∞). The expectation of X is
Z
2 ∞ x dx
µ = E(X ) =
π 0 1 + x2
1
ln(1 + R 2 ) = ∞.
= lim
R→∞ π
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