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447
CHAPTER 11
Analytic Geometry in Calculus
EXERCISE SET 11.
(1, 6 )
(3, 3 )
(4, e) (–1,^ r)
0
p / 2
(5, 8 )
(–6, – p )
( 32 , L) 0
p / 2
(–5, @)^ (–3,^ i)
(2, $)
(0, c) (2, g)
- (a) (
3 , 3) (b) (− 7 / 2 , 7
3 /2) (c) (
(d) (0, 0) (e) (− 7
3 / 2 , 7 /2) (f ) (− 5 , 0)
- (a) (− 4
- (b) (
2 /2) (c) (
(d) (5, 0) (e) (0, −2) (f ) (0, 0)
- (a) both (5, π) (b) (4, 11 π/6), (4, −π/6) (c) (2, 3 π/2), (2, −π/2) (d) (
2 , 5 π/4), (
2 , − 3 π/4) (e) both (6, 2 π/3) (f ) both (
2 , π/4)
- (a) (2, 5 π/6) (b) (− 2 , 11 π/6) (c) (2, − 7 π/6) (d) (− 2 , −π/6)
- (a) (5, 0 .6435) (b) (
29 , 5 .0929) (c) (1. 2716 , 0 .6658)
- (a) (5, 2 .2143) (b) (3. 4482 , 2 .6260) (c) (
4 + π^2 / 36 , 0 .2561)
- (a) r^2 = x^2 + y^2 = 4; circle (b) y = 4; horizontal line (c) r^2 = 3r cos θ, x^2 + y^2 = 3x, (x − 3 /2)^2 + y^2 = 9/4; circle (d) 3 r cos θ + 2r sin θ = 6, 3x + 2y = 6; line
- (a) r cos θ = 5, x = 5; vertical line (b) r^2 = 2r sin θ, x^2 + y^2 = 2y, x^2 + (y − 1)^2 = 1; circle (c) r^2 = 4r cos θ + 4r sin θ, x^2 + y^2 = 4x + 4y, (x − 2)^2 + (y − 2)^2 = 8; circle (d) r =
cos θ
sin θ cos θ , r cos^2 θ = sin θ, r^2 cos^2 θ = r sin θ, x^2 = y; parabola
- (a) r cos θ = 7 (b) r = 3 (c) r^2 − 6 r sin θ = 0, r = 6 sin θ (d) 4(r cos θ)(r sin θ) = 9, 4r^2 sin θ cos θ = 9, r^2 sin 2θ = 9/ 2
- (a) r sin θ = − 3 (b) r =
(c) r^2 + 4r cos θ = 0, r = −4 cos θ (d) r^4 cos^2 θ = r^2 sin^2 θ, r^2 = tan^2 θ, r = tan θ
448 Chapter 11
0
p / 2
3
3
r = 3 sin 2θ
-3 3
-2.
r = 2 cos 3θ
0
p / 2
-4 (^) -1 4
r = 3 − 4 sin 3θ
0
p / 2
r = 2 + 2 sin θ
- (a) r = 5
(b) (x − 3)^2 + y^2 = 9, r = 6 cos θ (c) Example 6, r = 1 − cos θ
- (a) From (8-9), r = a ± b sin θ or r = a ± b cos θ. The curve is not symmetric about the y-axis, so Theorem 11.2.1(a) eliminates the sine function, thus r = a ± b cos θ. The cartesian point (− 3 , 0) is either the polar point (3, π) or (− 3 , 0), and the cartesian point (− 1 , 0) is either the polar point (1, π) or (− 1 , 0). A solution is a = 1, b = −2; we may take the equation as r = 1 − 2 cos θ. (b) x^2 + (y + 3/2)^2 = 9/ 4 , r = −3 sin θ (c) Figure 11.1.18, a = 1, n = 3, r = sin 3θ
- (a) Figure 11.1.18, a = 3, n = 2, r = 3 sin 2θ
(b) From (8-9), symmetry about the y-axis and Theorem 11.1.1(b), the equation is of the form r = a ± b sin θ. The cartesian points (3, 0) and (0, 5) give a = 3 and 5 = a + b, so b = 2 and r = 3 + 2 sin θ. (c) Example 8, r^2 = 9 cos 2θ
- (a) Example 6 rotated through π/2 radian: a = 3, r = 3 − 3 sin θ
(b) Figure 11.1.18, a = 1, r = cos 5θ (c) x^2 + (y − 2)^2 = 4, r = 4 sin θ
450 Chapter 11
Spiral
2 p 4 p
6 p
8 p
- (^2) p
6 p
4 p
Spiral
1
Four-petal rose
3
Four-petal rose
9
Eight-petal rose
2
Three-petal rose
-1 1
-1 1
-3 3
-3 3
-1 1
- 0 ≤ θ ≤ 8 π 57. (a) − 4 π < θ < 4 π
- In I, along the x-axis, x = r grows ever slower with θ. In II x = r grows linearly with θ. Hence I: r =
θ; II: r = θ.
- (a) r = a/ cos θ, x = r cos θ = a, a family of vertical lines
(b) r = b/ sin θ, y = r sin θ = b, a family of horizontal lines
Exercise Set 11.1 451
- The image of (r 0 , θ 0 ) under a rotation through an angle α is (r 0 , θ 0 + α). Hence (f (θ), θ) lies on the original curve if and only if (f (θ), θ + α) lies on the rotated curve, i.e. (r, θ) lies on the rotated curve if and only if r = f (θ − α).
- (a) r = 1 + cos(θ − π/4) = 1 +
(cos θ + sin θ) (b) r = 1 + cos(θ − π/2) = 1 + sin θ (c) r = 1 + cos(θ − π) = 1 − cos θ (d) r = 1 + cos(θ − 5 π/4) = 1 −
(cos θ + sin θ)
- r^2 = 4 cos 2(θ − π/2) = −4 cos 2θ
- Either r − 1 = 0 or θ − 1 = 0, so the graph consists of the circle r = 1 and the line θ = 1. 0
p / 2
r = 1
u = 1
- (a) r^2 = Ar sin θ + Br cos θ, x^2 + y^2 = Ay + Bx, (x − B/2)^2 + (y − A/2)^2 = (A^2 + B^2 )/4, which
is a circle of radius
A^2 + B^2.
(b) Formula (4) follows by setting A = 0, B = 2a, (x − a)^2 + y^2 = a^2 , the circle of radius a about (a, 0). Formula (5) is derived in a similar fashion.
- y = r sin θ = (1 + cos θ) sin θ = sin θ + sin θ cos θ, dy/dθ = cos θ − sin^2 θ + cos^2 θ = 2 cos^2 θ + cos θ − 1 = (2 cos θ − 1)(cos θ + 1); dy/dθ = 0 if cos θ = 1/2 or if cos θ = −1; θ = π/3 or π (or θ = −π/3, which leads to the minimum point). If θ = π/ 3 , π, then y = 3
3 / 4 , 0 so the maximum value of y is 3
3 /4 and the polar coordinates of the highest point are (3/ 2 , π/3).
- x = r cos θ = (1 + cos θ) cos θ = cos θ + cos^2 θ, dx/dθ = − sin θ − 2 sin θ cos θ = − sin θ(1 + 2 cos θ), dx/dθ = 0 if sin θ = 0 or if cos θ = − 1 /2; θ = 0, 2π/3, or π. If θ = 0, 2π/3, π, then x = 2, − 1 / 4 , 0 so the minimum value of x is − 1 /4. The leftmost point has polar coordinates (1/ 2 , 2 π/3).
- (a) Let (x 1 , y 1 ) and (x 2 , y 2 ) be the rectangular coordinates of the points (r 1 , θ 1 ) and (r 2 , θ 2 ) then
d =
(x 2 − x 1 )^2 + (y 2 − y 1 )^2 =
(r 2 cos θ 2 − r 1 cos θ 1 )^2 + (r 2 sin θ 2 − r 1 sin θ 1 )^2
r^21 + r 22 − 2 r 1 r 2 (cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ) =
r^21 + r^22 − 2 r 1 r 2 cos(θ 1 − θ 2 ). An alternate proof follows directly from the Law of Cosines. (b) Let P and Q have polar coordinates (r 1 , θ 1 ), (r 2 , θ 2 ), respectively, then the perpendicular from OQ to OP has length h = r 2 sin(θ 2 − θ 1 ) and A = 12 hr 1 = 12 r 1 r 2 sin(θ 2 − θ 1 ). (c) From Part (a), d =
9 + 4 − 2 · 3 · 2 cos(π/ 6 − π/3) =
(d) A =^1 2 2 sin(5π/ 6 − π/3) = 1
- (a) 0 = (r^2 + a^2 )^2 − a^4 − 4 a^2 r^2 cos^2 θ = r^4 + a^4 + 2r^2 a^2 − a^4 − 4 a^2 r^2 cos^2 θ
= r^4 + 2r^2 a^2 − 4 a^2 r^2 cos^2 θ, so r^2 = 2a^2 (2 cos^2 θ − 1) = 2a^2 cos 2θ.
Exercise Set 11.2 453
- (a) dy/dx = (4 cos t)/(−3 sin t) = −(4/3) cot t; dy/dx
t=π/ 4 =^ −^4 /^3 , dy/dx
t=7π/ 4 = 4/^3 (b) (x/3)^2 + (y/4)^2 = 1, 2 x/9 + (2y/16)(dy/dx) = 0, dy/dx = − 16 x/ 9 y, dy/dx
x=3/√ 2 y=4/√ 2
= − 4 /3; dy/dx
x=3/√ 2 y=− 4 /√ 2
d^2 y dx^2 =^
d dx
dy dx =^
d dt
dy dx
dt dx =^ −^
4 t^2 (1/^2 t) =^ −^1 /(8t
(^3) ); positive when t = −1,
negative when t = 1
d^2 y dx^2 =^
d dt
dy dx
dt dx =^
−(4/3)(− csc^2 t) −3 sin t =^ −^
9 csc
(^3) t; negative at t = π/4, positive at t = 7π/4.
- dy/dx =
t)
t, d^2 y/dx^2 =
t 1 /(
t) = 4, dy/dx
t=1 = 4,^ d
(^2) y/dx 2 ∣∣ t=1 = 4
- dy/dx = t^2 t = t, d^2 y/dx^2 =
t , dy/dx
t=2 = 2,^ d
(^2) y/dx 2 ∣∣ t=2 = 1/^2
- dy/dx = sec
(^2) t sec t tan t = csc t, d^2 y/dx^2 = −^ csc^ t^ cot^ t sec t tan t = − cot^3 t,
dy/dx
t=π/ 3 = 2/
3, d^2 y/dx^2
t=π/ 3 =^ −^1 /(
- dy/dx = sinh^ t cosh t = tanh t, d
(^2) y dx^2 = sech^2 t/ cosh t = sech^3 t, dy/dx
t=0 = 0, d (^2) y/dx 2 ∣∣ t=0 = 1
dy dx
dy/dθ dx/dθ
− cos θ 2 − sin θ
d^2 y dx^2
d dθ
dy dx
dx dθ
(2 − sin θ)^2
2 − sin θ
(2 − sin θ)^3
dy dx
∣θ=π/ 3 =^
d^2 y dx^2
∣θ=π/ 3 =^
3 /2)^3
3)^3
- dy dx = 3 cos^ φ − sin φ = −3 cot φ; d
(^2) y dx^2 = d dφ (−3 cot φ) dφ dx = −3(− csc^2 φ)(− csc φ) = −3 csc^3 φ;
dy dx
φ=5π/ 6
d^2 y dx^2
φ=5π/ 6
- (a) dy/dx = −e
−t et^ = −e−^2 t; for t = 1, dy/dx = −e−^2 , (x, y) = (e, e−^1 ); y − e−^1 = −e−^2 (x − e), y = −e−^2 x + 2e−^1
(b) y = 1/x, dy/dx = − 1 /x^2 , m = − 1 /e^2 , y − e−^1 = − 1 e^2 (x − e), y = − 1 e^2 x +^2 e
- dy/dx =^16 t^ −^2 2 = 8t − 1; for t = 1, dy/dx = 7, (x, y) = (6, 10); y − 10 = 7(x − 6), y = 7x − 32
- dy/dx = 4 cos t −2 sin t = −2 cot t
(a) dy/dx = 0 if cot t = 0, t = π/2 + nπ for n = 0, ± 1 , · · ·
(b) dx/dy = −
tan t = 0 if tan t = 0, t = nπ for n = 0, ± 1 , · · ·
454 Chapter 11
- dy/dx = 2 t + 1 6 t^2 − 30 t + 24
2 t + 1 6(t − 1)(t − 4) (a) dy/dx = 0 if t = − 1 / 2
(b) dx/dy = 6(t − 1)(t − 4) 2 t + 1 = 0 if t = 1, 4
- x = y = 0 when t = 0, π; dy dx = 2 cos 2t cos t ; dy dx
t=
= 2, dy dx
t=π
= − 2 , the equations of the tangent lines are y = − 2 x, y = 2x.
- y(t) = 0 has three solutions, t = 0, ±π/2; the last two correspond to the crossing point.
For t = ±π/2, m = dy dx
±π ; the tangent lines are given by y = ±
π (x − 2).
- If y = 4 then t^2 = 4, t = ±2, x = 0 for t = ±2 so (0, 4) is reached when t = ±2. dy/dx = 2t/(3t^2 − 4). For t = 2, dy/dx = 1/2 and for t = −2, dy/dx = − 1 /2. The tangent lines are y = ±x/2 + 4.
- If x = 3 then t^2 − 3 t + 5 = 3, t^2 − 3 t + 2 = 0, (t − 1)(t − 2) = 0, t = 1 or 2. If t = 1 or 2 then y = 1 so (3, 1) is reached when t = 1 or 2. dy/dx = (3t^2 + 2t − 10)/(2t − 3). For t = 1, dy/dx = 5, the tangent line is y − 1 = 5(x − 3), y = 5x − 14. For t = 2, dy/dx = 6, the tangent line is y − 1 = 6(x − 3), y = 6x − 17.
(a) 1
-1 1
(b) dx dt = −3 cos^2 t sin t and dy dt = 3 sin^2 t cos t are both zero when t = 0, π/ 2 , π, 3 π/ 2 , 2 π, so singular points occur at these values of t.
- (a) when y = 0
(b) dx dy = a^ −^ a^ cos^ θ a sin θ = 0 when θ = 2nπ, n = 0, 1 ,... (which is when y = 0).
- Substitute θ = π/3, r = 1, and dr/dθ = −
3 in equation (7) gives slope m = 1/
- As in Exercise 21, θ = π/4, dr/dθ =
2 /2, r = 1 +
2 /2, m = − 1 −
- As in Exercise 21, θ = 2, dr/dθ = − 1 /4, r = 1/2, m = tan 2 − 2 2 tan 2 + 1
- As in Exercise 21, θ = π/6, dr/dθ = 4
3 a, r = 2a, m = 3
- As in Exercise 21, θ = 3π/4, dr/dθ = − 3
2 /2, r =
2 /2, m = − 2
- As in Exercise 21, θ = π, dr/dθ = 3, r = 4, m = 4/ 3
- m = dy dx
r cos θ + (sin θ)(dr/dθ) −r sin θ + (cos θ)(dr/dθ)
cos θ + 2 sin θ cos θ − sin θ + cos^2 θ − sin^2 θ
; if θ = 0, π/ 2 , π, then m = 1, 0 , − 1.
456 Chapter 11
- r^2 + (dr/dθ)^2 = a^2 + 0^2 = a^2 , L =
∫ (^2) π
0
adθ = 2πa
- r^2 + (dr/dθ)^2 = (2a cos θ)^2 + (− 2 a sin θ)^2 = 4a^2 , L =
∫ (^) π/ 2
−π/ 2
2 adθ = 2πa
- r^2 + (dr/dθ)^2 = [a(1 − cos θ)]^2 + [a sin θ]^2 = 4a^2 sin^2 (θ/2), L = 2
∫ (^) π
0
2 a sin(θ/2)dθ = 8a
- r^2 + (dr/dθ)^2 = [sin^2 (θ/2)]^2 + [sin(θ/2) cos(θ/2)]^2 = sin^2 (θ/2), L =
∫ (^) π
0
sin(θ/2)dθ = 2
- r^2 + (dr/dθ)^2 = (e^3 θ^ )^2 + (3e^3 θ^ )^2 = 10e^6 θ^ , L =
0
10 e^3 θ^ dθ =
10(e^6 − 1)/ 3
- r^2 + (dr/dθ)^2 = [sin^3 (θ/3)]^2 + [sin^2 (θ/3) cos(θ/3)]^2 = sin^4 (θ/3),
L =
∫ (^) π/ 2
0
sin^2 (θ/3)dθ = (2π − 3
- (a) From (3), dy dx
3 sin t 1 − 3 cos t (b) At t = 10, dy dx =^
3 sin 10 1 − 3 cos 10 ≈ −^0.^46402 , θ^ ≈^ tan
- (a) dy dx = 0 when dy dt = 2 sin t = 0, t = 0, π, 2 π, 3 π
(b) dx dt = 0 when 1 − 2 cos t = 0, cos t = 1/ 2 , t = π/ 3 , 5 π/ 3 , 7 π/ 3
- (a) r^2 + (dr/dθ)^2 = (cos nθ)^2 + (−n sin nθ)^2 = cos^2 nθ + n^2 sin^2 nθ = (1 − sin^2 nθ) + n^2 sin^2 nθ = 1 + (n^2 − 1) sin^2 nθ,
L = 2
∫ (^) π/(2n)
0
1 + (n^2 − 1) sin^2 nθdθ
(b) L = 2
∫ (^) π/ 4
0
1 + 3 sin^2 2 θdθ ≈ 2. 42
(c) n L
2
3
4
5
6
7
8
9
10
11
n L
12
13
14
15
16
17
18
19
20
(a)
0
p / 2
(b) r^2 + (dr/dθ)^2 = (e−θ^ )^2 + (−e−θ^ )^2 = 2e−^2 θ^ , L = 2
0
e−^2 θ^ dθ
(c) L = (^) θlim 0 →+∞^
∫ (^) θ 0
0
e−^2 θ^ dθ = (^) θlim 0 →+∞ (1 − e−^2 θ^0 ) = 1
Exercise Set 11.2 457
- x′^ = 2t, y′^ = 2, (x′)^2 + (y′)^2 = 4t^2 + 4
S = 2π
0
(2t)
4 t^2 + 4dt = 8π
0
t
t^2 + 1dt = 8 π 3 (t^2 + 1)^3 /^2
] 4
0
8 π 3
- x′^ = et(cos t − sin t), y′^ = et(cos t + sin t), (x′)^2 + (y′)^2 = 2e^2 t
S = 2π
∫ (^) π/ 2
0
(et^ sin t)
2 e^2 tdt = 2
2 π
∫ (^) π/ 2
0
e^2 t^ sin t dt
2 π
[
e^2 t(2 sin t − cos t)
]π/ 2
0
π(2eπ^ + 1)
- x′^ = −2 sin t cos t, y′^ = 2 sin t cos t, (x′)^2 + (y′)^2 = 8sin 2 t cos^2 t
S = 2π
∫ (^) π/ 2
0
cos^2 t
8sin 2 t cos^2 t dt = 4
2 π
∫ (^) π/ 2
0
cos^3 t sin t dt = −
2 π cos^4 t
]π/ 2
0
2 π
- x′^ = 1, y′^ = 4t, (x′)^2 + (y′)^2 = 1 + 16t^2 , S = 2π
0
t
1 + 16t^2 dt = π 24
- x′^ = −r sin t, y′^ = r cos t, (x′)^2 + (y′)^2 = r^2 , S = 2π
∫ (^) π
0
r sin t
r^2 dt = 2πr^2
∫ (^) π
0
sin t dt = 4πr^2
dx dφ =^ a(1^ −^ cos^ φ),^
dy dφ =^ a^ sin^ φ,
dx dφ
dy dφ
= 2a^2 (1 − cos φ)
S = 2π
∫ (^2) π
0
a(1 − cos φ)
2 a^2 (1 − cos φ) dφ = 2
2 πa^2
∫ (^2) π
0
(1 − cos φ)^3 /^2 dφ,
but 1 − cos φ = 2 sin^2 φ 2 so (1 − cos φ)^3 /^2 = 2
2 sin^3 φ 2 for 0 ≤ φ ≤ π and, taking advantage of the
symmetry of the cycloid, S = 16πa^2
∫ (^) π
0
sin^3 φ 2 dφ = 64πa^2 / 3
- (a) dr dt = 2 and dθ dt = 1 so dr dθ = dr/dt dθ/dt
= 2, r = 2θ + C, r = 10 when θ = 0 so 10 = C, r = 2θ + 10.
(b) r^2 + (dr/dθ)^2 = (2θ + 10)^2 + 4, during the first 5 seconds the rod rotates through an angle
of (1)(5) = 5 radians so L =
0
(2θ + 10)^2 + 4dθ, let u = 2θ + 10 to get
L =
10
u^2 + 4du =
[ (^) u 2
u^2 + 4 + 2 ln |u +
u^2 + 4|
] 20
10
=
[
104 + 2 ln
]
≈ 75 .7 mm
- x = r cos θ, y = r sin θ, dx dθ
dr dθ cos θ − r sin θ, dy dθ = r cos θ + dr dθ sin θ, ( dx dθ
dy dθ
= r^2 +
dr dθ
, and Formula (6) of Section 8.4 becomes
L =
∫ (^) β
α
r^2 +
dr dθ
dθ
Exercise Set 11.3 459
16. A = 2
∫ (^) π
0
[16 − (2 − 2 cos θ)^2 ]dθ = 10π 17. A = 2
∫ (^) π/ 3
0
[(2 + 2 cos θ)^2 − 9]dθ = 9
3 / 2 − π
18. A = 2
∫ (^) π/ 4
0
(16 sin^2 θ)dθ = 2π − 4
19. A = 2
[∫ (^2) π/ 3
0
(1/2 + cos θ)^2 dθ −
∫ (^) π
2 π/ 3
(1/2 + cos θ)^2 dθ
]
= (π + 3
20. A = 2
∫ (^) π/ 3
0
[
(2 + 2 cos θ)^2 − 9 4 sec^2 θ
]
dθ = 2π +^9 4
21. A = 2
∫ (^) cos−^1 (3/5)
0
(100 − 36 sec^2 θ)dθ = 100 cos−^1 (3/5) − 48
22. A = 8
∫ (^) π/ 8
0
(4a^2 cos^2 2 θ − 2 a^2 )dθ = 2a^2
- (a) r is not real for π/ 4 < θ < 3 π/4 and 5π/ 4 < θ < 7 π/ 4
(b) A = 4
∫ (^) π/ 4
0
a^2 cos 2θ dθ = a^2
(c) A = 4
∫ (^) π/ 6
0
[
4 cos 2θ − 2
]
dθ = 2
2 π 3
24. A = 2
∫ (^) π/ 2
0
sin 2θ dθ = 1 25. A =
∫ (^4) π
2 π
a^2 θ^2 dθ −
∫ (^2) π
0
a^2 θ^2 dθ = 8π^3 a^2
- (a) x = r cos θ, y = r sin θ,
(dx/dθ)^2 + (dy/dθ)^2 = (f ′(θ) cos θ − f (θ) sin θ)^2 + (f ′(θ) sin θ + f (θ) cos θ)^2 = f ′(θ)^2 + f (θ)^2 ;
S =
∫ (^) β
α
2 πf (θ) sin θ
f ′(θ)^2 + f (θ)^2 dθ if about θ = 0; similarly for θ = π/ 2
(b) f ′, g′^ are continuous and no segment of the curve is traced more than once.
- r^2 +
dr dθ
= cos^2 θ + sin^2 θ = 1,
so S =
∫ (^) π/ 2
−π/ 2
2 π cos^2 θ dθ = π^2.
28. S =
∫ (^) π/ 2
0
2 πeθ^ cos θ
2 e^2 θ^ dθ
2 π
∫ (^) π/ 2
0
e^2 θ^ cos θ dθ =
2 π 5 (e
π (^) − 2)
460 Chapter 11
29. S =
∫ (^) π
0
2 π(1 − cos θ) sin θ
1 − 2 cos θ + cos^2 θ + sin^2 θ dθ
2 π
∫ (^) π
0
sin θ(1 − cos θ)^3 /^2 dθ =
2 π(1 − cos θ)^5 /^2
π 0 = 32π/ 5
30. S =
∫ (^) π
0
2 πa(sin θ)a dθ = 4πa^2
- (a) r^3 cos^3 θ − 3 r^2 cos θ sin θ + r^3 sin^3 θ = 0, r = 3 cos^ θ^ sin^ θ cos^3 θ + sin^3 θ
- (a) A = 2
∫ (^) π/(2n)
0
a^2 cos^2 nθ dθ = πa^2 4 n (b) A = 2
∫ (^) π/(2n)
0
a^2 cos^2 nθ dθ = πa^2 4 n (c)
2 n ×^ total area =^
πa^2 4 n (d)^
n ×^ total area =^
πa^2 4 n
- If the upper right corner of the square is the point (a, a) then the large circle has equation r =
2 a and the small circle has equation (x − a)^2 + y^2 = a^2 , r = 2a cos θ, so
area of crescent = 2
∫ (^) π/ 4
0
[
(2a cos θ)^2 − (
2 a)^2
]
dθ = a^2 = area of square.
34. A =
∫ (^2) π
0
(cos 3θ + 2)^2 dθ = 9π/ 2
3
-3 3
35. A =
∫ (^) π/ 2
0
4 cos^2 θ sin^4 θ dθ = π/ 16
1
0 1
EXERCISE SET 11.
- (a) 4 px = y^2 , point (1, 1), 4 p = 1, x = y^2 (b) − 4 py = x^2 , point (3, −3), 12 p = 9, − 3 y = x^2
(c) a = 3, b = 2, x^2 9 +^
y^2 4 = 1^ (d)^ a^ = 3, b^ = 2,
x^2 4 +^
y^2 9 = 1
462 Chapter 11
- (a)
4
4
x
y 5 V (2, (^2) )
F (2, 2)
y = 3
(b)
2
4
x
y
V (–2, 2) (^7) F (–^4 , 2)
9 x = – 4
- (a)
x
9 y x = – 2
7 F (– 2 , 3)
V (–4, 3)
(b)
15 y = 16
17 F (–1, (^16) ) x
y
V (–1, 1)
- (a) c^2 = 16 − 9 = 7, c =
(4, 0)
(0, 3)
(0, – 3)
(–4, 0) (^) x
y
(b) x^2 1
y^2 9
c^2 = 9 − 1 = 8, c =
(0, 3)
(0, – 3)
(–1, 0) (1, 0)
x
y
(0, (^) √8)
- (a) c^2 = 25 − 4 = 21, c =
(0, 5)
(0, – 5)
(–2, 0) (2, 0)
x
y
(b) x^2 9
y^2 4
c^2 = 9 − 4 = 5, c =
(0, 2)
(0, – 2)
(–3, 0)
(3, 0)
x
y
Exercise Set 11.4 463
- (a) (x − 1)^2 16
(y − 3)^2 9
c^2 = 16 − 9 = 7, c =
(1, 6)
(1, 0)
x
y
(1 – √7, 3) (1 + √7, 3)
(–3, 3) (5, 3)
(b) (x + 2)^2 4
(y + 1)^2 3
c^2 = 4 − 3 = 1, c = 1
(–4, – 1) (–3, – 1) (–1, – 1) (0, – 1)
x
y (–2, – 1 + √3)
- (a) (x + 3)^2 16
(y − 5)^2 4
c^2 = 16 − 4 = 12, c = 2
(1, 5)
(–3, 7)
(–3, 3)
(–7, 5)
x
y (– 3 – (^2) √3, 5)
(b) x^2 4
(y + 2)^2 9
c^2 = 9 − 4 = 5, c =
(0, – 5)
(–2, – 2) (2, – 2)
x
y
- (a) (x + 1)^2 9
(y − 1)^2 1
c^2 = 9 − 1 = 8, c =
(–1, 2)
(–4, 1) (–1, 0)
(2, 1) x
y
(– 1 – (^) √8, 1)
(b) (x + 1)^2 4
(y − 5)^2 16
c^2 = 16 − 4 = 12, c = 2
(–3, 5) (1, 5)
(–1, 1)
(–1, 9)
x
y
- (a) (x^ + 1)
2 4
2 9
c^2 = 9 − 4 = 5, c =
(1, 3)
(–1, 6)
(–1, 0)
(–3, 3)
x
y
(b) (x^ −^ 2)
2 9
2 5
c^2 = 9 − 5 = 4, c = 2 (2, – 3 + (^) √5)
(4, – 3) (–1, – 3) (5, – 3) (0, – 3)
x
y
Exercise Set 11.4 465
- (a) (x + 1)^2 / 4 − (y − 1)^2 /1 = 1 c^2 = 4 + 1 = 5, c =
(–3, 1)
(1, 1)
x
y
y – 1 = – 12 ( x + 1)
y – 1 = 12 ( x + 1)
(b) (x − 1)^2 / 4 − (y + 3)^2 /64 = 1 c^2 = 4 + 64 = 68, c = 2
(–1, – 3) (3,^ – 3)^ x
y
y + 3 = – 4( x – 1)
y + 3 = 4( x – 1)
- (a) (y − 3)^2 / 4 − (x + 2)^2 /9 = 1 c^2 = 4 + 9 = 13, c =
(-2, 5)
(-2, 1)
x
y
2 y – 3 = – 3 ( x + 2)
2 y – 3 = 3 ( x + 2)
(b) (y + 5)^2 / 9 − (x + 2)^2 /36 = 1 c^2 = 9 + 36 = 45, c = 3
(–2, – 8)
(–2, – 2) (^) x
y
1 y + 5 = – 2 ( x + 2)
1 y + 5 = 2 ( x + 2) (–2, – 5 + 3√5)
- (a) y^2 = 4px, p = 3, y^2 = 12x (b) y^2 = − 4 px, p = 7, y^2 = − 28 x
- (a) x^2 = − 4 py, p = 4, x^2 = − 16 y (b) x^2 = − 4 py, p = 1/2, x^2 = − 2 y
- (a) x^2 = − 4 py, p = 3, x^2 = − 12 y
(b) The vertex is 3 units above the directrix so p = 3, (x − 1)^2 = 12(y − 1).
- (a) y^2 = 4px, p = 6, y^2 = 24x
(b) The vertex is half way between the focus and directrix so the vertex is at (2, 4), the focus is 3 units to the left of the vertex so p = 3, (y − 4)^2 = −12(x − 2)
- y^2 = a(x − h), 4 = a(3 − h) and 9 = a(2 − h), solve simultaneously to get h = 19/5, a = −5 so y^2 = −5(x − 19 /5)
- (x − 5)^2 = a(y + 3), (9 − 5)^2 = a(5 + 3) so a = 2, (x − 5)^2 = 2(y + 3)
- (a) x^2 /9 + y^2 /4 = 1
(b) a = 26/2 = 13, c = 5, b^2 = a^2 − c^2 = 169 − 25 = 144; x^2 /169 + y^2 /144 = 1
- (a) x^2 + y^2 /5 = 1
(b) b = 8 ,c = 6, a^2 = b^2 + c^2 = 64 + 36 = 100; x^2 /64 + y^2 /100 = 1
466 Chapter 11
- (a) c = 1, a^2 = b^2 + c^2 = 2 + 1 = 3; x^2 /3 + y^2 /2 = 1
(b) b^2 = 16 − 12 = 4; x^2 /16 + y^2 /4 = 1 and x^2 /4 + y^2 /16 = 1
- (a) c = 3, b^2 = a^2 − c^2 = 16 − 9 = 7; x^2 /16 + y^2 /7 = 1
(b) a^2 = 9 + 16 = 25; x^2 /25 + y^2 /9 = 1 and x^2 /9 + y^2 /25 = 1
- (a) a = 6, (2, 3) satisfies x^2 /36 + y^2 /b^2 = 1 so 4/36 + 9/b^2 = 1, b^2 = 8 1/8; x^2 /36 + y^2 /(81/8 ) = 1
(b) The center is midway between the foci so it is at (1, 3), thus c = 1, b = 1, a^2 = 1 + 1 = 2; (x − 1)^2 + (y − 3)^2 /2 = 1
- (a) Substitute (3, 2) and (1, 6) into x^2 /A + y^2 /B = 1 to get 9/A + 4/B = 1 and 1/A + 36/B = 1 which yields A = 10, B = 40; x^2 /10 + y^2 /40 = 1 (b) The center is at (2, −1) thus c = 2, a = 3, b^2 = 9 − 4 = 5; (x − 2)^2 /5 + (y + 1)^2 /9 = 1
- (a) a = 2, c = 3, b^2 = 9 − 4 = 5; x^2 / 4 − y^2 /5 = 1
(b) a = 1, b/a = 2, b = 2; x^2 − y^2 /4 = 1
- (a) a = 3, c = 5, b^2 = 25 − 9 = 16; y^2 / 9 − x^2 /16 = 1
(b) a = 3, a/b = 1, b = 3; y^2 / 9 − x^2 /9 = 1
- (a) vertices along x-axis: b/a = 3/2 so a = 8/3; x^2 /(64/9) − y^2 /16 = 1 vertices along y-axis: a/b = 3/2 so a = 6; y^2 / 36 − x^2 /16 = 1 (b) c = 5, a/b = 2 and a^2 + b^2 = 25, solve to get a^2 = 20, b^2 = 5; y^2 / 20 − x^2 /5 = 1
- (a) foci along the x-axis: b/a = 3/4 and a^2 + b^2 = 25, solve to get a^2 = 16, b^2 = 9; x^2 / 16 − y^2 /9 = 1 foci along the y-axis: a/b = 3/4 and a^2 + b^2 = 25 which results in y^2 / 9 − x^2 /16 = 1 (b) c = 3, b/a = 2 and a^2 + b^2 = 9 so a^2 = 9/5, b^2 = 36/5; x^2 /(9/5) − y^2 /(36/5) = 1
- (a) the center is at (6, 4), a = 4, c = 5, b^2 = 25 − 16 = 9; (x − 6)^2 / 16 − (y − 4)^2 /9 = 1
(b) The asymptotes intersect at (1/ 2 , 2) which is the center, (y − 2)^2 /a^2 − (x − 1 /2)^2 /b^2 = 1 is the form of the equation because (0, 0) is below both asymptotes, 4/a^2 − (1/4)/b^2 = 1 and a/b = 2 which yields a^2 = 3, b^2 = 3/4; (y − 2)^2 / 3 − (x − 1 /2)^2 /(3/4) = 1.
- (a) the center is at (1, −2); a = 2, c = 10, b^2 = 100 − 4 = 96; (y + 2)^2 / 4 − (x − 1)^2 /96 = 1
(b) the center is at (1, −1); 2a = 5 − (−3) = 8, a = 4, (x^ −^ 1)
2 16 − (y^ + 1)
2 16
- (a) y = ax^2 + b, (20, 0) and (10, 12) are on the curve so 400 a + b = 0 and 100a + b = 12. Solve for b to get b = 16 ft = height of arch.
(b) x^2 a^2
y^2 b^2 = 1, 400 = a^2 , a = 20;
b^2
b = 8
3 ft = height of arch. -20 -10 10 20
(10, 12)
x
y
