11
PARAMETRIC EQUATIONS AND
POLAR COORDINATES
11.1 Curves Defined by Parametric Equations
1. x = 1 +
0
1
x
1
y
0
2
2
√
1+ 2
3
√
1+ 3
−3
2.41
−4
2.73
−3
3. x = 5 sin t,
t
ET 10.1
√
t, y = t2 − 4t, 0 ≤ t ≤ 5
t
x
y
ET 10
3
5
√
1+ 5
0
3.24
5
y = t2 , −π ≤ t ≤ π
−π
−π/2
9.87
2.47
0
π2
5. x = 3t − 5,
4
−5
π2/4
0
π/2
π
0
0
5
π2/4
0
π2
2.47
9.87
y = 2t + 1
(a)
t
−2
−1
0
1
2
3
4
x
−11
−8
−5
−2
1
4
7
y
−3
−1
1
3
5
7
9
(b) x = 3t − 5 ⇒ 3t = x + 5 ⇒ t = 13 (x + 5) ⇒
y = 2 · 13 (x + 5) + 1, so y = 23 x +
7. x = t2 − 2,
(a)
13
.
3
y = 5 − 2t, −3 ≤ t ≤ 4
t
−3
−2
−1
0
1
2
3
4
x
7
2
−1
−2
−1
2
7
14
y
11
9
7
5
3
1
−1
−3
(b) y = 5 − 2t ⇒ 2t = 5 − y ⇒ t = 12 (5 − y) ⇒
2
x = 12 (5 − y) − 2, so x = 14 (5 − y)2 − 2, −3 ≤ y ≤ 11.
1
2
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CHAPTER 11
9. x =
(a)
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
√
t, y = 1 − t
(b) x =
t
0
1
2
3
4
x
0
1
1.414
1.732
2
y
1
0
−1
√
t ⇒ t = x2
−2
−3
⇒ y = 1 − t = 1 − x2 . Since t ≥ 0, x ≥ 0.
So the curve is the right half of the parabola y = 1 − x2 .
(b)
11. (a) x = sin θ, y = cos θ, 0 ≤ θ ≤ π. x2 + y 2 = sin2 θ + cos2 θ = 1. Since
0 ≤ θ ≤ π, we have sin θ ≥ 0, so x ≥ 0. Thus, the curve is the right half of
the circle x2 + y 2 = 1.
13. (a) x = sin t, y = csc t, 0 < t <
π
.
2
y = csc t =
1
1
= . For 0 < t <
sin t
x
(b)
π
,
2
we have 0 < x < 1 and y > 1. Thus, the curve is the portion
of the hyperbola y = 1/x with y > 1.
15. (a) x = e2t
⇒ 2t = ln x ⇒ t =
y =t+1 =
1
2
1
2
(b)
ln x.
ln x + 1.
17. (a) x = sinh t, y = cosh t
(b)
⇒ y2 − x2 = cosh2 t − sinh2 t = 1. Since
y = cosh t ≥ 1, we have the upper branch of the hyperbola y − x = 1.
2
19. x = 3 + 2 cos t, y = 1 + 2 sin t, π/2 ≤ t ≤ 3π/2.
2
By Example 4 with r = 2, h = 3, and k = 1, the motion of the particle
takes place on a circle centered at (3, 1) with a radius of 2. As t goes from
π
2
to
3π
2 ,
the particle starts at the point (3, 3) and
moves counterclockwise to (3, −1) [one-half of a circle].
21. x = 5 sin t, y = 2 cos t
⇒ sin t =
x 2 y 2
y
x
, cos t = . sin2 t + cos2 t = 1 ⇒
+
= 1. The motion of the
5
2
5
2
particle takes place on an ellipse centered at (0, 0). As t goes from −π to 5π, the particle starts at the point (0, −2) and moves
clockwise around the ellipse 3 times.
SECTION 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ET SECTION 10.1
¤
3
23. We must have 1 ≤ x ≤ 4 and 2 ≤ y ≤ 3. So the graph of the curve must be contained in the rectangle [1, 4] by [2, 3].
25. When t = −1, (x, y) = (0, −1). As t increases to 0, x decreases to −1 and y
increases to 0. As t increases from 0 to 1, x increases to 0 and y increases to 1.
As t increases beyond 1, both x and y increase. For t < −1, x is positive and
decreasing and y is negative and increasing. We could achieve greater accuracy
by estimating x- and y-values for selected values of t from the given graphs and
plotting the corresponding points.
27. When t = 0 we see that x = 0 and y = 0, so the curve starts at the origin. As t
increases from 0 to 12 , the graphs show that y increases from 0 to 1 while x
increases from 0 to 1, decreases to 0 and to −1, then increases back to 0, so we
arrive at the point (0, 1). Similarly, as t increases from
1
2
to 1, y decreases from 1
to 0 while x repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating x- and
y-values for selected values of t from the given graphs and plotting the corresponding points.
29. As in Example 6, we let y = t and x = t − 3t3 + t5 and use a t-interval of [−3, 3].
31. (a) x = x1 + (x2 − x1 )t, y = y1 + (y2 − y1 )t, 0 ≤ t ≤ 1. Clearly the curve passes through P1 (x1 , y1 ) when t = 0 and
through P2 (x2 , y2 ) when t = 1. For 0 < t < 1, x is strictly between x1 and x2 and y is strictly between y1 and y2 . For
every value of t, x and y satisfy the relation y − y1 =
P1 (x1 , y1 ) and P2 (x2 , y2 ).
Finally, any point (x, y) on that line satisfies
y2 − y1
(x − x1 ), which is the equation of the line through
x2 − x1
x − x1
y − y1
=
; if we call that common value t, then the given
y2 − y1
x2 − x1
parametric equations yield the point (x, y); and any (x, y) on the line between P1 (x1 , y1 ) and P2 (x2 , y2 ) yields a value of
t in [0, 1]. So the given parametric equations exactly specify the line segment from P1 (x1 , y1 ) to P2 (x2 , y2 ).
(b) x = −2 + [3 − (−2)]t = −2 + 5t and y = 7 + (−1 − 7)t = 7 − 8t for 0 ≤ t ≤ 1.
33. The circle x2 + (y − 1)2 = 4 has center (0, 1) and radius 2, so by Example 4 it can be represented by x = 2 cos t,
y = 1 + 2 sin t, 0 ≤ t ≤ 2π. This representation gives us the circle with a counterclockwise orientation starting at (2, 1).
(a) To get a clockwise orientation, we could change the equations to x = 2 cos t, y = 1 − 2 sin t, 0 ≤ t ≤ 2π.
(b) To get three times around in the counterclockwise direction, we use the original equations x = 2 cos t, y = 1 + 2 sin t with
the domain expanded to 0 ≤ t ≤ 6π.
4
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CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
(c) To start at (0, 3) using the original equations, we must have x1 = 0; that is, 2 cos t = 0. Hence, t =
x = 2 cos t, y = 1 + 2 sin t,
π
2
≤t≤
π
2.
So we use
3π
.
2
Alternatively, if we want t to start at 0, we could change the equations of the curve. For example, we could use
x = −2 sin t, y = 1 + 2 cos t, 0 ≤ t ≤ π.
35. Big circle: It’s centered at (2, 2) with a radius of 2, so by Example 4, parametric equations are
x = 2 + 2 cos t,
y = 2 + 2 sin t,
0 ≤ t ≤ 2π
Small circles: They are centered at (1, 3) and (3, 3) with a radius of 0.1. By Example 4, parametric equations are
and
(left)
x = 1 + 0.1 cos t,
y = 3 + 0.1 sin t,
(right)
0 ≤ t ≤ 2π
x = 3 + 0.1 cos t,
y = 3 + 0.1 sin t,
0 ≤ t ≤ 2π
Semicircle: It’s the lower half of a circle centered at (2, 2) with radius 1. By Example 4, parametric equations are
x = 2 + 1 cos t,
y = 2 + 1 sin t,
π ≤ t ≤ 2π
To get all four graphs on the same screen with a typical graphing calculator, we need to change the last t-interval to [0, 2π] in
order to match the others. We can do this by changing t to 0.5t. This change gives us the upper half. There are several ways to
get the lower half—one is to change the “+” to a “−” in the y-assignment, giving us
x = 2 + 1 cos(0.5t),
37. (a) x = t3
y = 2 − 1 sin(0.5t),
⇒ t = x1/3 , so y = t2 = x2/3 .
(b) x = t6
We get the entire curve y = x2/3 traversed in a left to
right direction.
(c) x = e−3t = (e−t )3
0 ≤ t ≤ 2π
⇒ t = x1/6 , so y = t4 = x4/6 = x2/3 .
Since x = t6 ≥ 0, we only get the right half of the
curve y = x2/3 .
[so e−t = x1/3 ],
y = e−2t = (e−t )2 = (x1/3 )2 = x2/3 .
If t < 0, then x and y are both larger than 1. If t > 0, then x and y are between 0
and 1. Since x > 0 and y > 0, the curve never quite reaches the origin.
39. The case
π
2
< θ < π is illustrated. C has coordinates (rθ, r) as in Example 7,
and Q has coordinates (rθ, r + r cos(π − θ)) = (rθ, r(1 − cos θ))
[since cos(π − α) = cos π cos α + sin π sin α = − cos α], so P has coordinates
(rθ − r sin(π − θ), r(1 − cos θ)) = (r(θ − sin θ), r(1 − cos θ))
[since sin(π − α) = sin π cos α − cos π sin α = sin α]. Again we have the
parametric equations x = r(θ − sin θ), y = r(1 − cos θ).
SECTION 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ET SECTION 10.1
¤
41. It is apparent that x = |OQ| and y = |QP | = |ST |. From the diagram,
x = |OQ| = a cos θ and y = |ST | = b sin θ. Thus, the parametric equations are
x = a cos θ and y = b sin θ. To eliminate θ we rearrange: sin θ = y/b ⇒
sin2 θ = (y/b)2 and cos θ = x/a ⇒ cos2 θ = (x/a)2 . Adding the two
equations: sin2 θ + cos2 θ = 1 = x2 /a2 + y 2 /b2 . Thus, we have an ellipse.
43. C = (2a cot θ, 2a), so the x-coordinate of P is x = 2a cot θ. Let B = (0, 2a).
Then ∠OAB is a right angle and ∠OBA = θ, so |OA| = 2a sin θ and
A = ((2a sin θ) cos θ, (2a sin θ) sin θ). Thus, the y-coordinate of P
is y = 2a sin2 θ.
45. (a)
There are 2 points of intersection:
(−3, 0) and approximately (−2.1, 1.4).
(b) A collision point occurs when x1 = x2 and y1 = y2 for the same t. So solve the equations:
3 sin t = −3 + cos t (1)
2 cos t = 1 + sin t
(2)
From (2), sin t = 2 cos t − 1. Substituting into (1), we get 3(2 cos t − 1) = −3 + cos t ⇒ 5 cos t = 0 ( ) ⇒
or
3π
2 .
We check that t =
π
2
does not. So the only collision point
π
2
occurs when t =
and this gives the point (−3, 0). [We could check our work by graphing x1 and x2 together as
3π
2 ,
3π
2
satisfies (1) and (2) but t =
cos t = 0 ⇒ t =
functions of t and, on another plot, y1 and y2 as functions of t. If we do so, we see that the only value of t for which both
pairs of graphs intersect is t =
3π
.]
2
(c) The circle is centered at (3, 1) instead of (−3, 1). There are still 2 intersection points: (3, 0) and (2.1, 1.4), but there are
no collision points, since ( ) in part (b) becomes 5 cos t = 6 ⇒ cos t =
6
5
> 1.
47. x = t2 , y = t3 − ct. We use a graphing device to produce the graphs for various values of c with −π ≤ t ≤ π. Note that all
the members of the family are symmetric about the x-axis. For c < 0, the graph does not cross itself, but for c = 0 it has a
cusp at (0, 0) and for c > 0 the graph crosses itself at x = c, so the loop grows larger as c increases.
5
6
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CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
49. Note that all the Lissajous figures are symmetric about the x-axis. The parameters a and b simply stretch the graph in the
x- and y-directions respectively. For a = b = n = 1 the graph is simply a circle with radius 1. For n = 2 the graph crosses
itself at the origin and there are loops above and below the x-axis. In general, the figures have n − 1 points of intersection,
all of which are on the y-axis, and a total of n closed loops.
a=b=1
n=2
n=3
11.2 Calculus with Parametric Curves
1. x = t sin t, y = t2 + t
⇒
ET 10.2
dx
dy
dy/dt
2t + 1
dy
= 2t + 1,
= t cos t + sin t, and
=
=
.
dt
dt
dx
dx/dt
t cos t + sin t
dy/dt
3t2 + 1
dy
dx
dy
= 3t2 + 1,
= 4t3 , and
=
=
. When t = −1,
dt
dt
dx
dx/dt
4t3
3. x = t4 + 1, y = t3 + t; t = −1.
(x, y) = (2, −2) and dy/dx =
4
−4
= −1, so an equation of the tangent to the curve at the point corresponding to t = −1
is y − (−2) = (−1)(x − 2), or y = −x.
√
t
5. x = e
√
, y = t − ln t2 ; t = 1.
When t = 1, (x, y) = (e, 1) and
2t
2t
e t
dy/dt
1 − 2/t
2t − 4
dy
2 dx
dy
=1− 2 =1− ,
= √ , and
=
= √t √ ·
= √ √t .
dt
t
t dt
dx
dx/dt
2 t
e / 2 t 2t
te
2
2
2
dy
= − , so an equation of the tangent line is y − 1 = − (x − e), or y = − x + 3.
dx
e
e
e
7. (a) x = 1 + ln t, y = t2 + 2; (1, 3).
dx
1
dy
dy/dt
2t
dy
= 2t,
= , and
=
=
= 2t2 .
dt
dt
t
dx
dx/dt
1/t
At (1, 3), x = 1 + ln t = 1 ⇒ ln t = 0 ⇒ t = 1 and
dy
= 2, so an equation of the tangent is y − 3 = 2(x − 1),
dx
or y = 2x + 1.
(b) x = 1 + ln t ⇒ x − 1 = ln t ⇒ t = ex−1 , so y = (ex−1 )2 + 2 = e2x−2 + 2 and
When x = 1,
dy
= 2e0 = 2, so an equation of the tangent is y = 2x + 1, as in part (a).
dx
9. x = 6 sin t, y = t2 + t; (0, 0).
dy
dy/dt
2t + 1
=
=
. The point (0, 0) corresponds to t = 0, so the
dx
dx/dt
6 cos t
slope of the tangent at that point is 16 . An equation of the tangent is therefore
y − 0 = 16 (x − 0), or y = 16 x.
dy
= 2e2x−2 .
dx
SECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ET SECTION 10.2
dy/dt
2t + 3t2
3
dy
=
=
= 1+ t ⇒
dx
dx/dt
2t
2
(d/dt) 1 + 32 t
d2 y
d(dy/dx)/dt
3/2
3
d dy
=
=
=
= .
=
dx2
dx dx
dx/dt
2t
2t
4t
11. x = 4 + t2 ,
y = t2 + t3
The curve is CU when
⇒
d2 y
> 0, that is, when t > 0.
dx2
13. x = t − et , y = t + e−t
dy/dt
1 − e−t
dy
=
=
dx
dx/dt
1 − et
⇒
1
et − 1
1− t
e = et = −e−t
=
1 − et
1 − et
⇒
The curve is CU when et < 1 [since e−t > 0] ⇒ t < 0.
d dy
d
(−e−t )
dt dx
e−t
d2 y
dt
=
=
=
.
dx2
dx/dt
dx/dt
1 − et
15. x = 2 sin t, y = 3 cos t, 0 < t < 2π.
d dy
− 3 sec2 t
dt dx
dy/dt
−3 sin t
3
d y
3
dy
=
=
= − tan t, so 2 =
= 2
= − sec3 t.
dx
dx/dt
2 cos t
2
dx
dx/dt
2 cos t
4
2
The curve is CU when sec3 t < 0 ⇒ sec t < 0 ⇒ cos t < 0 ⇒
π
2
<t<
3π
.
2
17. x = 10 − t2 , y = t3 − 12t.
dy
dy
= 3t2 − 12 = 3(t + 2)(t − 2), so
=0 ⇔
dt
dt
t = ±2 ⇔ (x, y) = (6, ∓16).
dx
dx
= −2t, so
= 0 ⇔ t = 0 ⇔ (x, y) = (10, 0).
dt
dt
The curve has horizontal tangents at (6, ±16) and a vertical
tangent at (10, 0).
19. x = 2 cos θ, y = sin 2θ.
dy
dy
= 2 cos 2θ, so
= 0 ⇔ 2θ =
dθ
dθ
[n an integer] ⇔ θ =
π
4
π
2
+ nπ
+ π2 n ⇔
√
dx
= −2 sin θ, so
(x, y) = ± 2, ±1 . Also,
dθ
dx
= 0 ⇔ θ = nπ
dθ
⇔ (x, y) = (±2, 0).
√
The curve has horizontal tangents at ± 2, ±1 (four points), and vertical tangents at (±2, 0).
21. From the graph, it appears that the rightmost point on the curve x = t − t6 , y = et
is about (0.6, 2). To find the exact coordinates, we find the value of t for which the
√
graph has a vertical tangent, that is, 0 = dx/dt = 1 − 6t5 ⇔ t = 1/ 5 6.
Hence, the rightmost point is
√
√
√ 5
−1/5
≈ (0.58, 2.01).
1/ 5 6 − 1/ 6 5 6 , e1/ 6 = 5 · 6−6/5 , e6
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7
8
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CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
23. We graph the curve x = t4 − 2t3 − 2t2 , y = t3 − t in the viewing rectangle [−2, 1.1] by [−0.5, 0.5]. This rectangle
corresponds approximately to t ∈ [−1, 0.8].
We estimate that the curve has horizontal tangents at about (−1, −0.4) and (−0.17, 0.39) and vertical tangents at
about (0, 0) and (−0.19, 0.37). We calculate
dy
dy/dt
3t2 − 1
=
= 3
. The horizontal tangents occur when
dx
dx/dt
4t − 6t2 − 4t
dy/dt = 3t2 − 1 = 0 ⇔ t = ± √13 , so both horizontal tangents are shown in our graph. The vertical tangents occur when
dx/dt = 2t(2t2 − 3t − 2) = 0 ⇔ 2t(2t + 1)(t − 2) = 0 ⇔ t = 0, − 12 or 2. It seems that we have missed one vertical
tangent, and indeed if we plot the curve on the t-interval [−1.2, 2.2] we see that there is another vertical tangent at (−8, 6).
25. x = cos t, y = sin t cos t.
dx/dt = − sin t, dy/dt = − sin2 t + cos2 t = cos 2t.
(x, y) = (0, 0) ⇔ cos t = 0 ⇔ t is an odd multiple of
dx/dt = −1 and dy/dt = −1, so dy/dx = 1. When t =
π
2.
3π
,
2
When t =
π
2,
dx/dt = 1 and
dy/dt = −1. So dy/dx = −1. Thus, y = x and y = −x are both tangent to the
curve at (0, 0).
27. x = rθ − d sin θ, y = r − d cos θ.
(a)
dy
dy
d sin θ
dx
= r − d cos θ,
= d sin θ, so
=
.
dθ
dθ
dx
r − d cos θ
(b) If 0 < d < r, then |d cos θ| ≤ d < r, so r − d cos θ ≥ r − d > 0. This shows that dx/dθ never vanishes,
so the trochoid can have no vertical tangent if d < r.
29. x = 2t3 , y = 1 + 4t − t2
dy/dt
4 − 2t
dy
dy
=
=
=1 ⇔
. Now solve
dx
dx/dt
6t2
dx
⇒
6t2 + 2t − 4 = 0 ⇔ 2(3t − 2)(t + 1) = 0 ⇔ t =
the point is (−2, −4).
2
3
4 − 2t
=1 ⇔
6t2
or t = −1. If t = 23 , the point is 16
, 29 , and if t = −1,
27 9
31. By symmetry of the ellipse about the x- and y-axes,
A=4
Ua
0
y dx = 4
= 2ab θ −
1
2
U0
π/2
sin 2θ
b sin θ (−a sin θ) dθ = 4ab
π/2
0
= 2ab π2 = πab
U π/2
0
sin2 θ dθ = 4ab
U π/2
0
1
2 (1
− cos 2θ) dθ
SECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ET SECTION 10.2
¤
33. The curve x = 1 + et , y = t − t2 = t(1 − t) intersects the x-axis when y = 0,
that is, when t = 0 and t = 1. The corresponding values of x are 2 and 1 + e.
The shaded area is given by
] x=1+e
]
(yT − yB ) dx =
x=2
t=1
t=0
[y(t) − 0] x0 (t) dt =
U1
0
(t − t2 )et dt
1
U1
tet dt − t2 et 0 + 2 0 tet dt
[Formula 97 or parts]
U1 t
1
= 3 0 te dt − (e − 0) = 3 (t − 1)et 0 − e
[Formula 96 or parts]
=
U1
0
tet dt −
U1
0
t2 et dt =
U1
0
= 3[0 − (−1)] − e = 3 − e
35. x = rθ − d sin θ, y = r − d cos θ.
U 2πr
U 2π
(r − d cos θ)(r − d cos θ) dθ = 0 (r2 − 2dr cos θ + d2 cos2 θ) dθ
2π
= r2 θ − 2dr sin θ + 12 d2 θ + 12 sin 2θ 0 = 2πr2 + πd2
A=
0
U 2π
y dx =
37. x = t − t2 , y =
0
4 3/2
t ,
3
1 ≤ t ≤ 2. dx/dt = 1 − 2t and dy/dt = 2t1/2 , so
(dx/dt)2 + (dy/dt)2 = (1 − 2t)2 + (2t1/2 )2 = 1 − 4t + 4t2 + 4t = 1 + 4t2 .
Ubs
U2√
Thus, L = a (dx/dt)2 + (dy/dt)2 dt = 1 1 + 4t2 dt ≈ 3.1678.
39. x = t + cos t, y = t − sin t, 0 ≤ t ≤ 2π.
dx 2
dt
+
dy 2
dt
Thus, L =
dx/dt = 1 − sin t and dy/dt = 1 − cos t, so
= (1 − sin t)2 + (1 − cos t)2 = (1 − 2 sin t + sin2 t) + (1 − 2 cos t + cos2 t) = 3 − 2 sin t − 2 cos t.
U 2π √
Ubs
(dx/dt)2 + (dy/dt)2 dt = 0
3 − 2 sin t − 2 cos t dt ≈ 10.0367.
a
x = 1 + 3t2 , y = 4 + 2t3 , 0 ≤ t ≤ 1.
41.
dx/dt = 6t and dy/dt = 6t2 , so (dx/dt)2 + (dy/dt)2 = 36t2 + 36t4 .
U1 √
U1√
Thus, L = 0 36t2 + 36t4 dt = 0 6t 1 + t2 dt
U2√ = 6 1 u 12 du [u = 1 + t2 , du = 2t dt]
l2
k
√
= 3 23 u3/2 = 2(23/2 − 1) = 2 2 2 − 1
1
43. x =
so
t
, y = ln (1 + t), 0 ≤ t ≤ 2.
1+t
dx
dt
2
+
dy
dt
2
=
(1 + t) · 1 − t · 1
dx
1
1
dy
=
=
,
=
and
dt
(1 + t)2
(1 + t)2
dt
1+t
t2 + 2t + 2
1
1
1
1 + (1 + t)2 =
+
=
. Thus,
4
2
4
(1 + t)
(1 + t)
(1 + t)
(1 + t)4
√
√ 2
] 3√ 2
3
s
t2 + 2t + 2
u +1
u +1
24
u = t + 1,
2 +1
+
ln
u
+
dt
=
du
=
−
u
du = dt
(1 + t)2
u2
u
0
1
1
√
√
√
√
10
= − 3 + ln 3 + 10 + 2 − ln 1 + 2
L=
]
2
9
10
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
45. x = et cos t, y = et sin t, 0 ≤ t ≤ π.
dx 2
+
dt
dy 2
dt
= [et (cos t − sin t)]2 + [et (sin t + cos t)]2
= (et )2 (cos2 t − 2 cos t sin t + sin2 t)
+ (et )2 (sin2 t + 2 sin t cos t + cos2 t
= e2t (2 cos2 t + 2 sin2 t) = 2e2t
√ π √
Uπ√
Uπ√
2e2t dt = 0 2 et dt = 2 et 0 = 2 (eπ − 1).
0
Thus, L =
x = et − t, y = 4et/2 , −8 ≤ t ≤ 3
47.
dx 2
dt
L=
+
dy 2
dt
= (et − 1)2 + (2et/2 )2 = e2t − 2et + 1 + 4et
= e2t + 2et + 1 = (et + 1)2
U3 s
U3
3t
(et + 1)2 dt = −8 (et + 1) dt = et + t −8
−8
= (e3 + 3) − (e−8 − 8) = e3 − e−8 + 11
49. x = t − et , y = t + et , −6 ≤ t ≤ 6.
dx 2
dt
+
dy 2
dt
Set f (t) =
= (1 − et )2 + (1 + et )2 = (1 − 2et + e2t ) + (1 + 2et + e2t ) = 2 + 2e2t , so L =
√
2 + 2e2t . Then by Simpson’s Rule with n = 6 and ∆t =
6−(−6)
6
= 2, we get
U6 √
2 + 2e2t dt.
−6
L ≈ 23 [f (−6) + 4f (−4) + 2f (−2) + 4f (0) + 2f (2) + 4f (4) + f (6)] ≈ 612.3053.
51. x = sin2 t, y = cos2 t, 0 ≤ t ≤ 3π.
(dx/dt)2 + (dy/dt)2 = (2 sin t cos t)2 + (−2 cos t sin t)2 = 8 sin2 t cos2 t = 2 sin2 2t ⇒
Distance =
lπ/2
√ U π/2
√ k
√
√
U 3π √
2 |sin 2t| dt = 6 2 0 sin 2t dt [by symmetry] = −3 2 cos 2t
= −3 2 (−1 − 1) = 6 2.
0
0
because the curve is the segment of x + y = 1 that lies in the first quadrant
√
U π/2
(since x, y ≥ 0), and this segment is completely traversed as t goes from 0 to π2 . Thus, L = 0 sin 2t dt = 2, as above.
The full curve is traversed as t goes from 0 to
π
2,
53. x = a sin θ, y = b cos θ, 0 ≤ θ ≤ 2π.
dx 2
dy 2
= (a cos θ)2 + (−b sin θ)2 = a2 cos2 θ + b2 sin2 θ = a2 (1 − sin2 θ) + b2 sin2 θ
c2
= a2 − (a2 − b2 ) sin2 θ = a2 − c2 sin2 θ = a2 1 − 2 sin2 θ = a2 (1 − e2 sin2 θ)
a
s
U
U π/2 t π/2
a2 1 − e2 sin2 θ dθ [by symmetry] = 4a 0
1 − e2 sin2 θ dθ.
So L = 4 0
dt
+
dt
55. (a) x = 11 cos t − 4 cos(11t/2), y = 11 sin t − 4 sin(11t/2).
Notice that 0 ≤ t ≤ 2π does not give the complete curve because
x(0) 6= x(2π). In fact, we must take t ∈ [0, 4π] in order to obtain the
complete curve, since the first term in each of the parametric equations has
period 2π and the second has period
2π
11/2
=
integer multiple of these two numbers is 4π.
4π
,
11
and the least common
SECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ET SECTION 10.2
¤
11
(b) We use the CAS to find the derivatives dx/dt and dy/dt, and then use Theorem 6 to find the arc length. Recent versions
√ U 4π s
(dx/dt)2 + (dy/dt)2 dt as 88E 2 2 i , where E(x) is the elliptic integral
of Maple express the integral 0
] 1√
√
1 − x2 t2
√
dt and i is the imaginary number −1.
2
1−t
0
Some earlier versions of Maple (as well as Mathematica) cannot do the integral exactly, so we use the command
evalf(Int(sqrt(diff(x,t)ˆ2+diff(y,t)ˆ2),t=0..4*Pi)); to estimate the length, and find that the arc
length is approximately 294.03. Derive’s Para_arc_length function in the utility file Int_apps simplifies the
U 4π t
− 4 sin t sin 11t
+ 5 dt.
−4 cos t cos 11t
integral to 11 0
2
2
57. x = 1 + tet , y = (t2 + 1)et , 0 ≤ t ≤ 1.
dx 2
+
dt
S=
U
U1
2πy ds =
0
dy 2
dt
]
2πy
dt
0
= 2π
]
13
4
=
=
π
81
k
u−4
9
2 5/2
u
5
2π
1215
s
√
U1
e2t (t + 1)2 (t2 + 2t + 2) dt = 0 2π(t2 + 1)e2t (t + 1) t2 + 2t + 2 dt ≈ 103.5999
dx 2
t dx 2
1
dt
+
+
U π/2
0
dt
dt
dt =
2
= 3t2 + (2t)2 = 9t4 + 4t2 .
]
− 83 u3/2
=
4
2πt2
0
√ 1
u 18 du
l13
1
2
2
u = 9t + 4, t = (u − 4)/9,
1
du = 18t dt, so t dt = 18
du
k
l13
2
· 15
3u5/2 − 20u3/2
π
81
π
.
2
1
, 1 ≤ t ≤ 2.
t2
2πy ds =
]
1
2
dx
dt
1
t2
0
s
t2 (9t2 + 4) dt
=
2π
9 · 18
]
13
4
(u3/2 − 4u1/2 ) du
4
dx 2
dθ
+
2π · a sin3 θ · 3a sin θ cos θ dθ = 6πa2
]
]
s
9t4 + 4t2 dt = 2π
√
√ 3 · 132 13 − 20 · 13 13 − (3 · 32 − 20 · 8) =
63. x = t + t3 , y = t −
and S =
dy 2
dy 2
61. x = a cos3 θ, y = a sin3 θ, 0 ≤ θ ≤
S=
so
= e2t (t + 1)2 + e2t (t + 1)4 = e2t (t + 1)2 [1 + (t + 1)2 ],
2π(t2 + 1)et
59. x = t3 , y = t2 , 0 ≤ t ≤ 1.
S=
= (tet + et )2 + [(t2 + 1)et + et (2t)]2 = [et (t + 1)]2 + [et (t2 + 2t + 1)]2
dy 2
dθ
U π/2
0
2π
1215
√
247 13 + 64
= (−3a cos2 θ sin θ)2 + (3a sin2 θ cos θ)2 = 9a2 sin2 θ cos2 θ.
π/2
sin4 θ cos θ dθ = 65 πa2 sin5 θ 0 = 65 πa2
= 1 + 3t2 and
dy
dt
=1+
2
dx 2 dy 2
2
2
2 2
,
so
+
=
(1
+
3t
)
+
1
+
dt
dt
t3
t3
v
2
1
2
2π t − 2
(1 + 3t2 )2 + 1 + 3 dt ≈ 59.101.
t
t
65. x = 3t2 , y = 2t3 , 0 ≤ t ≤ 5
⇒
dx 2
dt
+
dy 2
dt
= (6t)2 + (6t2 )2 = 36t2 (1 + t2 ) ⇒
s
√
U5
U5 √
(dx/dt)2 + (dy/dt)2 dt = 0 2π(3t2 )6t 1 + t2 dt = 18π 0 t2 1 + t2 2t dt
l26
k
U 26
U 26
√
u = 1 + t2 ,
= 18π 1 (u3/2 − u1/2 ) du = 18π 25 u5/2 − 23 u3/2
= 18π 1 (u − 1) u du
S=
U5
0
2πx
du = 2t dt
= 18π
2
5
√
· 676 26 −
2
3
√ · 26 26 − 25 − 23 =
1
24
5 π
√
949 26 + 1
12
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
67. If f 0 is continuous and f 0 (t) 6= 0 for a ≤ t ≤ b, then either f 0 (t) > 0 for all t in [a, b] or f 0 (t) < 0 for all t in [a, b]. Thus, f
is monotonic (in fact, strictly increasing or strictly decreasing) on [a, b]. It follows that f has an inverse. Set F = g ◦ f −1 ,
that is, define F by F (x) = g(f −1 (x)). Then x = f(t) ⇒ f −1 (x) = t, so y = g(t) = g(f −1 (x)) = F (x).
dy
d dy
dφ
d
1
dy
dy/dt
ẏ
=
tan−1
=
. But
=
=
⇒
dt
dt
dx
1 + (dy/dx)2 dt dx
dx
dx/dt
ẋ
1
dφ
d dy
ÿ ẋ − ẍẏ
ẋÿ − ẍẏ
d ẏ
ÿ ẋ − ẍẏ
=
⇒
. Using the Chain Rule, and the
= 2
=
=
dt dx
dt ẋ
ẋ2
dt
1 + (ẏ/ẋ)2
ẋ2
ẋ + ẏ 2
] tt
t dx 2 dy 2
1/2
2 2
dx 2
+ dt dt ⇒ ds
+ dy
= ẋ + ẏ 2
, we have that
fact that s =
dt
dt =
dt
dt
69. (a) φ = tan−1
dy
dx
⇒
0
dφ/dt
dφ
=
=
ds
ds/dt
ẋÿ − ẍẏ
ẋ2 + ẏ 2
dφ ẋÿ − ẍẏ |ẋÿ − ẍẏ|
ẋÿ − ẍẏ
1
=
=
.
So
κ
=
ds (ẋ2 + ẏ 2 )3/2 = (ẋ2 + ẏ 2 )3/2 .
(ẋ2 + ẏ 2 )1/2
(ẋ2 + ẏ 2 )3/2
d2 y
dy
, ÿ =
.
dx
dx2
2
1 · (d2 y/dx2 ) − 0 · (dy/dx)
d y/dx2 So κ =
=
.
[1 + (dy/dx)2 ]3/2
[1 + (dy/dx)2 ]3/2
(b) x = x and y = f (x) ⇒ ẋ = 1, ẍ = 0 and ẏ =
⇒ ẋ = 1 − cos θ ⇒ ẍ = sin θ, and y = 1 − cos θ ⇒ ẏ = sin θ ⇒ ÿ = cos θ. Therefore,
cos θ − (cos2 θ + sin2 θ)
cos θ − cos2 θ − sin2 θ
|cos θ − 1|
=
=
. The top of the arch is
κ=
(2 − 2 cos θ)3/2
[(1 − cos θ)2 + sin2 θ]3/2
(1 − 2 cos θ + cos2 θ + sin2 θ)3/2
71. x = θ − sin θ
characterized by a horizontal tangent, and from Example 2(b) in Section 11.2 [ ET 10.2], the tangent is horizontal when
θ = (2n − 1)π, so take n = 1 and substitute θ = π into the expression for κ: κ =
|cos π − 1|
|−1 − 1|
1
=
= .
4
(2 − 2 cos π)3/2
[2 − 2(−1)]3/2
73. The coordinates of T are (r cos θ, r sin θ). Since T P was unwound from
arc T A, T P has length rθ. Also ∠P T Q = ∠P T R − ∠QT R = 12 π − θ,
so P has coordinates x = r cos θ + rθ cos 12 π − θ = r(cos θ + θ sin θ),
y = r sin θ − rθ sin 12 π − θ = r(sin θ − θ cos θ).
11.3 Polar Coordinates
1. (a) 2,
π
3
ET 10.3
. The direction
we obtain the point 2, 7π
3
is a point that satisfies the r < 0
is 4π
, so −2, 4π
3
3
By adding 2π to
opposite
π
3
requirement.
π
,
3
SECTION 11.3 POLAR COORDINATES ET SECTION 10.3
(b) 1, − 3π
4
5π r > 0: 1, − 3π
4 + 2π = 1, 4
+ π = −1, π4
r < 0: −1, − 3π
4
(c) −1, π2
r > 0: −(−1), π2 + π = 1, 3π
2
r < 0: −1, π2 + 2π = −1, 5π
2
3. (a)
¤
x = 1 cos π = 1(−1) = −1 and
y = 1 sin π = 1(0) = 0 give us
the Cartesian coordinates (−1, 0).
x = 2 cos − 2π
= 2 − 12 = −1 and
3
√ √
y = 2 sin − 2π
= 2 − 23 = − 3
3
(b)
√ give us −1, − 3 .
√ √
2
= 2 and
x = −2 cos 3π
4 = −2 − 2
(c)
√ √
= −2 22 = − 2
y = −2 sin 3π
4
gives us
√ √
2, − 2 .
s
√
22 + (−2)2 = 2 2 and θ = tan−1 −2
= − π4 . Since (2, −2) is in the fourth
2
√
√
quadrant, the polar coordinates are (i) 2 2, 7π
and (ii) −2 2, 3π
.
4
4
5. (a) x = 2 and y = −2
(b) x = −1 and y =
⇒ r=
t
√ √
√ 2
3 ⇒ r = (−1)2 +
3 = 2 and θ = tan−1 −13 =
quadrant, the polar coordinates are (i) 2, 2π
and (ii) −2,
3
5π
3
.
2π
.
3
√ Since −1, 3 is in the second
13
14
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
7. The curves r = 1 and r = 2 represent circles with center
O and radii 1 and 2. The region in the plane satisfying
9. The region satisfying 0 ≤ r < 4 and −π/2 ≤ θ < π/6
does not include the circle r = 4 nor the line θ =
π
.
6
1 ≤ r ≤ 2 consists of both circles and the shaded region
between them in the figure.
11. 2 < r < 3,
5π
3
≤θ≤
7π
3
13. Converting the polar coordinates (2, π/3) and (4, 2π/3) to Cartesian coordinates gives us 2 cos π3 , 2 sin
√ , 4 sin 2π
4 cos 2π
= −2, 2 3 . Now use the distance formula.
3
3
d=
15. r = 2
⇔
17. r = 3 sin θ
π
3
√ = 1, 3 and
t
t
√ 2 √
√
√
√
(x2 − x1 )2 + (y2 − y1 )2 = (−2 − 1)2 + 2 3 − 3 = 9 + 3 = 12 = 2 3
s
x2 + y 2 = 2 ⇔ x2 + y 2 = 4, a circle of radius 2 centered at the origin.
⇒ r2 = 3r sin θ
⇔ x2 + y 2 = 3y
2 2
⇔ x2 + y − 32 = 32 , a circle of radius
The first two equations are actually equivalent since r2 = 3r sin θ
3
2
centered at 0, 32 .
⇒ r(r − 3 sin θ) = 0 ⇒ r = 0 or r = 3 sin θ. But
r = 3 sin θ gives the point r = 0 (the pole) when θ = 0. Thus, the single equation r = 3 sin θ is equivalent to the compound
condition (r = 0 or r = 3 sin θ).
19. r = csc θ
21. x = 3
⇔ r=
1
sin θ
⇔ r sin θ = 1 ⇔ y = 1, a horizontal line 1 unit above the x-axis.
⇔ r cos θ = 3 ⇔ r = 3/ cos θ
23. x = −y 2
⇔ r cos θ = −r2 sin2 θ
25. x2 + y 2 = 2cx
⇔ r2 = 2cr cos θ
⇔ r = 3 sec θ.
⇔ cos θ = −r sin2 θ
⇔ r=−
cos θ
= − cot θ csc θ.
sin2 θ
⇔ r2 − 2cr cos θ = 0 ⇔ r(r − 2c cos θ) = 0 ⇔ r = 0 or r = 2c cos θ.
r = 0 is included in r = 2c cos θ when θ =
π
2
+ nπ, so the curve is represented by the single equation r = 2c cos θ.
27. (a) The description leads immediately to the polar equation θ =
slightly more difficult to derive.
(b) The easier description here is the Cartesian equation x = 3.
π
6,
and the Cartesian equation y = tan π6 x =
√1
3
x is
SECTION 11.3 POLAR COORDINATES ET SECTION 10.3
29. θ = −π/6
⇔ r2 = r sin θ ⇔ x2 + y 2 = y ⇔
2 2
x2 + y − 12 = 12 . The reasoning here is the same
37. r = 4 sin 3θ
39. r = 2 cos 4θ
41. r = 1 − 2 sin θ
15
31. r = sin θ
as in Exercise 17. This is a circle of radius
33. r = 2(1 − sin θ). This curve is a cardioid.
¤
35. r = θ, θ ≥ 0
1
2
centered at 0, 12 .
16
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
43. r2 = 9 sin 2θ
3 θ
2
45. r = 2 cos
47. r = 1 + 2 cos 2θ
49. For θ = 0, π, and 2π, r has its minimum value of about 0.5. For θ =
π
2
and
3π
,
2
r attains its maximum value of 2.
We see that the graph has a similar shape for 0 ≤ θ ≤ π and π ≤ θ ≤ 2π.
51. x = (r) cos θ = (4 + 2 sec θ) cos θ = 4 cos θ + 2. Now, r → ∞
(4 + 2 sec θ) → ∞ ⇒ θ →
π −
2
consider 0 ≤ θ < 2π], so lim x =
r→∞
or θ →
3π +
2
lim x =
[since we need only
lim (4 cos θ + 2) = 2. Also,
θ→π/2−
r → −∞ ⇒ (4 + 2 sec θ) → −∞ ⇒ θ →
r→−∞
⇒
π +
2
or θ →
3π −
2
, so
lim (4 cos θ + 2) = 2. Therefore, lim x = 2 ⇒ x = 2 is a vertical asymptote.
θ→π/2+
r→±∞
SECTION 11.3 POLAR COORDINATES ET SECTION 10.3
¤
17
53. To show that x = 1 is an asymptote we must prove lim x = 1.
r→±∞
x = (r) cos θ = (sin θ tan θ) cos θ = sin2 θ. Now, r → ∞ ⇒ sin θ tan θ → ∞ ⇒
−
θ → π2 , so lim x = lim sin2 θ = 1. Also, r → −∞ ⇒ sin θ tan θ → −∞ ⇒
r→∞
θ→
π +
2
θ→π/2−
, so lim x =
r→−∞
lim
θ→π/2+
sin2 θ = 1. Therefore, lim x = 1 ⇒ x = 1 is
r→±∞
a vertical asymptote. Also notice that x = sin2 θ ≥ 0 for all θ, and x = sin2 θ ≤ 1 for all θ. And x 6= 1, since the curve is not
defined at odd multiples of
π
.
2
Therefore, the curve lies entirely within the vertical strip 0 ≤ x < 1.
55. (a) We see that the curve r = 1 + c sin θ crosses itself at the origin, where r = 0 (in fact the inner loop corresponds to
negative r-values,) so we solve the equation of the limaçon for r = 0 ⇔ c sin θ = −1 ⇔ sin θ = −1/c. Now if
|c| < 1, then this equation has no solution and hence there is no inner loop. But if c < −1, then on the interval (0, 2π)
the equation has the two solutions θ = sin−1 (−1/c) and θ = π − sin−1 (−1/c), and if c > 1, the solutions are
θ = π + sin−1 (1/c) and θ = 2π − sin−1 (1/c). In each case, r < 0 for θ between the two solutions, indicating a loop.
(b) For 0 < c < 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at θ =
for what c-values
d2 y
is negative at θ =
dθ2
y = r sin θ = sin θ + c sin2 θ
At θ =
3π
,
2
⇒
3π
,
2
dy
= cos θ + 2c sin θ cos θ = cos θ + c sin 2θ
dθ
⇒
d2 y
= − sin θ + 2c cos 2θ.
dθ2
this is equal to −(−1) + 2c(−1) = 1 − 2c, which is negative only for c > 12 . A similar argument shows that
⇒
π
2
(indicating a dimple) for c < − 12 .
x = r cos θ = 2 sin θ cos θ = sin 2θ, y = r sin θ = 2 sin2 θ
⇒
dy/dθ
2 · 2 sin θ cos θ
sin 2θ
dy
=
=
=
= tan 2θ
dx
dx/dθ
cos 2θ · 2
cos 2θ
π
√
π
π dy
,
= tan 2 ·
= tan = 3. [Another method: Use Equation 3.]
6 dx
6
3
When θ =
59. r = 1/θ
⇒ x = r cos θ = (cos θ)/θ, y = r sin θ = (sin θ)/θ
⇒
dy/dθ
sin θ(−1/θ2 ) + (1/θ) cos θ θ2
− sin θ + θ cos θ
dy
=
=
·
=
dx
dx/dθ
− cos θ − θ sin θ
cos θ(−1/θ2 ) − (1/θ) sin θ θ2
When θ = π,
61. r = cos 2θ
So we determine
since by the Second Derivative Test this indicates a maximum:
for −1 < c < 0, y only has a local minimum at θ =
57. r = 2 sin θ
3π
.
2
dy
−0 + π(−1)
−π
=
=
= −π.
dx
−(−1) − π(0)
1
⇒ x = r cos θ = cos 2θ cos θ, y = r sin θ = cos 2θ sin θ
⇒
dy/dθ
cos 2θ cos θ + sin θ (−2 sin 2θ)
dy
=
=
dx
dx/dθ
cos 2θ (− sin θ) + cos θ (−2 sin 2θ)
√
√
√
0 2/2 +
2/2 (−2)
π dy
− 2
= √
When θ = ,
√
= √ = 1.
4 dx
0 − 2/2 +
2/2 (−2)
− 2
18
¤
CHAPTER 11
63. r = 3 cos θ
dy
dθ
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
⇒ x = r cos θ = 3 cos θ cos θ, y = r sin θ = 3 cos θ sin θ
⇒
= −3 sin θ + 3 cos θ = 3 cos 2θ = 0 ⇒ 2θ = or
⇔ θ = or
k
l
So the tangent is horizontal at √32 , π4 and − √32 , 3π
same as √32 , − π4 .
4
dx
dθ
2
2
π
2
= −6 sin θ cos θ = −3 sin 2θ = 0 ⇒ 2θ = 0 or π
65. r = 1 + cos θ
3π
2
π
4
⇔ θ = 0 or
π
.
2
3π
.
4
So the tangent is vertical at (3, 0) and 0, π2 .
⇒ x = r cos θ = cos θ (1 + cos θ), y = r sin θ = sin θ (1 + cos θ) ⇒
dy
dθ
= (1 + cos θ) cos θ − sin2 θ = 2 cos2 θ + cos θ − 1 = (2 cos θ − 1)(cos θ + 1) = 0 ⇒ cos θ =
.
⇒ horizontal tangent at 32 , π3 , (0, π), and 32 , 5π
θ = π3 , π, or 5π
3
3
= −(1 + cos θ) sin θ − cos θ sin θ = − sin θ (1 + 2 cos θ) = 0 ⇒ sin θ = 0 or cos θ = − 12
, or 4π
⇒ vertical tangent at (2, 0), 12 , 2π
θ = 0, π, 2π
, and 12 , 4π
.
3
3
3
3
dx
dθ
Note that the tangent is horizontal, not vertical when θ = π, since lim
θ→π
67. r = 2 + sin θ
1
2
or −1 ⇒
⇒
dy/dθ
= 0.
dx/dθ
⇒ x = r cos θ = (2 + sin θ) cos θ, y = r sin θ = (2 + sin θ) sin θ
⇒
dy
dθ
= (2 + sin θ) cos θ + sin θ cos θ = cos θ · 2(1 + sin θ) = 0 ⇒ cos θ = 0 or sin θ = −1 ⇒
.
⇒ horizontal tangent at 3, π2 and 1, 3π
θ = π2 or 3π
2
2
= (2 + sin θ)(− sin θ) + cos θ cos θ = −2 sin θ − sin2 θ + 1 − sin2 θ = −2 sin2 θ − 2 sin θ + 1 ⇒
√
√
√
√
2±2 3
1− 3
2± 4+8
1+ 3
=
=
< −1
⇒
sin θ =
−4
−4
−2
−2
√ √
√
θ1 = sin−1 − 12 + 12 3 and θ2 = π − θ1 ⇒ vertical tangent at 32 + 12 3, θ1 and 32 + 12 3, θ2 .
√ √
√
Note that r(θ1 ) = 2 + sin sin−1 − 12 + 12 3 = 2 − 12 + 12 3 = 32 + 12 3.
dx
dθ
⇒ r2 = ar sin θ + br cos θ ⇒ x2 + y 2 = ay + bx ⇒
2
2 2 2
2 2
x2 − bx + 12 b + y 2 − ay + 12 a = 12 b + 12 a
⇒
x − 12 b + y − 12 a = 14 (a2 + b2 ), and this is a circle
√
with center 12 b, 12 a and radius 12 a2 + b2 .
69. r = a sin θ + b cos θ
Note for Exercises 71–76: Maple is able to plot polar curves using the polarplot command, or using the coords=polar option in a regular
plot command. In Mathematica, use PolarPlot. In Derive, change to Polar under Options State. If your graphing device cannot
plot polar equations, you must convert to parametric equations. For example, in Exercise 71, x = r cos θ = [1 + 2 sin(θ/2)] cos θ,
y = r sin θ = [1 + 2 sin(θ/2)] sin θ.
71. r = 1 + 2 sin(θ/2). The parameter interval is [0, 4π].
73. r = esin θ − 2 cos(4θ). The parameter interval is [0, 2π].
SECTION 11.3 POLAR COORDINATES ET SECTION 10.3
¤
75. r = 2 − 5 sin(θ/6). The parameter interval is [−6π, 6π].
77. It appears that the graph of r = 1 + sin θ −
π
6
is the same shape as
the graph of r = 1 + sin θ, but rotated counterclockwise about the
origin by π6 . Similarly, the graph of r = 1 + sin θ − π3 is rotated by
π
3.
In general, the graph of r = f (θ − α) is the same shape as that of
r = f (θ), but rotated counterclockwise through α about the origin.
That is, for any point (r0 , θ0 ) on the curve r = f (θ), the point
(r0 , θ0 + α) is on the curve r = f (θ − α), since r0 = f(θ0 ) = f ((θ0 + α) − α).
79. (a) r = sin nθ.
n=2
n=3
n=4
n=5
From the graphs, it seems that when n is even, the number of loops in the curve (called a rose) is 2n, and when n is odd,
the number of loops is simply n. This is because in the case of n odd, every point on the graph is traversed twice, due to
the fact that
r(θ + π) = sin[n(θ + π)] = sin nθ cos nπ + cos nθ sin nπ =
+
sin nθ
− sin nθ
if n is even
if n is odd
(b) The graph of r = |sin nθ| has 2n loops whether n is odd or even, since r(θ + π) = r(θ).
n=2
n=3
n=4
n=5
19
20
¤
81. r =
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
1 − a cos θ
. We start with a = 0, since in this case the curve is simply the circle r = 1.
1 + a cos θ
As a increases, the graph moves to the left, and its right side becomes flattened. As a increases through about 0.4, the right
side seems to grow a dimple, which upon closer investigation (with narrower θ-ranges) seems to appear at a ≈ 0.42 [the
√
actual value is 2 − 1]. As a → 1, this dimple becomes more pronounced, and the curve begins to stretch out horizontally,
until at a = 1 the denominator vanishes at θ = π, and the dimple becomes an actual cusp. For a > 1 we must choose our
parameter interval carefully, since r → ∞ as 1 + a cos θ → 0 ⇔ θ → ± cos−1 (−1/a). As a increases from 1, the curve
splits into two parts. The left part has a loop, which grows larger as a increases, and the right part grows broader vertically,
√
and its left tip develops a dimple when a ≈ 2.42 [actually, 2 + 1]. As a increases, the dimple grows more and more
pronounced. If a < 0, we get the same graph as we do for the corresponding positive a-value, but with a rotation through π
about the pole, as happened when c was replaced with −c in Exercise 80.
a=0
a = 0.42,|θ| ≤ 0.5
a = 0.3
a = 0.9, |θ| ≤ 0.5
a = 0.41, |θ| ≤ 0.5
a = 1, |θ| ≤ 0.1
a = 2.41, |θ − π| ≤ 0.2
a=2
a=4
a = 2.42, |θ − π| ≤ 0.2
¤
SECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES ET SECTION 10.4
21
dy/dθ
dy
− tan θ
− tan θ
dx/dθ
tan φ − tan θ
= dx
83. tan ψ = tan(φ − θ) =
=
dy
dy/dθ
1 + tan φ tan θ
tan θ
1+
tan θ
1+
dx
dx/dθ
dr
dr
dy
dx
sin2 θ
sin θ + r cos θ − tan θ
cos θ − r sin θ
−
tan θ
r cos θ + r ·
dθ
dθ
dθ
cos θ
=
= dθ
= dy
dx
dr
dr
dr sin2 θ
dr
+
tan θ
cos θ − r sin θ + tan θ
sin θ + r cos θ
cos θ +
·
dθ
dθ
dθ
dθ
dθ
dθ cos θ
=
r
r cos2 θ + r sin2 θ
=
dr
dr
dr/dθ
2
2
cos θ +
sin θ
dθ
dθ
11.4 Areas and Lengths in Polar Coordinates
1. r = θ 2 , 0 ≤ θ ≤
3. r = sin θ,
A=
π
3
]
≤θ≤
2π/3
π/3
=
5. r =
1
4
π
.
4
k
2π
3
1
2
−
A=
]
π/4
0
2π
3 .
sin2 θ dθ =
1
2
1 2
r
2
√ − 23 −
√
θ, 0 ≤ θ ≤ 2π. A =
]
]
π/2
=
1
2
=
1
2
=
]
]
1
2
√ l
3
2
dθ =
=
]
1
4
2π
1
2
0
π/2
π
3
+
]
π/4
0
1
4
θ−
√ 3
2
=
1
2
1 4
θ
2
sin 2θ
π
12
+
]
√ 2
θ dθ =
√
(16 + 9 sin2 θ) dθ
]
1
2
π/2
dθ =
2π/3
π/3
1 5 π/4
θ 0
10
=
1
4
2π
1
2 θ dθ
=
1
4θ
=9
0
U π/2
0
·2
0
]
0
π/2 π/2 41
2
16 + 9 · 12 (1 − cos 2θ) dθ
−
9
2
cos 2θ dθ = 41
2 θ−
U π/2
dθ =
(3 cos θ)2 dθ = 32
1
2 (1
+ cos 2θ) dθ =
9
2
θ+
1
2
U π/2
0
sin 2θ
0 ≤ θ ≤ π/4 and multiply it by 4.
U π/4
U π/4
A = 4 0 21 r2 dθ = 2 0 (4 cos 2θ) dθ
0
sin 4π
−
3
π
3
2 2π
0
= π2
9
4
[by Theorem 5.5.6(a) [ ET 5.5.7(a)] ]
sin 2θ
π/2
0
=
41π
4
− 0 − (0 − 0) =
cos2 θ dθ
π/2
0
=
9
2
π
2
3 2
2
+ 0 − (0 + 0) =
=
9π
.
4
11. The curve goes through the pole when θ = π/4, so we’ll find the area for
U π/4
=
[by Theorem 5.5.6(a) [ ET 5.5.7(a)] ]
Also, note that this is a circle with radius 32 , so its area is π
=8
1
2
4
−π/2
1 2
r
2
0
−
π 5
(16 + 24 sin θ + 9 sin2 θ) dθ
to θ = π/2 [not π]. By symmetry,
U π/2
3
1
10
−π/2
9. The area above the polar axis is bounded by r = 3 cos θ for θ = 0
A=2
2π
=
3
8
0
+ 3 sin θ)2 dθ =
1
2 ((4
−π/2
dθ =
(1 − cos 2θ) dθ =
1 2
2r
0
1
(θ2 )2
2
2π/3
+
2π
π/4
0
π/3
π
3
]
π
.
2
7. r = 4 + 3 sin θ, − π2 ≤ θ ≤
A=
]
1
4
dθ =
ET 10.4
π/4
cos 2θ dθ = 4 sin 2θ 0 = 4
9π
4
41π
4
1
π5
10,240
+
1
2
sin 2π
3
22
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
13. One-sixth of the area lies above the polar axis and is bounded by the curve
r = 2 cos 3θ for θ = 0 to θ = π/6.
U π/6
U π/6
A = 6 0 12 (2 cos 3θ)2 dθ = 12 0 cos2 3θ dθ
U π/6
= 12
(1 + cos 6θ) dθ
2 0
15. A=
=
=
=6 θ+
U 2π
1
2
sin 6θ
+ 2 sin 6θ)2 dθ =
π/6
0
U 2π
= 6 π6 = π
(1 + 4 sin 6θ + 4 sin2 6θ) dθ
U 2π 1 + 4 sin 6θ + 4 · 12 (1 − cos 12θ) dθ
0
U 2π
(3 + 4 sin 6θ − 2 cos 12θ) dθ
0
1
2 (1
0
1
2
1
6
1
2
0
2π
3θ − 23 cos 6θ − 16 sin 12θ 0
= 12 (6π − 23 − 0) − 0 − 23 − 0 = 3π
=
1
2
17. The shaded loop is traced out from θ = 0 to θ = π/2.
=
U π/2
=
1
4
A=
0
1
2
1 2
2r
U π/2
0
π
2
U π/2
sin2 2θ dθ
1
1
θ−
2 (1 − cos 4θ) dθ = 4
=
1
2
dθ =
0
π
8
1
4
sin 4θ
π/2
0
π
⇒ 3 cos 5θ = 0 ⇒ 5θ = π2 ⇒ θ = 10
.
U π/10
U π/10
U π/10
A = −π/10 21 (3 cos 5θ)2 dθ = 0
9 cos2 5θ dθ = 92 0
(1 + cos 10θ) dθ = 92 θ +
19. r = 0
21.
1
10
sin 10θ
π/10
0
=
9π
20
This is a limaçon, with inner loop traced
out between θ =
7π
6
and
11π
6
[found by
solving r = 0].
A= 2
]
3π/2
7π/6
1
2 (1
+ 2 sin θ)2 dθ =
]
3π/2 7π/6
1 + 4 sin θ + 4 sin2 θ dθ =
√
3π/2 7π
= θ − 4 cos θ + 2θ − sin 2θ 7π/6 = 9π
− 2 +2 3−
2
23. 2 cos θ = 1
⇒ cos θ =
A= 2
=
U π/3
0
1
2
⇒ θ=
1
[(2 cos θ)2
2
π
3
or
− 12 ] dθ =
√
3
2
=
0
(4 cos2 θ − 1) dθ
U π/3 1
U π/3
4 2 (1 + cos 2θ) − 1 dθ = 0 (1 + 2 cos 2θ) dθ
0
π/3
= θ + sin 2θ 0 =
π
3
+
√
3
2
3π/2 1 + 4 sin θ + 4 · 12 (1 − cos 2θ) dθ
7π/6
√
π − 323
5π
3 .
U π/3
]
SECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES ET SECTION 10.4
25. To find the area inside the leminiscate r2 = 8 cos 2θ and outside the circle r = 2,
we first note that the two curves intersect when r2 = 8 cos 2θ and r = 2,
that is, when cos 2θ = 12 . For −π < θ ≤ π, cos 2θ =
1
2
⇔ 2θ = ±π/3
or ±5π/3 ⇔ θ = ±π/6 or ±5π/6. The figure shows that the desired area is
4 times the area between the curves from 0 to π/6. Thus,
U π/6 1
U π/6
A= 4 0
(8 cos 2θ) − 12 (2)2 dθ = 8 0 (2 cos 2θ − 1) dθ
2
k
lπ/6
√
√
= 8 sin 2θ − θ
= 8 3/2 − π/6 = 4 3 − 4π/3
0
27. 3 cos θ = 1 + cos θ
1
2
⇔ cos θ =
⇒ θ=
π
3
or − π3 .
U π/3
A = 2 0 21 [(3 cos θ)2 − (1 + cos θ)2 ] dθ
U π/3
U π/3
= 0 (8 cos2 θ − 2 cos θ − 1) dθ = 0 [4(1 + cos 2θ) − 2 cos θ − 1] dθ
π/3
(3 + 4 cos 2θ − 2 cos θ) dθ = 3θ + 2 sin 2θ − 2 sin θ 0
√
√
=π+ 3− 3=π
=
29.
U π/3
0
√
√
sin θ
⇒ tan θ = 3 ⇒ θ =
3=
cos θ
2
U π/2 √
1
(sin θ)2 dθ + π/3 12
3 cos θ dθ
2
√
3 cos θ = sin θ
A=
=
=
U π/3
0
U π/3
0
1
4
1
2
θ−
k
=
1
4
π
3
=
π
12
−
· 12 (1 − cos 2θ) dθ +
π
8
√
3
16
⇒
A= 8·2
+
π
8
−
1
π/3 2
U π/8
0
33. sin 2θ = cos 2θ
U π/8
0
1
4
√
3 3
16
=
5π
24
−
1
2
√
3
4
sin 2 2θ dθ = 8
sin 4θ
π/8
0
U π/8
0
= 4 π8 −
1
4
1
(1
2
sin 2θ dθ
U π/8
0
π
2
π
4
[since r2 = sin 2θ]
π/8
2 sin 2θ dθ = − cos 2θ 0
√
√
= − 12 2 − (−1) = 1 − 12 2
=
√ l
3
4
− cos 4θ) dθ
·1 =
⇒ tan 2θ = 1 ⇒ 2θ =
1
2
π
.
3
· 3 · 12 (1 + cos 2θ) dθ
sin 2θ
= 1 ⇒ tan 2θ = 1 ⇒ 2θ =
cos 2θ
⇒
=4 θ−
A= 4
U π/2
π/3
π/2
sin 2θ 0 + 34 θ + 12 sin 2θ π/3
l
k
√ − 43 − 0 + 34 π2 + 0 − π3 +
1
2
31. sin 2θ = cos 2θ
θ=
⇒
−1
⇒ θ=
π
8
π
4
⇒
¤
23
24
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
35. The darker shaded region (from θ = 0 to θ = 2π/3) represents
From this area, we’ll subtract
1
2
1
2
of the desired area plus
37. The pole is a point of intersection.
θ=
π
6
or
⇒ 1 = 2 sin θ
1
2
⇒ sin θ =
⇒
5π
.
6
The other two points of intersection are
39. 2 sin 2θ = 1
⇒ sin 2θ =
1
2
3
2
,
π
6
⇒ 2θ =
and
3
2
.
, 5π
6
π 5π 13π
, 6, 6 ,
6
or
17π
.
6
By symmetry, the eight points of intersection are given by
(1, θ), where θ =
π 5π 13π
12 , 12 , 12 ,
(−1, θ), where θ =
and
7π 11π 19π
, 12 , 12 ,
12
17π
12 ,
and
and
23π
.
12
[There are many ways to describe these points.]
41. The pole is a point of intersection. sin θ = sin 2θ = 2 sin θ cos θ
sin θ (1 − 2 cos θ) = 0 ⇔ sin θ = 0 or cos θ =
θ = 0, π,
and
√
π
3,
3 2π
, 3
2
or − π3
of the area of the inner loop.
of the area of the inner loop (the lighter shaded region from θ = 2π/3 to θ = π), and then
double that difference to obtain the desired area.
kU
2
Uπ
2 l
2π/3 1 1
A=2 0
dθ − 2π/3 12 12 + cos θ dθ
2 2 + cos θ
Uπ U 2π/3 1
+ cos θ + cos2 θ dθ − 2π/3 14 + cos θ + cos2 θ dθ
= 0
4
U 2π/3 1
1
= 0
4 + cos θ + 2 (1 + cos 2θ) dθ
Uπ − 2π/3 14 + cos θ + 12 (1 + cos 2θ) dθ
2π/3 π
θ
θ
θ
sin 2θ
θ
sin 2θ
+ sin θ + +
+ sin θ + +
−
=
4
2
4
4
2
4
0
2π/3
√
√ √
√
π
π
3
π
3
π
π
3
π
3
= 6 + 2 +3 − 8 − 4 +2 + 6 + 2 +3 − 8
√ √
= π4 + 34 3 = 14 π + 3 3
1 + sin θ = 3 sin θ
1
2
1
2
⇒
⇒ the other intersection points are
[by symmetry].
√
3 π
2 , 3
⇔
SECTION 11.4 AREAS AND LENGTHS IN POLAR COORDINATES ET SECTION 10.4
¤
25
43.
From the first graph, we see that the pole is one point of intersection. By zooming in or using the cursor, we find the θ-values
of the intersection points to be α ≈ 0.88786 ≈ 0.89 and π − α ≈ 2.25. (The first of these values may be more easily
estimated by plotting y = 1 + sin x and y = 2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant, that is,
A= 2
]
α
0
=
45. L =
3 α
3θ 0
4
]
b
a
=3
1
(2θ)2
2
]
dθ + 2
]
π/2
1
(1
2
α
+ θ − 2 cos θ + 12 θ −
]
s
r2 + (dr/dθ)2 dθ =
0
]
α
4θ2 dθ +
0
1
4
π/3
0
π/3
+ sin θ)2 dθ =
sin 2θ
π/2
α
=
4 3
3α
+
]
b
a
=
]
0
2π
0
1
(3π)
3
]
s
r2 + (dr/dθ)2 dθ =
]
t
θ2 (θ2 + 4) dθ =
0
2π
0
2π
θ
]
s
θ2 + 4 dθ =
4
4π 2 +4
1
2
3
2
+
is 2π
= π.
2π
]
t
(θ2 )2 + (2θ)2 dθ =
0
s
θ θ2 + 4 dθ
Now let u = θ2 + 4, so that du = 2θ dθ
]
α
π
4
1 + 2 sin θ + 12 (1 − cos 2θ) dθ
− α − 2 cos α + 12 α −
π/3
0
As a check, note that the circumference of a circle with radius
47. L =
2
π/2
]
s
(3 sin θ)2 + (3 cos θ)2 dθ =
π/3
dθ = 3 θ 0 = 3 π3 = π.
circle (from θ = 0 to θ = π),
π
]
θ dθ =
√
u du =
1
2
·
2
3
1
2
2π
3
2
1
4
t
9(sin2 θ + cos2 θ) dθ
sin 2α ≈ 3.4645
= 3π, and since θ = 0 to π =
π
3
traces out
s
θ4 + 4θ2 dθ
du and
k
l4(π2 +1)
= 13 [43/2 (π 2 + 1)3/2 − 43/2 ] = 83 [(π2 + 1)3/2 − 1]
u3/2
4
49. The curve r = 3 sin 2θ is completely traced with 0 ≤ θ ≤ 2π. r2 +
U 2π s
9 sin2 2θ + 36 cos2 2θ dθ ≈ 29.0653
L= 0
dr 2
dθ
= (3 sin 2θ)2 + (6 cos 2θ)2
⇒
1
3
of the
26
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
θ
51. The curve r = sin
L=
]
0
2
is completely traced with 0 ≤ θ ≤ 4π. r2 +
4π t
sin2 θ2 +
1
4
cos2
θ
2
dθ ≈ 9.6884
dr 2
dθ
= sin2
θ
2
+
1
2
2
cos θ2
⇒
53. The curve r = cos4 (θ/4) is completely traced with 0 ≤ θ ≤ 4π.
2
r2 + (dr/dθ)2 = [cos4 (θ/4)]2 + 4 cos3 (θ/4) · (− sin(θ/4)) · 14
= cos8 (θ/4) + cos6 (θ/4) sin2 (θ/4)
= cos6 (θ/4)[cos2 (θ/4) + sin2 (θ/4)] = cos6 (θ/4)
U 4π s
U 4π cos6 (θ/4) dθ = 0 cos3 (θ/4) dθ
0
U 2π
U π/2
= 2 0 cos3 (θ/4) dθ [since cos3 (θ/4) ≥ 0 for 0 ≤ θ ≤ 2π] = 8 0 cos3 u du
L=
π/2
= 8 13 (2 + cos2 u) sin u 0 = 83 [(2 · 1) − (3 · 0)] =
68
16
3
u = 14 θ
55. (a) From (11.2.7) [ET (10.2.7)],
s
(dx/dθ)2 + (dy/dθ)2 dθ
s
Ub
[from the derivation of Equation 11.4.5 [ ET 10.4.5] ]
= a 2πy r2 + (dr/dθ)2 dθ
t
Ub
= a 2πr sin θ r2 + (dr/dθ)2 dθ
S=
Ub
a
2πy
(b) The curve r2 = cos 2θ goes through the pole when cos 2θ = 0 ⇒
2θ =
π
2
⇒ θ=
π
.
4
We’ll rotate the curve from θ = 0 to θ =
π
4
and double
this value to obtain the total surface area generated.
2
dr
dr
sin2 2θ
sin2 2θ
r2 = cos 2θ ⇒ 2r
= −2 sin 2θ ⇒
.
=
=
dθ
dθ
r2
cos 2θ
S=2
]
π/4
0
= 4π
]
t
√
2π cos 2θ sin θ cos 2θ + sin2 2θ /cos 2θ dθ = 4π
0
]
0
π/4
√
1
dθ = 4π
cos 2θ sin θ √
cos 2θ
]
0
π/4
π/4
√
cos 2θ sin θ
u
cos2 2θ + sin2 2θ
dθ
cos 2θ
√
√ π/4
sin θ dθ = 4π − cos θ 0 = −4π 22 − 1 = 2π 2 − 2
¤
SECTION 11.5 CONIC SECTIONS ET SECTION 10.5
11.5 Conic Sections
⇒ y 2 = 12 x. 4p = 12 , so p = 18 . The vertex
is (0, 0), the focus is 18 , 0 , and the directrix is x = − 18 .
1. x = 2y 2
5. (x + 2)2 = 8 (y − 3). 4p = 8, so p = 2. The vertex is
27
ET 10.5
1
⇒ x2 = − 14 y. 4p = − 14 , so p = − 16
.
1
The vertex is (0, 0), the focus is 0, − 16
, and the
3. 4x2 = −y
directrix is y =
1
.
16
7. y 2 + 2y + 12x + 25 = 0
(−2, 3), the focus is (−2, 5), and the directrix is y = 1.
2
⇒
y + 2y + 1 = −12x − 24 ⇒
(y + 1)2 = −12(x + 2). 4p = −12, so p = −3.
The vertex is (−2, −1), the focus is (−5, −1), and the
directrix is x = 1.
9. The equation has the form y 2 = 4px, where p < 0.
Since the parabola passes through (−1, 1), we have
12 = 4p(−1), so 4p = −1 and an equation is y 2 = −x
or x = −y 2 . 4p = −1, so p = − 14 and the focus is
1 − 4 , 0 while the directrix is x = 14 .
11.
√
√
x2
y2
+
= 1 ⇒ a = 9 = 3, b = 5,
9
5
√
√
c = a2 − b2 = 9 − 5 = 2. The ellipse is centered at
(0, 0), with vertices at (±3, 0). The foci are (±2, 0).
28
¤
CHAPTER 11
13. 4x2 + y 2 = 16
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
y2
x2
+
=1 ⇒
4
16
⇒
15. 9x2 − 18x + 4y 2 = 27
√
√
16 = 4, b = 4 = 2,
√
√
√
c = a2 − b2 = 16 − 4 = 2 3. The ellipse is
9(x2 − 2x + 1) + 4y 2 = 27 + 9 ⇔
a=
9(x − 1)2 + 4y 2 = 36 ⇔
centered at (0, 0), with vertices at (0, ±4). The foci
√ are 0, ±2 3 .
17. The center is (0, 0), a = 3, and b = 2, so an equation is
19.
√
5 ⇒ center (1, 0),
√ vertices (1, ±3), foci 1, ± 5
√
√ √
y2
x2
+
= 1. c = a2 − b2 = 5, so the foci are 0, ± 5 .
4
9
5
center (0, 0), vertices (±12, 0), foci (±13, 0), asymptotes y = ± 12
x.
Note: It is helpful to draw a 2a-by-2b rectangle whose center is the center
of the hyperbola. The asymptotes are the extended diagonals of the
rectangle.
⇔
√
x2
y2
−
= 1 ⇒ a = 4 = 2 = b,
4
4
√
√
4 + 4 = 2 2 ⇒ center (0, 0), vertices (0, ±2),
√ foci 0, ±2 2 , asymptotes y = ±x
c=
23. 4x2 − y 2 − 24x − 4y + 28 = 0
2
⇔
2
4(x − 6x + 9) − (y + 4y + 4) = −28 + 36 − 4 ⇔
(y + 2)2
(x − 3)2
−
=1 ⇒
1
4
√
√
√
√
a = 1 = 1, b = 4 = 2, c = 1 + 4 = 5 ⇒
√
center (3, −2), vertices (4, −2) and (2, −2), foci 3 ± 5, −2 ,
4(x − 3)2 − (y + 2)2 = 4 ⇔
asymptotes y + 2 = ±2(x − 3).
y2
(x − 1)2
+
=1 ⇒
4
9
a = 3, b = 2, c =
√
y2
x2
−
= 1 ⇒ a = 12, b = 5, c = 144 + 25 = 13 ⇒
144
25
21. y 2 − x2 = 4
⇔
SECTION 11.5 CONIC SECTIONS ET SECTION 10.5
¤
29
⇔ x2 = 1(y + 1). This is an equation of a parabola with 4p = 1, so p = 14 . The vertex is (0, −1) and the
focus is 0, − 34 .
25. x2 = y + 1
27. x2 = 4y − 2y 2
⇔ x2 + 2y 2 − 4y = 0 ⇔ x2 + 2(y 2 − 2y + 1) = 2 ⇔ x2 + 2(y − 1)2 = 2 ⇔
√ √
x2
(y − 1)2
+
= 1. This is an equation of an ellipse with vertices at ± 2, 1 . The foci are at ± 2 − 1, 1 = (±1, 1).
2
1
(y + 1)2
− x2 = 1. This is an equation
4
√ √
of a hyperbola with vertices (0, −1 ± 2) = (0, 1) and (0, −3). The foci are at 0, −1 ± 4 + 1 = 0, −1 ± 5 .
29. y 2 + 2y = 4x2 + 3
⇔ y 2 + 2y + 1 = 4x2 + 4 ⇔ (y + 1)2 − 4x2 = 4 ⇔
31. The parabola with vertex (0, 0) and focus (0, −2) opens downward and has p = −2, so its equation is x2 = 4py = −8y.
33. The distance from the focus (−4, 0) to the directrix x = 2 is 2 − (−4) = 6, so the distance from the focus to the vertex is
1
(6)
2
= 3 and the vertex is (−1, 0). Since the focus is to the left of the vertex, p = −3. An equation is y 2 = 4p(x + 1) ⇒
y 2 = −12(x + 1).
35. A parabola with vertical axis and vertex (2, 3) has equation y − 3 = a(x − 2)2 . Since it passes through (1, 5), we have
5 − 3 = a(1 − 2)2
⇒ a = 2, so an equation is y − 3 = 2(x − 2)2 .
37. The ellipse with foci (±2, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2,
so b2 = a2 − c2 = 25 − 4 = 21. An equation is
y2
x2
+
= 1.
25
21
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6)
are 2 units from the center, so c = 2 and b =
√
√
√
(x − 0)2
(y − 4)2
a2 − c2 = 42 − 22 = 12. An equation is
+
=1 ⇒
2
b
a2
x2
(y − 4)2
+
= 1.
12
16
41. An equation of an ellipse with center (−1, 4) and vertex (−1, 0) is
from the center, so c = 2. Thus, b2 + 22 = 42
⇒ b2 = 12, and the equation is
43. An equation of a hyperbola with vertices (±3, 0) is
b2 = 25 − 9 = 16, so the equation is
(x + 1)2
(y − 4)2
+
= 1. The focus (−1, 6) is 2 units
2
b
42
(y − 4)2
(x + 1)2
+
= 1.
12
16
x2
y2
− 2 = 1. Foci (±5, 0) ⇒ c = 5 and 32 + b2 = 52
2
3
b
⇒
y2
x2
−
= 1.
9
16
45. The center of a hyperbola with vertices (−3, −4) and (−3, 6) is (−3, 1), so a = 5 and an equation is
(y − 1)2
(x + 3)2
−
= 1. Foci (−3, −7) and (−3, 9) ⇒ c = 8, so 52 + b2 = 82
52
b2
equation is
(x + 3)2
(y − 1)2
−
= 1.
25
39
⇒ b2 = 64 − 25 = 39 and the
30
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
47. The center of a hyperbola with vertices (±3, 0) is (0, 0), so a = 3 and an equation is
Asymptotes y = ±2x ⇒
x2
y2
− 2 = 1.
2
3
b
b
x2
y2
= 2 ⇒ b = 2(3) = 6 and the equation is
−
= 1.
a
9
36
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
a − c from it) while the farthest point is the other vertex (at a distance of a + c). So for this lunar orbit,
(a − c) + (a + c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a + c) − (a − c) = 2c = 314 − 110,
or c = 102. Thus, b2 = a2 − c2 = 3,753,196, and the equation is
y2
x2
+
= 1.
3,763,600
3,753,196
51. (a) Set up the coordinate system so that A is (−200, 0) and B is (200, 0).
|P A| − |P B| = (1200)(980) = 1,176,000 ft =
b2 = c2 − a2 =
(b) Due north of B
3,339,375
121
⇒
2450
11
mi = 2a ⇒ a =
1225
,
11
and c = 200 so
121x2
121y 2
−
= 1.
1,500,625
3,339,375
⇒ x = 200 ⇒
121y 2
133,575
(121)(200)2
−
=1 ⇒ y=
≈ 248 mi
1,500,625
3,339,375
539
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
u
x2
a√ 2
a
=
b + x2 , so y 0 = x(b2 + x2 )−1/2 and
b2
b
b
k
l
a
(b2 + x2 )−1/2 − x2 (b2 + x2 )−3/2 = ab(b2 + x2 )−3/2 > 0 for all x, and so f is concave upward.
y 00 =
b
y = f(x) = a
1+
55. (a) If k > 16, then k − 16 > 0, and
y2
x2
+
= 1 is an ellipse since it is the sum of two squares on the left side.
k
k − 16
(b) If 0 < k < 16, then k − 16 < 0, and
left side.
y2
x2
+
= 1 is a hyperbola since it is the difference of two squares on the
k
k − 16
(c) If k < 0, then k − 16 < 0, and there is no curve since the left side is the sum of two negative terms, which cannot equal 1.
(d) In case (a), a2 = k, b2 = k − 16, and c2 = a2 − b2 = 16, so the foci are at (±4, 0). In case (b), k − 16 < 0, so a2 = k,
b2 = 16 − k, and c2 = a2 + b2 = 16, and so again the foci are at (±4, 0).
57. x2 = 4py
y−
⇒ 2x = 4py 0
⇒ y0 =
x
, so the tangent line at (x0 , y0 ) is
2p
x0
x20
=
(x − x0 ). This line passes through the point (a, −p) on the
4p
2p
directrix, so −p −
x0
x20
=
(a − x0 ) ⇒ −4p2 − x20 = 2ax0 − 2x20
4p
2p
x20 − 2ax0 − 4p2 = 0 ⇔ x20 − 2ax0 + a2 = a2 + 4p2
⇔
⇔
SECTION 11.5 CONIC SECTIONS ET SECTION 10.5
¤
31
s
s
(x0 − a)2 = a2 + 4p2 ⇔ x0 = a ± a2 + 4p2 . The slopes of the tangent lines at x = a ± a2 + 4p2
s
a ± a2 + 4p2
, so the product of the two slopes is
are
2p
s
s
a + a2 + 4p2 a − a2 + 4p2
a2 − (a2 + 4p2 )
−4p2
·
=
=
= −1,
2p
2p
4p2
4p2
showing that the tangent lines are perpendicular.
59. For x2 + 4y 2 = 4, or x2/4 + y 2 = 1, use the parametrization x = 2 cos t, y = sin t, 0 ≤ t ≤ 2π to get
L=4
U π/2 s
U π/2 s
U π/2 s
(dx/dt)2 + (dy/dt)2 dt = 4 0
4 sin2 t + cos2 t dt = 4 0
3 sin2 t + 1 dt
0
Using Simpson’s Rule with n = 10, ∆t =
L≈
61.
4
3
π/2 − 0
10
=
π
,
20
and f (t) =
s
3 sin2 t + 1, we get
π π
f (0) + 4f 20
+ 2f 2π
+ · · · + 2f 8π
+ 4f 9π
+ f π2 ≈ 9.69
20
20
20
20
x2
y2
y2
x2 − a2
b√ 2
− 2 =1 ⇒
=
⇒ y=±
x − a2 .
2
2
2
a
b
b
a
a
s
] c s
c
s
b
a2 39 2b x
A= 2
ln x + x2 − a2 x2 − a2 dx =
x2 − a2 −
a 2
2
a a
a
√
b √ 2
c c − a2 − a2 ln c + c2 − a2 + a2 ln |a|
a
√
Since a2 + b2 = c2 , c2 − a2 = b2 , and c2 − a2 = b.
b
b
= cb − a2 ln(c + b) + a2 ln a = cb + a2 (ln a − ln(b + c))
a
a
=
= b2 c/a + ab ln[a/(b + c)], where c2 = a2 + b2 .
63. Differentiating implicitly,
x2
y2
+ 2 =1 ⇒
a2
b
2x 2yy 0
b2 x
+ 2 = 0 ⇒ y0 = − 2
a2
b
a y
[y 6= 0]. Thus, the slope of the tangent
b2 x1
y1
y1
and of F2 P is
. By the formula from Problems Plus, we have
. The slope of F1 P is
a2 y1
x1 + c
x1 − c
b2 x1
y1
+ 2
using b2 x21 + a2 y12 = a2 b2 ,
a2 y12 + b2 x1 (x1 + c)
a2 b2 + b2 cx1
x1 + c
a y1
=
tan α =
=
and a2 − b2 = c2
a2 y1 (x1 + c) − b2 x1 y1
c2 x1 y1 + a2 cy1
b2 x1 y1
1− 2
a y1 (x1 + c)
2
b cx1 + a2
b2
=
=
cy1 (cx1 + a2 )
cy1
line at P is −
and
y1
b2 x1
−
b2 cx1 − a2
−a2 y12 − b2 x1 (x1 − c)
b2
−a2 b2 + b2 cx1
a2 y1
x1 − c
=
=
=
=
tan β =
2
2
2
2
2
2
a y1 (x1 − c) − b x1 y1
c x1 y1 − a cy1
cy1 (cx1 − a )
cy1
b x1 y1
1− 2
a y1 (x1 − c)
Thus, α = β.
−
32
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
11.6 Conic Sections in Polar Coordinates
ET 10.6
1. The directrix y = 6 is above the focus at the origin, so we use the form with “ + e sin θ” in the denominator. [See Theorem 6
and Figure 2(c).] r =
7
·6
42
ed
4
=
=
1 + e sin θ
4 + 7 sin θ
1 + 74 sin θ
3. The directrix x = −5 is to the left of the focus at the origin, so we use the form with “ − e cos θ” in the denominator.
r=
3
·5
15
ed
4
=
=
3
1 − e cos θ
4
−
3 cos θ
1 − 4 cos θ
5. The vertex (4, 3π/2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus (d = 8), and we use the
form with “−e sin θ ” in the denominator. e = 1 for a parabola, so an equation is r =
1(8)
8
ed
=
=
.
1 − e sin θ
1 − 1 sin θ
1 − sin θ
7. The directrix r = 4 sec θ (equivalent to r cos θ = 4 or x = 4) is to the right of the focus at the origin, so we will use the form
with “+e cos θ” in the denominator. The distance from the focus to the directrix is d = 4, so an equation is
r=
9. r =
1
2
ed
4
2 (4)
· =
=
.
1 + e cos θ
2 + cos θ
1 + 12 cos θ 2
ed
1
=
, where d = e = 1.
1 + sin θ
1 + e sin θ
(a) Eccentricity = e = 1
(b) Since e = 1, the conic is a parabola.
(c) Since “+e sin θ ” appears in the denominator, the directrix is above the
focus at the origin. d = |F l| = 1, so an equation of the directrix is y = 1.
(d) The vertex is at 12 , π2 , midway between the focus and the directrix.
11. r =
1/4
3
12
, where e =
·
=
4 − sin θ 1/4
1 − 14 sin θ
(a) Eccentricity = e =
(b) Since e =
1
4
1
4
and ed = 3 ⇒ d = 12.
1
4
< 1, the conic is an ellipse.
(c) Since “−e sin θ ” appears in the denominator, the directrix is below the focus
at the origin. d = |F l| = 12, so an equation of the directrix is y = −12.
(d) The vertices are 4, π2 and 12
, so the center is midway between them,
, 3π
5
2
that is, 45 , π2 .
SECTION 11.6 CONIC SECTIONS IN POLAR COORDINATES ET SECTION 10.6
13. r =
1/6
3/2
9
·
=
, where e =
6 + 2 cos θ 1/6
1 + 13 cos θ
(a) Eccentricity = e =
(b) Since e =
1
3
1
3
and ed =
3
2
⇒ d = 92 .
1
3
< 1, the conic is an ellipse.
(c) Since “+e cos θ ” appears in the denominator, the directrix is to the right of
the focus at the origin. d = |F l| = 92 , so an equation of the directrix is
x = 92 .
(d) The vertices are
9 that is, 16
,π .
15. r =
9
8
, 0 and 94 , π , so the center is midway between them,
1/4
3/4
3
·
=
, where e = 2 and ed =
4 − 8 cos θ 1/4
1 − 2 cos θ
3
4
⇒ d = 38 .
(a) Eccentricity = e = 2
(b) Since e = 2 > 1, the conic is a hyperbola.
(c) Since “−e cos θ ” appears in the denominator, the directrix is to the left of
the focus at the origin. d = |F l| = 38 , so an equation of the directrix is
x = − 38 .
(d) The vertices are − 34 , 0 and 14 , π , so the center is midway between them,
that is, 12 , π .
17. (a) r =
1
, where e = 2 and ed = 1 ⇒ d = 12 . The eccentricity
1 − 2 sin θ
e = 2 > 1, so the conic is a hyperbola. Since “−e sin θ ” appears in the
denominator, the directrix is below the focus at the origin. d = |F l| = 12 ,
so an equation of the directrix is y = − 12 . The vertices are −1, π2 and
1 3π , so the center is midway between them, that is, 23 , 3π
.
3, 2
2
(b) By the discussion that precedes Example 4, the equation
is r =
1
1 − 2 sin θ −
3π
4
.
¤
33
34
¤
CHAPTER 11
PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10
19. For e < 1 the curve is an ellipse. It is nearly circular when e is close to 0. As e
increases, the graph is stretched out to the right, and grows larger (that is, its
right-hand focus moves to the right while its left-hand focus remains at the
origin.) At e = 1, the curve becomes a parabola with focus at the origin.
21. |P F | = e |P l|
⇒ r = e[d − r cos(π − θ)] = e(d + r cos θ) ⇒
r(1 − e cos θ) = ed ⇒ r =
23. |P F | = e |P l|
ed
1 − e cos θ
⇒ r = e[d − r sin(θ − π)] = e(d + r sin θ) ⇒
ed
1 − e sin θ
r(1 − e sin θ) = ed ⇒ r =
25. We are given e = 0.093 and a = 2.28 × 108 . By (7), we have
r=
2.28 × 108 [1 − (0.093)2 ]
2.26 × 108
a(1 − e2 )
=
≈
1 + e cos θ
1 + 0.093 cos θ
1 + 0.093 cos θ
27. Here 2a = length of major axis = 36.18 AU
⇒ a = 18.09 AU and e = 0.97. By (7), the equation of the orbit is
18.09 1 − (0.97)2
1.07
≈
. By (8), the maximum distance from the comet to the sun is
r=
1 − 0.97 cos θ
1 − 0.97 cos θ
18.09(1 + 0.97) ≈ 35.64 AU or about 3.314 billion miles.
29. The minimum distance is at perihelion, where 4.6 × 107 = r = a(1 − e) = a(1 − 0.206) = a(0.794)
⇒
a = 4.6 × 107/0.794. So the maximum distance, which is at aphelion, is
r = a(1 + e) = 4.6 × 107/0.794 (1.206) ≈ 7.0 × 107 km.
31. From Exercise 29, we have e = 0.206 and a(1 − e) = 4.6 × 107 km. Thus, a = 4.6 × 107/0.794. From (7), we can write the
equation of Mercury’s orbit as r = a
dr
−a(1 − e2 )e sin θ
=
dθ
(1 − e cos θ)2
r2 +
dr
dθ
1 − e2
. So since
1 − e cos θ
⇒
2
=
a2 (1 − e2 )2
a2 (1 − e2 )2 e2 sin2 θ
a2 (1 − e2 )2
+
=
(1 − 2e cos θ + e2 )
(1 − e cos θ)2
(1 − e cos θ)4
(1 − e cos θ)4
¤
CHAPTER 11 REVIEW ET CHAPTER 10
the length of the orbit is
]
L=
2π
0
]
s
r2 + (dr/dθ)2 dθ = a(1 − e2 )
0
2π
35
√
1 + e2 − 2e cos θ
dθ ≈ 3.6 × 108 km
(1 − e cos θ)2
This seems reasonable, since Mercury’s orbit is nearly circular, and the circumference of a circle of radius a
is 2πa ≈ 3.6 × 108 km.
11 Review
ET 10
1. (a) A parametric curve is a set of points of the form (x, y) = (f (t), g(t)), where f and g are continuous functions of a
variable t.
(b) Sketching a parametric curve, like sketching the graph of a function, is difficult to do in general. We can plot points on the
curve by finding f(t) and g(t) for various values of t, either by hand or with a calculator or computer. Sometimes, when
f and g are given by formulas, we can eliminate t from the equations x = f (t) and y = g(t) to get a Cartesian equation
relating x and y. It may be easier to graph that equation than to work with the original formulas for x and y in terms of t.
2. (a) You can find
dy
dy/dt
dy
as a function of t by calculating
=
[if dx/dt 6= 0].
dx
dx
dx/dt
(b) Calculate the area as
than (f(α), g(α))].
3. (a) L =
(b) S =
Ub
a
y dx =
Uβ
α
g(t) f 0 (t)dt [or
Uα
β
g(t) f 0 (t)dt if the leftmost point is (f (β), g(β)) rather
Uβ s
Uβs
(dx/dt)2 + (dy/dt)2 dt = α [f 0 (t)]2 + [g 0 (t)]2 dt
α
Uβ
α
2πy
s
s
Uβ
(dx/dt)2 + (dy/dt)2 dt = α 2πg(t) [f 0 (t)]2 + [g0 (t)]2 dt
4. (a) See Figure 5 in Section 11.3 [ ET 10.3].
(b) x = r cos θ, y = r sin θ
(c) To find a polar representation (r, θ) with r ≥ 0 and 0 ≤ θ < 2π, first calculate r =
cos θ = x/r and sin θ = y/r.
s
x2 + y 2 . Then θ is specified by
dr
dy
d
d
sin θ + r cos θ
(y)
(r
sin
θ)
dθ
dy
= dθ = dθ
, where r = f(θ).
= dθ
= 5. (a) Calculate
dx
d
d
dx
dr
(x)
(r cos θ)
cos θ − r sin θ
dθ
dθ
dθ
dθ
(b) Calculate A =
(c) L =
Ub
1 2
r
a 2
dθ =
Ub
1
[f(θ)]2
a 2
dθ
Ubs
Ubs
Ubs
(dx/dθ)2 + (dy/dθ)2 dθ = a r2 + (dr/dθ)2 dθ = a [f (θ)]2 + [f 0 (θ)]2 dθ
a
6. (a) A parabola is a set of points in a plane whose distances from a fixed point F (the focus) and a fixed line l (the directrix)
are equal.
(b) x2 = 4py; y 2 = 4px
36
¤
CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES
ET CHAPTER 10
7. (a) An ellipse is a set of points in a plane the sum of whose distances from two fixed points (the foci) is a constant.
(b)
y2
x2
+ 2
= 1.
2
a
a − c2
8. (a) A hyperbola is a set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant.
This difference should be interpreted as the larger distance minus the smaller distance.
x2
y2
−
=1
a2
c2 − a2
√
c2 − a2
x
(c) y = ±
a
(b)
9. (a) If a conic section has focus F and corresponding directrix l, then the eccentricity e is the fixed ratio |P F | / |P l| for points
P of the conic section.
(b) e < 1 for an ellipse; e > 1 for a hyperbola; e = 1 for a parabola.
(c) x = d: r =
1. False.
ed
ed
ed
ed
. x = −d: r =
. y = d: r =
. y = −d: r =
.
1 + e cos θ
1 − e cos θ
1 + e sin θ
1 − e sin θ
Consider the curve defined by x = f (t) = (t − 1)3 and y = g(t) = (t − 1)2 . Then g0 (t) = 2(t − 1), so g0 (1) = 0,
but its graph has a vertical tangent when t = 1. Note: The statement is true if f 0 (1) 6= 0 when g0 (1) = 0.
3. False.
For example, if f (t) = cos t and g(t) = sin t for 0 ≤ t ≤ 4π, then the curve is a circle of radius 1, hence its length
U 4π s
U 4π
U 4π s
[f 0 (t)]2 + [g 0 (t)]2 dt = 0
(− sin t)2 + (cos t)2 dt = 0 1 dt = 4π, since as t increases
is 2π, but 0
from 0 to 4π, the circle is traversed twice.
5. True.
The curve r = 1 − sin 2θ is unchanged if we rotate it through 180◦ about O because
1 − sin 2(θ + π) = 1 − sin(2θ + 2π) = 1 − sin 2θ. So it’s unchanged if we replace r by −r. (See the discussion
after Example 8 in Section 11.3 [ ET 10.3].) In other words, it’s the same curve as r = −(1 − sin 2θ) = sin 2θ − 1.
7. False.
The first pair of equations gives the portion of the parabola y = x2 with x ≥ 0, whereas the second pair of equations
traces out the whole parabola y = x2 .
9. True.
By rotating and translating the parabola, we can assume it has an equation of the form y = cx2 , where c > 0.
The tangent at the point a, ca2 is the line y − ca2 = 2ca(x − a); i.e., y = 2cax − ca2 . This tangent meets
the parabola at the points x, cx2 where cx2 = 2cax − ca2 . This equation is equivalent to x2 = 2ax − a2
[since c > 0]. But x2 = 2ax − a2 ⇔ x2 − 2ax + a2 = 0 ⇔ (x − a)2 = 0 ⇔ x = a ⇔
x, cx2 = a, ca2 . This shows that each tangent meets the parabola at exactly one point.
CHAPTER 11 REVIEW ET CHAPTER 10
¤
37
1. x = t2 + 4t, y = 2 − t, −4 ≤ t ≤ 1. t = 2 − y, so
x = (2 − y)2 + 4(2 − y) = 4 − 4y + y 2 + 8 − 4y = y 2 − 8y + 12 ⇔
x + 4 = y 2 − 8y + 16 = (y − 4)2 . This is part of a parabola with vertex
(−4, 4), opening to the right.
3. y = sec θ =
1
1
= . Since 0 ≤ θ ≤ π/2, 0 < x ≤ 1 and y ≥ 1.
cos θ
x
This is part of the hyperbola y = 1/x.
5. Three different sets of parametric equations for the curve y =
(i) x = t, y =
√
x are
√
t
(ii) x = t4 , y = t2
(iii) x = tan2 t, y = tan t, 0 ≤ t < π/2
There are many other sets of equations that also give this curve.
The Cartesian coordinates are x = 4 cos 2π
= 4 − 12 = −2 and
3
√ √ √
3
=
4
y = 4 sin 2π
= 2 3, that is, the point −2, 2 3 .
3
2
7. (a)
s
√
√
3
y
⇒ tan θ =
, and since
(−3)2 + 32 = 18 = 3 2. Also, tan θ =
x
−3
√
(−3, 3) is in the second quadrant, θ = 3π
. Thus, one set of polar coordinates for (−3, 3) is 3 2, 3π
, and two others are
4
4
√ 7π √ 11π 3 2, 4 and −3 2, 4 .
(b) Given x = −3 and y = 3, we have r =
9. r = 1 − cos θ. This cardioid is
symmetric about the polar axis.
38
¤
CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES
ET CHAPTER 10
11. r = cos 3θ. This is a
three-leaved rose. The curve is
traced twice.
13. r = 1 + cos 2θ. The curve is
symmetric about the pole and
both the horizontal and vertical
axes.
15. r =
3
1 + 2 sin θ
⇒ e = 2 > 1, so the conic is a hyperbola. de = 3 ⇒
and the form “+2 sin θ” imply that the directrix is above the focus at
.
the origin and has equation y = 32 . The vertices are 1, π2 and −3, 3π
2
d=
3
2
17. x + y = 2
⇔ r cos θ + r sin θ = 2 ⇔ r(cos θ + sin θ) = 2 ⇔ r =
2
cos θ + sin θ
19. r = (sin θ)/θ. As θ → ±∞, r → 0.
As θ → 0, r → 1. In the first figure,
there are an infinite number of
x-intercepts at x = πn, n a nonzero
integer. These correspond to pole
points in the second figure.
21. x = ln t, y = 1 + t2 ; t = 1.
dx
1
dy
dy/dt
2t
dy
= 2t and
= , so
=
=
= 2t2 .
dt
dt
t
dx
dx/dt
1/t
When t = 1, (x, y) = (0, 2) and dy/dx = 2.
23. r = e−θ
⇒ y = r sin θ = e−θ sin θ and x = r cos θ = e−θ cos θ
dy
dy/dθ
=
=
dx
dx/dθ
When θ = π,
dr
dθ
dr
dθ
⇒
sin θ + r cos θ
−e−θ sin θ + e−θ cos θ −eθ
sin θ − cos θ
=
·
.
=
−e−θ cos θ − e−θ sin θ −eθ
cos θ + sin θ
cos θ − r sin θ
dy
0 − (−1)
1
=
=
= −1.
dx
−1 + 0
−1
CHAPTER 11 REVIEW ET CHAPTER 10
25. x = t + sin t, y = t − cos t
⇒
dy/dt
1 + sin t
dy
=
=
dx
dx/dt
1 + cos t
¤
39
⇒
d dy
(1 + cos t) cos t − (1 + sin t)(− sin t)
dt dx
d 2y
(1 + cos t)2
cos t + cos2 t + sin t + sin2 t
1 + cos t + sin t
=
=
=
=
2
dx
dx/dt
1 + cos t
(1 + cos t)3
(1 + cos t)3
27. We graph the curve x = t3 − 3t, y = t2 + t + 1 for −2.2 ≤ t ≤ 1.2.
By zooming in or using a cursor, we find that the lowest point is about
(1.4, 0.75). To find the exact values, we find the t-value at which
3
.
dy/dt = 2t + 1 = 0 ⇔ t = − 12 ⇔ (x, y) = 11
8 , 4
29. x = 2a cos t − a cos 2t
sin t = 0 or cos t =
1
2
⇒
⇒ t = 0,
y = 2a sin t − a sin 2t ⇒
t = 0,
2π
3 ,
or
dx
= −2a sin t + 2a sin 2t = 2a sin t(2 cos t − 1) = 0 ⇔
dt
4π
3 .
π
,
3
π, or
5π
.
3
dy
= 2a cos t − 2a cos 2t = 2a 1 + cos t − 2 cos2 t = 2a(1 − cos t)(1 + 2 cos t) = 0 ⇒
dt
Thus the graph has vertical tangents where
t=
π
,
3
π and
t=
2π
3
and
5π
,
3
4π
.
3
t
and horizontal tangents where
To determine what the slope is
where t = 0, we use l’Hospital’s Rule to evaluate
lim
t→0
dy/dt
= 0, so there is a horizontal tangent
dx/dt
there.
x
y
0
a
π
3
2π
3
3
a
2
− 12 a
3
a
2
√
3 3
a
2
− 12 a
3
a
2
√
−323a
√
− 23 a
π
4π
3
5π
3
0
√
−3a
π
31. The curve r2 = 9 cos 5θ has 10 “petals.” For instance, for − 10
≤θ≤
0
π
,
10
there are two petals, one with r > 0 and one
with r < 0.
A = 10
U π/10
1 2
r
−π/10 2
dθ = 5
33. The curves intersect when 4 cos θ = 2
U π/10
−π/10
9 cos 5θ dθ = 5 · 9 · 2
U π/10
⇒ cos θ = 12 ⇒ θ = ± π3
for −π ≤ θ ≤ π. The points of intersection are 2, π3 and 2, − π3 .
0
π/10
cos 5θ dθ = 18 sin 5θ 0
= 18
40
¤
CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES
35. The curves intersect where 2 sin θ = sin θ + cos θ
sin θ = cos θ
⇒θ=
π
,
4
⇒
and also at the origin (at which θ =
ET CHAPTER 10
3π
4
on the second curve).
U 3π/4
U π/4
A = 0 21 (2 sin θ)2 dθ + π/4 12 (sin θ + cos θ)2 dθ
U π/4
U 3π/4
= 0 (1 − cos 2θ) dθ + 12 π/4 (1 + sin 2θ) dθ
π/4 3π/4
= θ − 12 sin 2θ 0 + 12 θ − 14 cos 2θ π/4 = 12 (π − 1)
37. x = 3t2 , y = 2t3 .
√
U2s
U2√
U2√
U2s
(dx/dt)2 + (dy/dt)2 dt = 0 (6t)2 + (6t2 )2 dt = 0 36t2 + 36t4 dt = 0 36t2 1 + t2 dt
0
√
U2
U2 √
U5
u = 1 + t2 , du = 2t dt
= 0 6 |t| 1 + t2 dt = 6 0 t 1 + t2 dt = 6 1 u1/2 12 du
k
l5
√
= 6 · 12 · 23 u3/2 = 2(53/2 − 1) = 2 5 5 − 1
L=
1
U 2π s
U 2π s
39. L = π
r2 + (dr/dθ)2 dθ = π
(1/θ)2 + (−1/θ2 )2 dθ =
]
2π
π
s
θ2 + 1
dθ
θ2
% s
&
√
√
√
2π
s
2π + 4π2 + 1
θ2 + 1
π2 + 1
4π2 + 1
24
2
√
+ ln θ + θ + 1
−
+ ln
= −
=
θ
π
2π
π + π2 + 1
π
√
√
√
2π + 4π2 + 1
2 π 2 + 1 − 4π2 + 1
√
+ ln
=
2π
π + π2 + 1
41. x = 4
S=
√
1
t3
+ 2, 1 ≤ t ≤ 4 ⇒
t, y =
3
2t
U4
1
= 2π
2πy
s
t √ 2
U4 2/ t + (t2 − t−3 )2 dt
(dx/dt)2 + (dy/dt)2 dt = 1 2π 13 t3 + 12 t−2
U 4 1
1
3
t3 + 12 t−2
s
U4
(t2 + t−3 )2 dt = 2π 1 13 t5 +
5
6
1 6 5
4
+ 12 t−5 dt = 2π 18
t + 6 t − 18 t−4 1 =
43. For all c except −1, the curve is asymptotic to the line x = 1. For
c < −1, the curve bulges to the right near y = 0. As c increases, the
bulge becomes smaller, until at c = −1 the curve is the straight line x = 1.
As c continues to increase, the curve bulges to the left, until at c = 0 there
is a cusp at the origin. For c > 0, there is a loop to the left of the origin,
whose size and roundness increase as c increases. Note that the x-intercept
of the curve is always −c.
471,295
π
1024
CHAPTER 11 REVIEW ET CHAPTER 10
45.
x2
y2
+
= 1 is an ellipse with center (0, 0).
9
8
√
a = 3, b = 2 2, c = 1 ⇒
47. 6y 2 + x − 36y + 55 = 0
2
¤
41
⇔
6(y − 6y + 9) = −(x + 1) ⇔
(y − 3)2 = − 16 (x + 1), a parabola with vertex (−1, 3),
1
opening to the left, p = − 24
⇒ focus − 25
, 3 and
24
foci (±1, 0), vertices (±3, 0).
directrix x = − 23
24 .
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 4,
y2
x2
+
= 1.
25
9
so b2 = a2 − c2 = 52 − 42 = 9. An equation is
51. The center of a hyperbola with foci (0, ±4) is (0, 0), so c = 4 and an equation is
The asymptote y = 3x has slope 3, so
10b2 = 16 ⇒ b2 =
8
5
a
3
=
b
1
and so a2 = 16 −
8
5
⇒ a = 3b and a2 + b2 = c2
=
72
.
5
Thus, an equation is
y2
x2
− 2 = 1.
2
a
b
⇒ (3b)2 + b2 = 42
⇒
x2
5y 2
5x2
y2
−
= 1, or
−
= 1.
72/5
8/5
72
8
53. x2 = −(y − 100) has its vertex at (0, 100), so one of the vertices of the ellipse is (0, 100). Another form of the equation of a
parabola is x2 = 4p(y − 100) so 4p(y − 100) = −(y − 100) ⇒ 4p = −1 ⇒ p = − 14 . Therefore the shared focus is
so 2c = 399
. So a = 100 − 399
found at 0, 399
− 0 ⇒ c = 399
and the center of the ellipse is 0, 399
= 401
and
4
4
8
8
8
8
2
y − 399
4012 − 3992
x2
8
= 25. So the equation of the ellipse is 2 +
=1 ⇒
b =a −c =
82
b
a2
2
or
2
2
(8y − 399)2
x2
+
= 1.
25
160,801
55. Directrix x = 4
⇒ d = 4, so e =
1
3
⇒ r=
2
y − 399
x2
8
+ 401 2 = 1,
25
8
4
ed
=
.
1 + e cos θ
3 + cos θ
57. In polar coordinates, an equation for the circle is r = 2a sin θ. Thus, the coordinates of Q are x = r cos θ = 2a sin θ cos θ
and y = r sin θ = 2a sin2 θ. The coordinates of R are x = 2a cot θ and y = 2a. Since P is the midpoint of QR, we use the
midpoint formula to get x = a(sin θ cos θ + cot θ) and y = a(1 + sin2 θ).
PROBLEMS PLUS
1. x =
]
t
1
cos u
du, y =
u
]
t
1
sin u
dx
cos t
dy
sin t
du, so by FTC1, we have
=
and
=
. Vertical tangent lines occur when
u
dt
t
dt
t
dx
= 0 ⇔ cos t = 0. The parameter value corresponding to (x, y) = (0, 0) is t = 1, so the nearest vertical tangent
dt
occurs when t =
L=
π
.
2
]
1
Therefore, the arc length between these points is
π/2
v
dx
dt
2
+
dy
dt
2
dt =
]
u
π/2
1
sin2 t
cos2 t
+ 2 dt =
2
t
t
]
1
π/2
π/2
dt = ln t 1 = ln π2
t
3. In terms of x and y, we have x = r cos θ = (1 + c sin θ) cos θ = cos θ + c sin θ cos θ = cos θ + 12 c sin 2θ and
y = r sin θ = (1 + c sin θ) sin θ = sin θ + c sin2 θ. Now −1 ≤ sin θ ≤ 1 ⇒ −1 ≤ sin θ + c sin2 θ ≤ 1 + c ≤ 2, so
−1 ≤ y ≤ 2. Furthermore, y = 2 when c = 1 and θ =
π
,
2
while y = −1 for c = 0 and θ =
3π
.
2
Therefore, we need a viewing
rectangle with −1 ≤ y ≤ 2.
To find the x-values, look at the equation x = cos θ + 12 c sin 2θ and use the fact that sin 2θ ≥ 0 for 0 ≤ θ ≤
π
2
and
sin 2θ ≤ 0 for − π2 ≤ θ ≤ 0. [Because r = 1 + c sin θ is symmetric about the y-axis, we only need to consider
So for − π2 ≤ θ ≤ 0, x has a maximum value when c = 0 and then x = cos θ has a maximum value
of 1 at θ = 0. Thus, the maximum value of x must occur on 0, π2 with c = 1. Then x = cos θ + 12 sin 2θ ⇒
− π2 ≤ θ ≤
dx
dθ
π
2 .]
= − sin θ + cos 2θ = − sin θ + 1 − 2 sin2 θ
⇒
dx
dθ
= −(2 sin θ − 1)(sin θ + 1) = 0 when sin θ = −1 or
1
2
[but sin θ 6= −1 for 0 ≤ θ ≤ π2 ]. If sin θ = 12 , then θ = π6 and
√
√
x = cos π6 + 12 sin π3 = 34 3. Thus, the maximum value of x is 34 3, and,
√
by symmetry, the minimum value is − 34 3. Therefore, the smallest
viewing rectangle that contains every member of the family of polar curves
√ √
r = 1 + c sin θ, where 0 ≤ c ≤ 1, is − 34 3, 34 3 × [−1, 2].
5. (a) If (a, b) lies on the curve, then there is some parameter value t1 such that
3t1
3t21
= a and
= b. If t1 = 0,
1 + t31
1 + t31
the point is (0, 0), which lies on the line y = x. If t1 6= 0, then the point corresponding to t =
x=
1
is given by
t1
3(1/t1 )2
3t2
3t1
3(1/t1 )
= a. So (b, a) also lies on the curve. [Another way to see
= 3 1 = b, y =
= 3
3
1 + (1/t1 )
t1 + 1
1 + (1/t1 )3
t1 + 1
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when
t = t2
⇒ t = 0 or 1, so the points are (0, 0) and
3
3
2, 2
.
3t
3t2
=
3
1+t
1 + t3
⇒
43
44
¤
(b)
PROBLEMS PLUS
√
(1 + t3 )(6t) − 3t2 (3t2 )
dy
6t − 3t4
=
=
= 0 when 6t − 3t4 = 3t(2 − t3 ) = 0 ⇒ t = 0 or t = 3 2, so there are
3
2
3
2
dt
(1 + t )
(1 + t )
√ √
3
2, 3 4 . Using the symmetry from part (a), we see that there are vertical tangents at
horizontal tangents at (0, 0) and
√ √ (0, 0) and 3 4, 3 2 .
(c) Notice that as t → −1+ , we have x → −∞ and y → ∞. As t → −1− , we have x → ∞ and y → −∞. Also
y − (−x − 1) = y + x + 1 =
slant asymptote.
(d)
(t + 1)3
(t + 1)2
3t + 3t2 + (1 + t3 )
→ 0 as t → −1. So y = −x − 1 is a
=
= 2
3
3
1+t
1+t
t −t+1
dx
(1 + t3 )(3) − 3t(3t2 )
6t − 3t4
dy/dt
t(2 − t3 )
3 − 6t3
dy
dy
=
=
=
=
=
and from part (b) we have
. So
.
3
2
3
2
3
2
dt
(1 + t )
(1 + t )
dt
(1 + t )
dx
dx/dt
1 − 2t3
d dy
dt dx
2(1 + t3 )4
1
d2 y
=
>0 ⇔ t< √
Also 2 =
.
3
dx
dx/dt
3(1 − 2t3 )3
2
So the curve is concave upward there and has a minimum point at (0, 0)
√ √ and a maximum point at 3 2, 3 4 . Using this together with the
information from parts (a), (b), and (c), we sketch the curve.
3 3
3t
3t2
27t3 + 27t6
27t3 (1 + t3 )
27t3
+
=
=
=
and
3
3
3
3
3
3
1+t
1+t
(1 + t )
(1 + t )
(1 + t3 )2
3t2
27t3
3t
, so x3 + y 3 = 3xy.
=
3xy = 3
3
3
1+t
1+t
(1 + t3 )2
(e) x3 + y 3 =
(f ) We start with the equation from part (e) and substitute x = r cos θ, y = r sin θ. Then x3 + y 3 = 3xy
r3 cos3 θ + r3 sin3 θ = 3r2 cos θ sin θ. For r 6= 0, this gives r =
⇒
3 cos θ sin θ
. Dividing numerator and denominator
cos3 θ + sin3 θ
1
sin θ
3
cos
θ
cos
θ
3 sec θ tan θ
.
=
by cos3 θ, we obtain r =
3
1 + tan3 θ
sin θ
1+
cos3 θ
π
(g) The loop corresponds to θ ∈ 0, 2 , so its area is
2
]
U π/2 r2
1 U π/2 3 sec θ tan θ
9 U π/2 sec2 θ tan2 θ
9 ∞ u2 du
dθ
=
dθ
=
dθ
=
0
2
2 0
1 + tan3 θ
2 0 (1 + tan3 θ)2
2 0 (1 + u3 )2
b
= lim 92 − 13 (1 + u3 )−1 0 = 32
A=
[let u = tan θ]
b→∞
(h) By symmetry, the area between the folium and the line y = −x − 1 is equal to the enclosed area in the third quadrant,
plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is 12 , and since y = −x − 1 ⇒
1
, the area in the fourth quadrant is
sin θ + cos θ
%
2 2 &
]
3 sec θ tan θ
1 −π/4
1
CAS 1
−
−
dθ = . Therefore, the total area is
2 −π/2
sin θ + cos θ
1 + tan3 θ
2
r sin θ = −r cos θ − 1 ⇒ r = −
1
2
+ 2 12 = 32 .