(2610 means problem 26 in 10-th edition textbook, 309 means problem 30 in 9-th edition textbook )
1
12.3
2610 = 309 : Note the acceleration is ~a = −32~j. The path of the projectile is
described by the vector-valued function:
√
√
~r(t) = h450 2 t, 3 + 450 2 t − 16t2 i.
√
√
2. So the maximal height
1. Maximal height: 450 2 − 32t1 = 0 ⇒ t1 = 450
32
is:
√ 450 √
4502
hmax = y(t1 ) = 3 + 450 2
2 − 16 2 2 ≈ 6331.13f t.
32
32
√
√
2. Range: 3 + 450 2t2 − 16t22 = 0 ⇒ t2 = 39.78. The range is 450 2t2 =
25315.5 ft.
12
π
3610 = 429 : Note that 12◦ = 180
π = 15
. The corresponding vector-valued
function is
π
π
~r(t) = hv0 cos
t, v0 sin
t − 16t2 i.
15
15
The time when the projectile hits the ground is obtained by
v0 (sin(π/15))t2 − 16t22 = 0 =⇒ t2 =
v0
sin(π/15).
16
The range for the projectile is
R = v 0 t2 =
v02
π
π
v2
2π
(sin )(cos ) = 0 sin
.
16
15
15
32
15
So
v2
2π
R = 200 = 0 sin
=⇒ v0 =
32
15
2
s
200 · 32
≈ 125.44f t/sec.
sin 2π
15
12.4
√
2 −1/2
1010 = 149 : ~r 0 (t) = h1, 1, −t(4 −
i. ~r(t1 ) = h1, 1, 3i ⇒ t1 = 1. So the
√t )
tangent vector at P is h1, 1, −1/ 3i. The unit tangent vector is
√
p
p
√
~r 0 (1)
h1, 1, −1/ 3i
~
p
T (1) =
=
=
h
3/7,
3/7,
−1/
7i.
k~r 0 (1)k
7/3
√
The tangent line at point P = (1, 1, 3) is given by
r
r
√
3
3
1
x=1+
t, y = 1 +
t, z = 3 − √ t.
7
7
7
1
1910 = 299 : ~r 0 (t) = −6 sin t~i + 6 cos t~j. T~ = − sin t~i + cos t~j. T~ 0 (t) =
− cos t~i − sin t~j. The principal normal vector is
√
√
3π
T~ 0 (t)
2~
2~
~
~
~
~
= − cos t i − sin t j, N
i−
j.
N (t) =
=
0
~
4
2
2
kT (t)k
2210 = 369 : ~r 0 (t) = 2t~i + 2~j. The unit tangent vector is
T~ (t)
t~i + ~j
T~ (t) =
=√
.
t2 + 1
kT~ (t)k
T~ 0 (t)
=
=
~i(t2 + 1) − (t~i + ~j)t
+ (t~i + ~j)(−1/2)(t2 + 1)−3/2 (2t) =
(t2 + 1)3/2
t2 + 1
~i − t~j
.
2
(t + 1)3/2
√
~i
So the principal normal vector is
~ 0
~ ~
~ (t) = T (t) = √i − tj .
N
0
1 + t2
kT~ (t)k
Note that another way to get principal normal vector for a plane curve is to use
the following trick.
~r 0 (t) = h2t, 2i ⇒ n1 = h−2, 2ti, n2 = h2, −2ti.
n1 , n2 are normal vectors in opposite directions. From the geometry of the
3
2
1
0.5
1.0
1.5
2.0
-1
-2
-3
curve, at t = 1, the turning direction is downward-right, so we need to choose
~ ~
~ (t) = n2 (t) = √i − tj .
n2 = h2, −2ti =⇒ N
kn2 (t)k
1 + t2
The acceleration vector is:
~a(t) = ~r 00 (t) ≡ 2~i.
aN
t~i + ~j
2t
aT = ~a · T~ = 2~i · √
=√
.
2
1+t
1 + t2
r
q
4t2
2
2
2
= kak − aT = 4 −
=√
.
2
1+t
1 + t2
2
When t = 1, we get
~i + ~j
T~ (1) = √ ,
2
~ ~
~ (1) = i − j ,
N
2
aT =
√
2,
aN =
√
2.
2910 − 3210 = 459 − 489 :
29: ~r 0 (t) = −aω sin(ωt)~i + aω cos(ωt)~j. Unit tangent vector:
T~ (t) = − sin(ωt)~i + cos(ωt)~j.
T~ 0 (t) = −ω cos(ωt)~i − ω sin(ωt)~j. The principal normal vector
~ 0
~ (t) = T (t) = − cos(ωt)~i − sin(ωt)~j.
N
kT~ 0 (t)k
00
= −aω 2 cos(ωt)~i − aω 2 sin(ωt)~j.
q
aT = ~a(t) · T~ = 0. aN = k~ak2 − a2T = aω 2 .
~a(t) = ~r
~ (t) = −ω 2~r(t) is in the
30: T~ (t) is orthogonal to the position vector ~r(t). N
opposite direction of ~r(t).
31: The speed of the object at any time t is a constant: k~v (t)k = aω. So
d
aT = dt
k~v k = 0 as seen above.
32: If ω is halved, i.e. ω
3
ω/2, then aN
aN /4.
12.5
1010 = 109 : ~r 0 (t) = 2t~j + 3t2~k. k~r 0 (t)k =
Z
2
Z
0
k~r (t)kdt =
L =
0
=
1
18
2
t
p
4+
9t2 dt
0
Z
4
40
√
√
√
4t2 + 9t4 = |t| 4 + 9t2 .
1
=
18
Z
2
p
4 + 9t2 d(4 + 9t2 )
0
√
1 2 3/2 i40
8
udu =
u
=
(10 10 − 1).
18 3
27
4
3
√
1210 = 129 : ~r 0 (t) = h2 cos t, 5, −2 sin ti. k~r 0 (t)k = 29.
Z π
√
k~r 0 (t)kdt = 29π.
L=
0
1710 = 199 : (a): ~r 0 (t) = h−2 sin t, 2 cos t, 1i. k~r 0 (t)k =
Z
s(t) =
t
k~r 0 (u)kdu =
√
√
5. So
5t.
0
√
(b): t = s/ 5. So the arc length parametrization is:
s
s
s
~r(s) = 2 cos √ , 2 sin √ , √
.
5
5
5
√
(c): ~r( 5) = h2 cos 1, 2 sin
√ 1, 1i. The coordinate of the point=(2 cos 1, 2 sin 1, 1)
for the arc length s = √5. Similarly,
of the point for the arc
√ the coordinate
√
length s = 4 is (2 cos(4/ 5), 2 sin(4/ 5), 4/ 5).
(d):
2
s
2
s
1
~r 0 (s) = − √ sin √ , √ cos √ , √
, k~r 0 (s)k = 1.
5
5
5
5
5
4
13.1
1610 = 169 :
y
y
g(x, y) = ln |t||x = ln .
x
So (a): g(4, 1) = ln(1/4) = − ln 4. (b): g(6, 3) = − ln 2. (c): g(2, 5) = ln(5/2).
(d): g(1/2, 7) = ln 14.
2410 = 249 : Domain={(x, y) ∈ R2 ; x 6= y}. Range=R = (−∞, +∞)
4