Section 13.3
4.
0
r(t) = cos
pti + sin tj + ln cos tk ⇒ r (t) = − sin ti + cos tj − tan tk,
0
2
2
|r (t)| = (− sin t) + cos t + (− tan t)2 = | sec t|.
Since sec t > 0 for 0 ≤ t ≤ π4 , here we can say |r0 (t)| = sec t. Then
π
π
√
√
R4
L = sec tdt = [ln | sec t + tan t|]04 = ln | 2 + 1| − ln |1 + 0| = ln( 2 + 1).
0
13.
0
r(t) = 2ti + (1 − 3t)j + (5 + 4t)k ⇒ r0 (t) = 2i − 3j + 4k and ds
dt = |r (t)| =
t
t
√
√
√
R
R
√
4 + 9 + 16 = 29.Then s = s(t) = |r0 (u)|du =
29du = 29t. There0
0
fore , t = √129 s , and substituting for t in the origin equation , we have
r(t(s)) = √229 si + (1 − √329 s)j + (5 + √429 s)k
16.
−4t
−2t2 +2
0
r(t) = ( t22+1 − 1)i + t22t
+1 j ⇒ r (t) = (t2 +1)2 i + (t2 +1)2 j,
q
q
ds
−2t2 +2 2
4
2
0
2
[ (t2−4t
dt = |r (t)| =
+1)2 ] + [ (t2 +1)2 ] =
(t2 +1)2 = t2 +1 .
Since the initial point (1,0) corresponds to t=0 , the arc function s(t) =
Rt
|r0 (u)|du =
0
Rt
0
2
u2 +1 du
1
2s
= 2 arctan t. Then arctan t =
⇒ t = tan
1
2 s.
Substituting , we have
1−tan2 ( 1 s)
2 tan( 1 s)
2 tan( 1 s)
2
r(t(s)) = [ tan2 (21 s)+1 − 1]i + tan2 ( 1 s)+1
j= r(t(s)) = 1+tan2 ( 12 s) i + 1+tan2 2( 1 s) j
2
2
2
2
=cos si + sin sj
With this parametrization , we recognize the function is representing the unit
circle. Note here that the curve approachs , but doesn’t include , the point (1,0)
, since cos s = −1 for s = π +2kπ (k an integer) but then t = tan 12 s is undefined
.
18.
2
(a)
t + t sin t) ⇒ r0 (t) = (2t, t sin t, t cos t) ⇒ |r0 (t)| =
p r(t) = (t , sin t − t cos t, cos
√
0
√1 (2, sin t, cos t).
4t2 + t2 sin2 t + t2 cos2 t = 5t [since t > 0]. Then T (t) = |rr0 (t)
(t)| =
5
T 0 (t) =
(b)κ(t)
√1 (0, cos t, − sin t). ⇒
5
√
0
(t)|
1/ 5
1
√
= |T
= 5t
|r 0 (t)| =
5t
|T 0 (t)| =
√1 .
5
1
Thus N (t) =
T 0 (t)
|T 0 (t)|
= (0, cos t, − sin t).
29.
5
3
1
f (x) = 4x 2 , f 0 (x) = 10x 2 , f√00 (x) = 15x 2 ,
|f 00 (x)|
15 x
κ(x) =
3 =
3
0
2
3
[1+(f (x)) ] 2
(1+100x ) 2
40.
Here r(t) = p
(f (t), g(t)), r0 (t) = (f 0 (t), g 0 (t)), r00 (t) = (f 00 (t), g 00 (t)),
3
3
0
3
|r (t)| = [ (f 0 (t))2 + (g 0 (t))2 ]3 = [(f 0 (t))2 + (g 0 (t))2 ] 2 = (ẋ2 + ẏ 2 ) 2 and
1
|r0 (t) × r00 (t)| = |(0, 0, f 0 (t)g 00 (t) − f 00 (t)g 0 (t))| = [(ẋÿ − ẍẏ)2 ] 2 = |ẋÿ − ẍẏ|.
|ẋÿ−ẍẏ|
Thus κ(t) = 2 2 3
(ẋ +ẏ ) 2
41.
x = exp t cos t ⇒ ẋ = exp t(cos t−sin t) ⇒ ẍ = exp t(− sin t−cos t)+exp t(cos t−
sin t) = −2 exp t sin t
y = exp t sin t ⇒ ẏ = exp t(sin t+cos t) ⇒ ÿ = exp t(− sin t+cos t)+exp t(cos t+
sin t) = 2 exp t cos t
ẏ|
√ 1
∴ κ(t) = |ẋÿ−ẍ
3 =
2 exp t
2
2
(ẋ +ẏ ) 2
43.
0
2
(t)
,1)
2 2 1
(1, 23 , 1) corresponds to t = 1. T (t) = |rr0 (t)|
= (2t,2t
2t2 +1 ,so T (1) = ( 3 , 3 , 3 ).
T 0 (t) = −4t(2t2 + 1)−2 (2t, 2t2 , 1) + (2t2 + 1)−1 (2, 4t, 0) = 2(2t2 + 1)−2 (1 −
2
0
,2t,−2t)
2(2t2 +1)−2 (1−2t2 ,2t,−2t)
√
= (1−2t1+2t
.
2t2 , 2t, −2t). N (t) = |TT 0 (t)
2
(t)| =
2
−2
2 2
2
2
2(2t +1)
(1−2t ) +(2t) +(−2t)
N (1) = (− 13 , 23 , − 23 ) and B(1) = T (1) × N (1) = (− 23 , 13 , 32 ).
45.
0
(0, π, −2) corresponds to t = π.r(t) = (2 sin 3t, t, 2 cos 3t) ⇒ T (t) = |rr0 (t)
(t)| =
√1 (6 cos 3t, 1, −6 sin 3t)
37
T (π) = √137 (−6, 1, 0) is normal vector for normal plane , and so (-6,1,0) is also
normal. Thus the equation for the plane is −6(x − 0) + 1(y − π) + 0(z + 2) = 0
or y − 6x = π
T 0 (t) = √137 (−18 sin 3t, 0, −18 cos 3t) ⇒ |T 0 (t)| = √1837 ⇒ N (t) = |TT0(t)
(t)| =
(− sin 3t, 0, − cos 3t). so N (π) = (0, 0, 1) and B(π) = √137 (1, 6, 0). Since B(π) is
normal to the osculating plane , an equation foe the plane is 1(x − 0) + 6(y −
π) + 0(z + 2) = 0 or x + 6y = 6π.
49.
The tangent vector is normal to the normal plane , and the vector (6,6,-8) is
normal to the given plane. But T (t)||r0 (t) and (6, 6, −8)||(3, 3, −4) , so we need
to find t such that r0 (t)||(3, 3, −4).
2
r(t) = (t3 , 3t, t4 ) ⇒ r0 (t) = (3t2 , 3, 4t3 )||(3, 3, −4) when t = −1. So the planes
are parallel at the point (-1,-3,1)
3