Thomas Calculus 13th [Solutions]

Section 13.5 Tangential and Normal Components of Acceleration 957
20. a( t) aT
T aN
N , where d d
2
aT v (10) 0 and a 100 0 100 .
dt dt
N v a T N Now, from
2
f ( x) Exercise 5(a) Section 13.4, we find for y f ( x)
x that 2 2 ; also,
2
3 2
2
3 2
2
3 2
1 f ( x)
1 (2 x) 1 4x
2
2
r( t)
ti t j is the position vector of the moving mass v i 2tj v 1 4t T 1 i 2 tj
.
2
1 4t
At (0, 0): T(0) i, N(0)
j and (0) 2 F ma m(100 ) N 200 m j;
2 2 2 2
At 2, 2 : T 2 1 i 2 2 j 1 i j, N 2 i 1 j , and 2 2
3 3 3 3 3
27
200 2 2 400 2
F ma m(100 ) N m i 1 j mi 200 mj
27 3 3 81 81
21. r x0 At i y0 Bt j z0 Ct k v Ai Bj Ck a 0 v a 0 0. Since the curve is a
plane curve, 0.
22.
2
aN
0 v 0 0 (since the particle is moving, we cannot have zero speed) the curvature is
zero so the particle is moving along a straight line
23. From Example 1, v t and aN
t so that a 2 aN
t 1, 0 1
N v
2 2
t t
t t
v
24. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows:
f ( t) g ( t) 0
f ( t) g ( t) 0
r f ( t) i g( t) j v f ( t) i g ( t) j a f ( t) i g ( t) j
d a
f ( t) g ( t) 0
f ( t) i g ( t) j
dt
2
v a
0
25. r( t) f ( t) i g( t) j h( t) k v f ( t) i g ( t) j h ( t) k ; v k 0 h ( t) 0 h( t)
C
r( t) f ( t) i g( t)
j Ck and r( a) f ( a) i g( a) j Ck 0 f ( a) 0, g( a) 0 and C 0 h( t) 0.
26. From Example 2,
2 2
v a sin t i a cos t j bk v a b
T v 1 a sin t i a cos t j bk
;
v 2 2
a b
d T
dt
1 a cos t i asin
t
2 2
a b
j
d T
dt
N cos t i sin t j ;
d T
dt
i j k
B T N asin t a cost b bsin t i b cos t j a k
2 2 2 2 2 2 2 2 2 2 2 2
a b a b a b a b a b a b
cost
sin t 0
d B 1 b cos t i bsin t j d B N b 1 d B N 1
b b ,
dt 2 2 dt 2 2 dt 2 2 2 2 2 2
a b a b v
a b a b a b
which is consistent with the result in Example 2.
Copyright
2014 Pearson Education, Inc.