Thomas Calculus 13th [Solutions]

Section 13.3 Arc Length in Space 943
12. r cos t t sin t i sin t t cost j v sin t sin t t cost i cost cost t sin t j
2 2 2
t t t t t t t t t t
cos i sin j v cos cos , since
Length
s
s
2
2 2
2 3
2 2 2 8
t
2
t s t d t
2 0 2
t t t t t t t t
13. r e cos t i e sin t j e k v e cos t e sin t i e sin t e cos t j e k
v
Length
t t
2
t t
2
t
2
2t t t
t
e t e t e t e t e e e s t e d e
0
cos sin sin cos 3 3 3 3 3
ln 4 3 3
s 0 s ln 4 0 3e
3
4
14.
2 2 2
t
(1 2 t) (1 3 t) (6 6 t) 2 3 6 2 3 ( 6) 7 s( t) 7 d 7t
0
Length s(0) s( 1) 0 ( 7) 7
r i j k v i j k v
15.
2
2 2 2 2
t t t t t t
r 2 i 2 j 1 k v 2i 2j 2 k v 2 2 2 4 4
2
2 1 t Length
1 2 2 1
2
t dt t t t t
0
2 2
2 1 2 1 ln 1 2 ln 1 2
1
0
16. Let the helix make one complete turn from t 0 to
t 2 . Note that the radius of the cylinder is 1
the circumference of the base is 2 . When
t 2 , the point P is
cos 2 , sin 2 , 2 1, 0, 2 the cylinder is
2 units high. Cut the cylinder along PQ and
flatten. The resulting rectangle has a width equal to
the circumference of the cylinder 2 and a height
equal to 2 , the height of the cylinder. Therefore,
the rectangle is a square and the portion of the helix
from t 0 to t 2 is its diagonal.
17. (a) r cos t i sin t j 1 cos t k, 0 t 2 x cos t, y sin t, z 1 cos t
2 2 2 2
x y cos t sin t 1, a right circular cylinder with the z -axis as the axis and radius 1.
Therefore P cos t, sin t, 1 cost lies on the cylinder
2 2
x y 1; t 0 P 1, 0, 0 is on the curve;
t
2
Q 0, 1,1 is on the curve; t R 1, 0, 2 is on the curve. Then PQ i j k and
i j k
PR 2i 2k PQ PR 1 1 1 2i 2k is a vector normal to the plane of P, Q, and R. Then
2 0 2
the plane containing P, Q, and R has an equation 2x 2z 2(1) 2(0) or x z 1. Any point on the
curve will satisfy this equation since x z cos t (1 cos t ) 1. Therefore, any point on the curve lies on
2
the intersection of the cylinder x
2
y 1 and the plane x z 1 the curve is an ellipse.
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