math192_hw3sp06 - HW3 Solutions 13.3 Arc Length and the...

HW3 Solutions
13.3 Arc Length and the Unit Tangent Vector T
8. T =
v
| v |
= (cos t ) i - (sin t ) j and Length =
R
2
√
2
tdt =
h
t
2
2
i
2
√
2
10. Let P( t
0
) denote the point. Then v = (12 cos t ) i + (12 sin t ) j + 5 k and
- 13 π =
R
t
0
0
p
144 cos 2 t + 144 sin
2
t + 25 t dt =
R
t
0
0
13 dt = 13 t
0
⇒ t
0
= - π , and the point is
P ( - π ) = (12 sin( - π ) , - 12 cos( - π ) , - 5 π ) = (0 , 12 , - 5 π )
14. r (1 + 2 t ) i + (1 + 3 t ) j + (6 - 6 t ) k ⇒ v = 2 i + 3 j - 6 k ⇒ | v | =
p
2
2
+ 3
3
+ ( - 6)
2
= 7 ⇒ s ( t ) =
R
6
0
7 dτ = 7 t ⇒ Length = s (0) - s ( - 1) = 0 - ( - 7) = 7
14.1 Functions of Several Variables
8.
(a) Domain: all ( x, y ) satisfying x
2
+ y
2
≤ 9
(b) Range: 0 ≤ z ≤ 3
(c) level curves are circles centered at the origin with radii r ≤ 3
(d) boundary is the cirle x 2 + y 2 = 9
(e) closed
(f) bounded
13. f
14. e
15. a
16. c
17. d
18. b

20.
-6
-4
-2
0
2
4
6
-5
0
5
-40
-30
-20
-10
0
10
x
f(x,y) = 4-y
2
y
Figure 1: (a)
x
y
z = 4-y
2
-6 -4 -2 0 2 4 6
-6
-4
-2
0
2
4
6
Figure 2: (b)
42. f ( x, y, z ) = ln( x
2
+ y + z
2
) at ( - 1 , 2 , 1) ⇒ w = ln( x
2
+ y + z
2
); at ( - 1 , 2 , 1) ⇒ w = ln(1+2+1) =
ln 4 ⇒ ln 4 = ln( x
2
+ y + z
2
) ⇒ 4 = x
2
+ y + z
2
46. Minimum of f along the line is f (1 , 0 , 9) = (1)(0)