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86. Repeating the steps of Eq. 1647
→
Eq. 1653, but applying
cos
cos
2cos
cos
22
α
β
αβ
+−
§·
+=
¨¸
©¹
(see Appendix E) instead of Eq. 1650, we obtain
[0.10cos
]cos4
y
xt
′
=π
π
, with SI units
understood.
(a) For nonnegative
x
, the smallest value to produce cos
π
x
= 0 is
x
= 1/2, so the answer
is
x
= 0.50 m.
(b) Taking the derivative,
[]
()
0.10cos
4 sin4
dy
ux
t
dt
′
′
==
π
−
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics

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