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35.
Since
}
d
d
i
t
}52
2 sin
t
1
2 cos
t
, the largest magnitude of the
current occurs when
2
2 sin
t
1
2 cos
t
5
0, or
sin
t
5
cos
t
. Squaring both sides gives sin
2
t
5
cos
2
t
, and
we know that sin
2
t
1
cos
2
t
5
1, so sin
2
t
5
cos
2
t
5 }
1
2
}
.
Thus the possible values of
t
are
}
p
4
}
,
}
3
4
p
}
,
}
5
4
p
}
, and so on.
Eliminating extraneous solutions, the solutions of
sin
t
5
cos
t
are
t
5 }
p
4
}
1
k
p
for integers
k
, and at these
times
)
i
)
5
)
2 cos
t
1
2 sin
t
)
5
2
ˇ
2
w
. The peak current is
2
ˇ
2
w
amps.
36.
The square of the distance is
D
(
x
)
5
1
x
2 }
3
2
}
2
2
1
(
ˇ
x
w
1
0)
2
5
x
2
2
2
x
1 }
9
4
}
,
so
D
9
(
x
)
5
2
x
2
2 and the critical point occurs at
x
5
1.
Since
D
9
(
x
)
,
0 for
x
,
1 and
D
9
(
x
)
.
0 for
x
.
1, the
critical point corresponds to the minimum distance. The
minimum distance is
ˇ
D
w
(1
w
)
w
5 }
ˇ
2
5
w
}
.
37.
Calculus method:
The square of the distance from the point (1,
ˇ
3
w
) to
(
x
,
ˇ
1
w
6
w
2
w
x
w
2
w
) is given by
D
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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