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Math 265 (Butler)
Practice Midterm I — A (Solutions)
1.
A particle moves through three dimensional space with velocity
v
(
t
) =
h
sec
2
t,
2 sec
t
tan
t,
tan
2
t
i
.
At time
t
= 0 the particle is at
h
0
,
1
,
2
i
, ﬁnd the position function of the particle for

π/
4
≤
t
≤
π/
4.
If
r
(
t
) is the position function then
r
0
(
t
) =
v
(
t
). So taking antiderivatives we
have
r
(
t
) =
Z
v
(
t
)
dt
=
±Z
sec
2
tdt,
Z
2 sec
t
tan
Z
tan
2
tdt
²
=
±
tan
t
+
C,
2 sec
t
+
D,
Z
(sec
2
t

1)
dt
²
=
±
tan
t
+
C,
2 sec
t
+
D,
tan
t

t
+
E
²
.
Two of the three integrals are straightforward. The last one is the trickiest
but this follows by relating tan
2
t
(something which we cannot directly inte
grate) to sec
2
t
(something which is easy to integrate). Now all that is left is
to determine the constants
C,D,E
. We have
r
(0) =
h
C,
2 +
D,E
i
=
h
0
,
1
,
2
i
giving the constants we need. So we have that the position function of the
particle is
r
(
t
) =
±
tan
t,
2 sec
t

1
,
tan
t

t
+ 2
²
.
(The condition for

π/
4
≤
t
≤
π/
4 is not needed directly, it only is used to
guarantee that we stay away from the vertical asymptotes (solutions cannot
be pushed past vertical asymptotes, something you will learn in your future
math classes).)
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View Full Document 2.
Find the point on the plane
x
+2
y
= 5+3
z
which is
closest
to the point (4
,
4
,

7).
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This note was uploaded on 02/06/2012 for the course MATH 265 taught by Professor Gregorac during the Spring '08 term at Iowa State.
 Spring '08
 Gregorac
 Math

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