Ams 261 - Applied Calculus Iii Cheat Sheet
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3AMS 261 - Applied Calculus III
Spring 2013, Sample Midterm 1
brie y
2i
j + k
i
2j + k
False. The vectors do not dier by a scalar multiple.
v
v
u = v
w
u = w
False For example, if
,
and
. It is true that
v = 1, 1, 1
u = 1, 0, 1
w = 2, 1, 2
v
u = v
w
but
.
u = w
2
2
2
2
f (x, y) = x
+y
g(x, y, z) = x
+y
z =
0
True, because
.
g(x, y, z) = f (x, y)
z = 0
z = f (x, y)
True,
. At constant speed
so
.
a(t) = a
T (t) + a
N (t)
a
T (t) = 0
a(t) = a
N (t)
T
N
T
N
f (x, y)
f
(x, y)
f
(x, y)
f
x
y
True, Let
and
, where
and
are constants. Then,
f
(x, y) = a
f
(x, y) = b
a
b
f
(x, y)dx =
x
y
x
and
, so
. Thus, f is linear.
ax + g(y) + c
f
(x, y)dy = by + h(x) + c
f (x, y) = ax + by + C
1
y
2
Note: You do not need to give the reasoning in
the test for multiple choice questions.
(2, 8, 5)
( 4, 2, 3)
y 8
x 2
z 5
=
=
4
2
7
y 5
x 2
z 8
=
=
4
2
7
y+8
x 2
z 5
=
=
3
4
4
y+8
x+2
z 5
=
=
3
3
4
y 8
x 2
z 5
=
=
3
3
4
Answer (e), The vector component of the line connecting
and
is
(2, 8, 5)
( 4, 2, 3)
. The line through the 2 points in parametric form is
v = 2
( 4), 8
2, 5
( 3) = 6, 6, 8
,
,
. Solve for
and set the resulting equations equal to each other.
x = 2 + 6t
y = 8 + 6t
z = 5 + 8t
t
So
, which can be simpli ed to (e) after multiplying everything by 2.
y 8
x 2
x 5
=
=
6
6
8
z =
6x
xz
x
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