Math 1920
Workshop 1: Work and Flux Solutions
Problem 1: Work
a)
We know that
~u
·
~v
=
k
kk
k
cos
θ
. Hence if the angle
θ
is acute the cosine will be
positive and so the dot product would be positive too, because magnitudes are always
positive. In reverse if the angle is obtuse then the cosine and thus the dot product
would be negative. So we should look at the angles for each segment.
From A to D, the path vector points straight down, while the force points upward
and to the right; the angle between them is obtuse, so the dot product (work done) is
negative. From D to C, the path and the force are perpendicular, so their dot product,
and therefore the work done, is zero. Thus the work done along this path from A to C
is negative.
b)
~
F
is not conservative. Consider if we traveled from A to C via point B instead– the
work done along this path is positive (why?), so the work cannot be path–independent.
Problem 2: Flux through a Pipe
a)
Assuming that the ﬂuid velocity is the same over the entire crosssection, the volumetric
ﬂow rate is then 5 m/s
×
3 m
2
= 15 m
3
/s.
Why? Consider a column of ﬂuid of area
A
= 3 m
2
and width
v
= 5 m, as shown in
the part (a) of ﬁgure below. This is the amount of liquid that will ﬂow through
A
in
1 second. The volume of this liquid is
vA
= 15 m
3
, and the ﬂow rate is
vA
m
3
/s.
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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 Spring '06
 PANTANO
 Math, Multivariable Calculus, Dot Product, Flux Solutions

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