Quiz 8
Math 1321 - Accelerated Engineering Calc II
April 15, 2016
Name:
Quiz Score:
/10
Answer each question completely in the area below. Show all work and explain your reasoning.
If the work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are
allowed. Anyone found violating these rules will be asked to leave immediately. Point values
are in the square to the left of the question. If there are any other issues, please ask the
instructor.
4
1. Consider the following four vector fields:
(i) F = hx + y , 1i,
(ii) F = hy , xi,
(iii) F = hy , 1i,
(iv) F = h2/y , xi.
Match each vector field description with the appropriate plot below. Justify your answer.
-4
-4
4
4
2
2
-2
2
4
-4
-2
-2
-2
-4
-4
(a)
(b)
4
4
2
2
-2
2
4
-4
-2
-2
-2
-4
-4
(c)
2
4
2
4
(d)
Solution: There isn’t a definite way to do this, but I’ll explain some of the basic logic that I
would use. We see that for (i) and (iii), the i component is always 1, meaning that this narrows
it down to the vector fields always pointing up: (a) and (b). From these two, we see that if we
change x, (a) does not change, meaning (a) definitely goes with (iii) and (b).
To narrow down the remaining two, we see that the values of the vector field are small near the
origin, which clearly corresponds to (ii) as (iv) becomes large near the origin. Thus, (b) goes
with (ii) and (d) goes with (iv).
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Quiz 8
6
Math 1321 - Accelerated Engineering Calc II
2. Compute the line integral
Z
F · dr,
April 15, 2016
F(x, y ) = x 2 i − xy j,
where
C
and C is the quarter-circle in the top-right half-plane, oriented counter-clockwise.
Hint: parameterize this curve!
Solution: As the hint suggests, we have to parameterize the curve C. We know the parameterization of a FULL circle is:
r(t) = cos(t)i + sin(t)j,
0 ≤ t ≤ 2π,
however, we only have a quarter circle. Thus, we take the modification
r(t) = cos(t)i + sin(t)j,
0 ≤ t ≤ π/2.
We can compute the derivative of this
r0 (t) = − sin(t)i + cos(t)j,
and we know the relationship between the infinitesimals is
dr = r0 dt = [− sin(t)i + cos(t)j]dt,
so our integral becomes
Z
Z
Z π
2
0
F(r) · r dt =
F · dr =
C
0
π
2
hcos2 (t), − cos(t) sin(t)i · h− sin(t), cos(t)i dt,
0
where, the dot product evaluates to
hcos2 (t), − cos(t) sin(t)i · h− sin(t), cos(t)i = −2 cos2 (t) sin(t).
so our integral becomes a u-substitution, where u = cos(t)
Z
Z π
2
2 3 t=π/2
2
F · dr = −2
cos2 (t) sin(t) dt =
cos (t) t=0 = − .
3
3
C
0
Note it makes perfect sense we get a negative answer when we notice the line integral goes
against the “circulation” of the vector field:
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
2/2