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 areallowed. Anyone found violating these rules will be asked to leave immediately. Point valuesare in the square to the left of the question. If there are any other issues, please ask theinstructor.

1.4 Consider the following four vector fields:

(i) F = 〈x + y , 1〉, (ii) F = 〈y , x〉, (iii) F = 〈y , 1〉, (iv) F = 〈2/y, x〉.Match each vector field description with the appropriate plot below. Justify your answer.

-4 -2 2 4

-4

-2

2

4

(a)

-4 -2 2 4

-4

-2

2

4

(b)

-4 -2 2 4

-4

-2

2

4

(c)

-4 -2 2 4

-4

-2

2

4

(d)

Solution: There isn’t a definite way to do this, but I’ll explain some of the basic logic that Iwould use. We see that for (i) and (iii), the i component is always 1, meaning that this narrowsit down to the vector fields always pointing up: (a) and (b). From these two, we see that if wechange 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 theorigin, which clearly corresponds to (ii) as (iv) becomes large near the origin. Thus, (b) goeswith (ii) and (d) goes with (iv).

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Quiz 8 Math 1321 - Accelerated Engineering Calc II April 15, 2016

2.6 Compute the line integral∫C

F · dr, where F(x, y) = x2i− xy j,

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 parameter-ization 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

r′(t) = − sin(t)i+ cos(t)j,

and we know the relationship between the infinitesimals is

dr = r′ dt = [− sin(t)i+ cos(t)j]dt,

so our integral becomes∫C

F · dr =∫ π

2

0

F(r) · r′ dt =∫ π

2

0

〈cos2(t),− cos(t) sin(t)〉 · 〈− sin(t), cos(t)〉 dt,

where, the dot product evaluates to

〈cos2(t),− cos(t) sin(t)〉 · 〈− sin(t), cos(t)〉 = −2 cos2(t) sin(t).

so our integral becomes a u-substitution, where u = cos(t)∫C

F · dr = −2∫ π

2

0

cos2(t) sin(t) dt =2

3

[cos3(t)

]t=π/2t=0

= −2

3.

Note it makes perfect sense we get a negative answer when we notice the line integral goesagainst the “circulation” of the vector field:

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

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