# SM221 Final Exam – May 7, 2012 – Answer Key 1.

```SM221 Final Exam – May 7, 2012 – Answer Key
Part I
1.
c
6.
c
11.
d
16.
a
2.
a
7.
b
12.
a
17.
d
3.
b
8.
c
13.
b
18.
d
4.
e
9.
c
14.
b
19.
b
5.
b
10.
a
15.
a
20.
b
SM221 Final Exam – May 7, 2012 – Answer Key
Part II
21. See Theorem 15, Section 14.6, p. 939 and Theorem 3, Section 16.5, p. 1092 of text.
Part III
22. π (π₯, π¦) = π₯ 3 − 3π₯π¦ + π¦ 3. ππ₯ = 3π₯ 2 − 3π¦, ππ¦ = −3π₯ + 3π¦ 2
Setting ππ₯ = 0 and ππ¦ = 0 gives two critical points, (0,0) and (1,1).
2
The discriminant is π· = ππ₯π₯ ππ¦π¦ − ππ₯π¦
= (6π₯)(6π¦) − (−3)2 .
At (0,0), we have π· = −9 &lt; 0 so (0,0) is a saddle point of π.
At (1,1), we have π· = 27 &gt; 0 and ππ₯π₯ = 6 &gt; 0 so π has a local minimum at (1,1).
23.
οΏ½
2π
0
2
οΏ½ οΏ½
0
5
1+π 2
π ππ§ ππ ππ = 8π
24. a. π (π₯, π¦) = 3π₯π¦ + 2π¦ 2 .
b. π (0,2) − π(1,0) = 8 − 0 = 8
c. π₯ = 1 − π‘, π¦ = 2π‘, 0 ≤ π‘ ≤ 1, ππ₯ = −ππ‘, ππ¦ = 2ππ‘
οΏ½ π­ β ππ = οΏ½ 3π¦ ππ₯ + (3π₯ + 4π¦)ππ¦
πΆ
1
πΆ
1
= ∫0 3(2π‘)(−1) + οΏ½3(1 − π‘) + 4(2π‘)οΏ½2 ππ‘ = ∫0 6 + 4π‘ ππ‘ = 8
25.
π(π₯, π¦, π§) =
30
1+π₯ 2 +π¦ 2 +π§ 2
, ππ₯ =
−60π₯
(1+π₯ 2 +π¦ 2 +π§2 )2
, with similar formulas for ππ¦ and ππ§ .
The rate of change of temperature at the point (1,1,1) in the direction of ⟨2, −2,1⟩ is
−60
1
5
1
⟨1,1,1⟩ β ⟨2, −2,1⟩ = −
∇π(1,1,1) β ⟨2, −2,1⟩ =
16
3
4
3
i.e., the temperature is decreasing at 1.25o C per meter.
26. a.
b. ∫πΆ 2π¦ ππ₯ − 2π₯ ππ¦
2π
5
=β¬interior of πΆ −4 ππ΄ = ∫0 ∫0 −4π ππ ππ =
−4(Area of circle of radius 5)= -100π.
c. No, π­ is not conservative because the
integral of π­ around the closed curve πΆ is not
zero.
27. The mass of the wire in grams is
π
π
π = ∫0 0.1√1 + π‘ 2 οΏ½(−2 sin(π‘))2 + (2 cos(π‘))2 + 12 ππ‘ = ∫0 0.1√1 + π‘ 2 √5 ππ‘ ≅ 1.366.
28. The mass of the solid in grams is
2π
π
3
π = ∫0 ∫02 ∫0 2π π2 sin(π) ππ ππ ππ = 81π.
29. Let π be the given surface. A parametrization of π is π(π₯, π¦) = ⟨π₯, π¦, 3 − π₯ − π¦⟩, for (π₯, π¦) in
the region π· given by 0 ≤ π₯ ≤ 3 and 0 ≤ π¦ ≤ 3 − π₯.
ππ₯ = ⟨1,0, −1⟩, ππ¦ = ⟨0,1, −1⟩, ππ₯ &times; ππ¦ = ⟨1,1,1⟩ (which is upward).
3
3−π₯
β¬π π­ β ππΊ = ∫0 ∫0
3
3−π₯
⟨π¦, 2π₯, 3 − π₯ − π¦⟩ β ⟨1,1,1⟩ ππ¦ ππ₯ = ∫0 ∫0
3 + π₯ ππ¦ ππ₯ = 18.
30. Let π· be the region enclosed by π. By the Divergence Theorem, the charge π is
2
2
2
π0 β­π· div π¬ ππ = π0 ∫0 ∫0 ∫0 2 + 1 + 3 ππ§ ππ¦ ππ₯ = 48 π0 .
```