2020-21 2nd MTH004 Final Exercise Solutions
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2020-21 2nd MTH004 Final Exercise Solutions
I. Choose the correct answer.
1. Which one of the following series converges? [ C ]
(A) n#1(n) . (B) n . (C) n sin ╱ 、. (D) n .
2. The series (_1)n伞1 A (A is a positive constant) [ B ] n#1 ^n^n + 1
(A) converges absolutely. (B) converges conditionally.
(C) diverges. (D) converges or diverges depending on the value of A. ~
3. Suppose that the series an converges. Which one of the following series must converge? [ D ]
n#1
~ ~ ~ ~
4. Given power series n2 zn , its convergence set is [ D ]
(A) ╱ _ ← 、. (B) ← _ ← 、. (C) ╱ _ ← ]. (D) ← _ ← ].
~
5. Suppose that the power series an (z _ 3)n converges conditionally at z = _1. Which one of the
n#1
following statements is always TRUE? [ C ]
~ ~
(A) an 4n is absolutely convergent. (B) an 4n is conditionally convergent.
n#1 n#1
~ ~
(C) an 5n is divergent. (D) an 5n is convergent.
n#1 n#1
~ n
6. The sum of the series 2n is [ C ] n#1
(A) 8. (B) 4. (C) 2. (D) 1.
7. Suppose that both o- and u- are nonzero vectors. If o-!u- = |o- × u-|, then the angle between o- and u- is [ A ]
π π 5π 7π
8. Suppose that o- and u- are nonzero vectors and o- is not parallel to u-. Which one of the following statements is NOT TRUE? [ D ]
(A) (o- _ u-) . (o- × u-) = 0 .
(C) (o- × u-) × (u- × o-) = .
(B) (o- × u-) . (u- × o-) ← 0 .
(D) (o- _ u-) × (o- × u-) = .
z + 1 y + 1 3 + 1
(A) The Line L lies in the plane Π –
(B) L is parallel to Π, but does not lie in Π –
(C) L is perpendicular to Π –
(D) L and Π are intersecting but not perpendicular.
10. Find the area of the part of the plane 2z + 3y + 3 = 6 that lies in the first octant? [ A ]
(A) 3^14. (B)6^14. (C) 6. (D) 9.
11. Which one of the following is the equation of the plane that contains the line z = 3u← y = 1 + u← 3 = 2u and parallel to the intersection of the planes y + 3 = _1 and 2z _ y + 3 = 0? [ A ]
(A) 3z _ 5y _ 23 + 5 = 0.
(C) 3z _ 5y _ 23 _ 5 = 0.
(B) 3z + 5y _ 23 _ 5 = 0.
(D) 3z + 5y _ 23 + 5 = 0.
z2 y2
(A) (B) (C) (D)
13. Which of the following limits does not exist? [ D ]
cos(zy) _ 1 ^1 + zy _ 1
^z2 + y2 z2 + y2 .
14. Let f be a differentiable function with f (1← 1) = 2 and fα (1 ← 1) = 6← fg (1 ← 1) = 4. Use the linear approximation of f (z← y) to approximate f (0 – 98 ← 1 – 01). [ B ]
(A) 1.84. (B) 1.92. (C) 2.08. (D) 2.16.
zy
(A) z _ 4y + 3 = 0. (B) z _ 4y _ 3 + 8 = 0.
(C) z + 4y _ 3 _ 8 = 0. (D) z + 4y + 3 _ 16 = 0.
16. If f (z← y) = z3 _ 4z2 + 2zy _ y2 , then which one of the following statements is TRUE? [ C ] (A) f (2← 2) is a local minimum value.
(B) f (0← 0) is a local minimum value.
(C) The point (2 ← 2) is a stationary point but f (2← 2) is not a local extreme value.
(D) The point (0 ← 0) is a stationary point but f (0← 0) is not a local extreme value.
17. If R = {(z← y) : 1 ≤ z ≤ 2 ← 0 ≤ y ≤ 2}, then (z + y)dǎ= [ B ]
R
(A) 4. (B) 5. (C) 6. (D) 8.
3 g
18. (y2 _ z2 )dzdy= [ A ] 1 ó
(A) . (B) _ . (C) 10. (D) _10.
1 ^g
19. Change the order of integration of the iterated integral I = [ f (z← y)dz]dy . [ D ]
1 ^α 1 1 ó ó
(A) I = [ f (z← y)dy]dz. (B) I = [ f (z← y)dy]dz.
ó ó ó ^α
1 α2 1 1
(C) I = [ f (z← y)dy]dz. (D) I = [ f (z← y)dy]dz. ó ó ó α2
20. Find the surface area of the part of the paraboloid 3 = z2 + y2 that is below the plane 3 = 1.
[ C ]
(5^5 _ 1)π (5^5 _ 1)π
(A) 3π . (B) 6π . (C) 6 . (D) 12 .
II. Calculations and comprehensive problems.
21.
(1) Let f (z← y) = ln ^z2 + y2 . Find fα (z← y), fg (z← y) and fαg (1 ← 1).
(2) Let 3 = 3 (z← y) be an implicit function defined by the equation 32 y _ z33 _ 1 = 0. Find |)ó ,1 ,1〉 and |)ó ,1 ,1〉.
Solution.
(1) fα (z← y) = z2 + y2 ← fg (z← y) = z2 + y2 ← fαg (z← y) = _ (z2 + y2 )2 fαg (1 ← 1) = _ 2 .
(2) Let F (z← y← 3) = 32 y _ z33 _ 1← Fα = _33 ← Fg = 32 ← F之 = 23y _ 332 z.
= _F(F)α之 = _ = ← |)ó ,1 ,1〉 = –
= _F(F)g之 = _ = _ ← |)ó ,1 ,1〉 = _ –
22. The temperature at (z← y← 3) of a solid sphere centered at the origin is T (z← y← 3) = e_ )α2 伞g2 伞之2〉.
(1) Find the directional derivative of T at (1 ← 1 ← _1) in the direction of u- = (1← 2 ← 2) .
(2) In which direction at (1 ← 1 ← _1), the temperature increases most rapidly? Find the maximum rate of change.
Solution.
(1) Tα = e_ )α2 伞g2 伞之2〉(_2z)← Tg = e_ )α2 伞g2 伞之2〉(_2y)← T之 = e_ )α2 伞g2 伞之2〉(_23) 9T (1← 1 ← _1) = e_3 (_2 ← _2 ← 2) ← o- = = ( ← ← )
DU→T (1← 1 ← _1) = 9T (1← 1 ← _1) ! o- = _
(2) In 9T (1← 1 ← _1) = e_3 (_2 ← _2 ← 2), the temperature increases most rapidly.
The maximum rate of change is |9T (1← 1 ← _1)| = 2e_3 ^3.
23. Find the global maximum value and the global minimum value of f (z← y) = 2z2 + y2 + zy on the closed bounded set s = {(z← y) : z2 + y2 ≤ 1}.
Solution. Method I
From , we get the stationary point (0 ← 0).
On the boundary of s, let z = cos u← y = sin u← u e [0 ← 2π]. Then
F (u) := f (z(u)← y(u)) = 2 cos2 u + sin2 u + cos u sin u = cos 2u + sin 2u + –
Solving F\ (u) = _ sin 2u + cos 2u = 0 < tan 2u = 1 < 2u = 4 ← 2u = 4 ← 2u = 4 or 2u = 4 , we get
π 5π 9π 13π
Comparing with the values of function f (z← y) = 2z2 + y2 + zy on the following points
π π 9π 9π 3 +^2
8 8 8 8 2 ←
5π 5π 13π 13π 3 _ ^2
f (z( 8 )← y( 8 )) = f (z( 8 )← y( 8 )) = 2 ← f (z(0)← y(0)) = f (z(2π)← y(2π)) = 2←
3 +^2
2
Method II
From , we get the stationary point (0 ← 0).
On the boundary of s: (z← y)← z2 + y2 = 1.
Let F (z← y← A) = 2z2 + y2 + zy + A(z2 + y2 _ 1). Then
,.Fα = 4z + y + 2Az = 0 (1)
.
.
.
.‘Fλ = z2 + y2 _ 1 = 0 (3)
(1)y _ (2)z, we have 4zy + y2 _ (2zy + z2 ) = 0 ÷ (z + y)2 _ 2z2 = 0, then
,.z + y = ^2z
.
. or
.
,.y = (^2 _ 1)z
.
÷ .or
.
Substituting y = (^2 _ 1)z into (3), we have
z2 + (^2 _ 1)2 z2 = 1 ÷ z2 = ÷ z = ±
which corresponds to
(z1 ← y1 ) = ← 、← (z2 ← y2 ) = ╱ _ ← _ ← 、←
Substituting y = _(^2 + 1)z into (3), we have
z2 + (^2 + 1)2 z2 = 1 ÷ z2 = ÷ z = ±
which corresponds to
(z3 ← y3 ) = ← 、← (z4 ← y4 ) = ╱ _ ← ← 、←
Comparing the following function values
3 +^2 3 _ ^2
f (0← 0) = 0← f (z1 ← y1 ) = f (z2 ← y2 ) = 2 ← f (z3 ← y3 ) = f (z4 ← y4 ) = 2 ←
3 +^2
2
24. Find the volume of the solid in the first octant that is inside the sphere z2 + y2 + 32 = 1 and outside the cylinder z2 + y2 = z.
Solution.
The trace of the sphere on zy _plane is
that is, The trace of the cylinder on zy _plane is , that is,.
3 = ^1 _ z2 _ y2 , we have
1
4
.
0
s = {(z← y) : 0 ≤ z ≤ 1← 0 ≤ y ≤ (1/4 _ (z _ )2 } = {(t← θ) : 0 ≤ t ≤ cos θ← 0 ≤ θ ≤ π/2} –
V =
π
=
6
π
=
6
2022-07-21