**Unformatted text preview: **Math 241: Exam 2, March 13, 2018
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• ADA: 8am, Nam • [email protected]: 1pm, Zhang, N. • ADB: 9am, Block Gorman • BDA: 8am, Huynh • ADC: 10am, Block Gorman • BDB: 9am, Huynh • ADD: 11am, Shin • BDC: 10am, Park • ADE: Noon, Shin • BDD: 11am, Han • ADF: 1pm, Mousley • BDE: Noon, Park • ADG: 2pm, Okano • BDF: 1pm, Han • ADH: 3pm, Wojtalewicz • BDG: 2pm, Drake • ADI: 4pm, Wojtalewicz • BDH: 3pm, Zhang, Y. • ADK: 9am, Christenson • BDI: 4pm, Zhang, Y. • ADL: 10am, Field • BDJ: 9am, Field • ADM: 2pm, Gao • BDK: 10am, Christenson • ADN: 3pm, Gao • BDL: noon, Huang • ADO: noon, Bavisetty • BDM: 2pm, Mousley • ADP: 1pm, Bavisetty • BDN: 3pm, Okano • AD1: 11am, Weigandt • BDO: 4pm, Drake • AD2: 1pm, Rennie • BDR: 11am, Huang 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 Instructions: You have 75 minutes to complete this exam. There are 70 points
available and not all problems are weighted equally. Calculators, books, notes,
and suchlike aids are not permitted. It is not necessary to show work for multiplechoice questions. For all other questions, show work that justifies your answer
as in those problems credit will not be given for correct answers without proper
justification. Work written outside of the space provided for a problem will not be
graded. The last page of the exam contains a table of trigonometric identities. Do not open exam until instructed.
Question: 1 2 3 4 5 6 7 8 9 Total
Points:
9 7 7 9 7 8 7 8 8 70
Score: 1. (a) (3 points) Find the tangent plane to the surface
z = 2(x − 1)2 + 3(y − 3)2 + 2
at the point (2, 4, 7).
Solution: z − 7 = 4(x − 2) + 6(y − 4). An equation of this plane is (circle one):
(A) z − 7 = 4(x − 2) + 6(y − 4); (B ) z − 7 = 2(x − 2) + 3(y − 4); (C ) z − 7 = 4(x − 1) + 6(y − 3); (D) z − 7 = 2(x − 1) + 3(y − 3); (b) (3 points) Use the linear approximation to f (x, y) = 2xe sin(x y) at the point (2, 0) to
find the approximate value of f (2.1, 0.1).
Solution: L(x, y) = 4 + 2(x − 2) + 8(y − 0), so f (2.1, .1) ≈ L(2.1, .1) = 5.0 (circle one): (A) 5.2;
(c) (3 points) Find
Solution: ∂y
∂z ∂y
∂z (B) 5.0; (C) 4.4; (D) 4.8; at the point (1, 1, 1) on the level surface 3x y 3 + z y − 4xz = 0. 4x−y = 9x y 2 +z , at (1, 1, 1) this is 3
.
10 (circle one): (A) 3
;
10 (B) 12 ; (C) −10
;
3 (D) −3
10 ; 2. (a) (4 points) Use the Chain Rule to find ∂w
∂x at the point (x, y, t ) = (2, 1, π), where w = 3r 2 + 3θ 2 , r = y + x cos t , θ = x + y sin t
Solution: ∂w
∂x = 6r cos t + 6θ and at the point (x, y, t ) = (2, 1, π) we have (circle one): (A) −6; (B) 18; (C) 6; ∂w
∂x = 18. (D) 0; (b) (3 points) The tangent plane to the ellipsoid, 2x 2 + 4y 2 + 3z 2 = 6, is parallel to the
plane, 4x + 4y − 6z = 9, at which of the following points?
Solution: The normal vector for the tangent plane to this ellipsoid at (x, y, z) is 〈4x, 8y, 6z〉,
which is parallel to the normal vector of the plane 4x + 4y − 6z = 9 at (1, 12 , −1). (circle one): (A) (−1, 12 , 1); (B) (1, −1
, 1);
2 (C) (1, 12 , 1); (D) (1, 21 , −1); 3. (a) (3 points) Find the directional derivative of f (x, y, z) = 9x y +3z 2 at the point (1, −2, 2)
in the direction of the vector from the point (1, −2, 2) to the origin.
Solution: The vector from that point to the origin in ~
v = 〈−1, 2, −2〉, the unit vector
−1 2 −2
in that direction is ~
u = 〈 3 , 3 , 3 〉. D ~u f (1, −2, 2) = ~
u · ∇ f (1, −2, 2) = 4. (circle one): (A) 43 ; (B) 4; (C) −4; (D) −4
;
3 (b) (4 points) Find the unit vector that minimizes the directional derivative D ~u f (x, y, z)
at the point (1, 1, 3) where
f (x, y, z) = x y z + e 3−y z + x 2 .
Solution: This minimum occurs in the direction of −∇ f (1, 1, 3) = 〈−5, 0, 0〉. The unit
vector in this direction is 〈−1, 0, 0〉. (circle one):
(A) 〈−1, 0, 0〉; (B ) 〈 p−5
, p−6
, p−2
〉;
65
65
65 (C ) 〈1, 0, 0〉; (D) 〈 p565 , p665 , p265 〉; 4. Let f (x, y) = x 3 + 2y 3 − 3x 2 − 3y 2 − 9x.
(a) (3 points) How many critical points does f (x, y) have?
Solution: Solving the system f x (x, y) = 3x 2 − 6x − 9 = 0 and f y (x, y) = 6y 2 − 6y = 0,
gives critical points of (3, 0), (3, 1), (−1, 0), and (−1, 1). So f (x, y) has 4 critical points. (circle one): (A) 1; (B) 2; (C) 3; (D) 4; (b) (2 points) At how many of these critical points does f (x, y) have a local minimum?
Solution: f xx = 6x − 6, f y y = 12y − 6, f x y = f y x = 0, and D = (6x − 6)(12y − 6). Using
the 2nd Derivative test gives that f (x, y) has one local minimum at (3, 1). (circle one): (A) 0; (B) 1; (C) 2; (D) 3; (c) (2 points) At how many of these critical points does f (x, y) have a local maximum?
Solution: f xx = 6x − 6, f y y = 12y − 6, f x y = f y x = 0, and D = (6x − 6)(12y − 6). Using
the 2nd Derivative test gives that f (x, y) has one local maximum at (−1, 0). (circle one): (A) 0; (B) 1; (C) 2; (D) 3; (d) (2 points) At how many of these critical points does f (x, y) have a saddle point?
Solution: f xx = 6x − 6, f y y = 12y − 6, f x y = f y x = 0, and D = (6x − 6)(12y − 6). Using
the 2nd Derivative test gives that f (x, y) has two saddle points at (3, 0) and (−1, 1). (circle one): (A) 0; (B) 1; (C) 2; (D) 3; 5. Let f (x, y) = x 2 + y 2 and D = { (x, y) | x 2 + y 2 + 6x − 27 ≤ 0 }.
(a) (4 points) What is the absolute maximum value of f (x, y) on the boundary of D?
Solution: Using Lagrange Multipliers with g (x, y) = x 2 + y 2 +6x −27 = 0, we solve the
system:
2x = λ2(x + 3)
2y = λ2y
2 2 x + y + 6x − 27 = 0
The solutions to this system, (λ, x, y), are ( 12 , 3, 0) and ( 23 , −9, 0). Since f (3, 0) = 9 and
f (−9, 0) = 81. The answer is 81. (circle one): (A) 25; (B) 49; (C) 121; (D) 81; (b) (3 points) What is the absolute minimum value of f (x, y) on D?
Solution: Using calculations from part (a) and noting the only critical point of f (x, y)
is (0, 0) along with the Extreme Value Theorem gives the answer of 0. (circle one): (A) 16; (B) 4; (C) 9; (D) 0; 6. (8 points) Find the point on the ellipsoid x 2 + 2y 2 + 3z 2 = 54 where the function
f (x, y, z) = 2x + 4y + 6z is minimized. Solution: Letting g (x, y, z) = x 2 + 2y 2 + 3z 2 , we use Lagrange multipliers to find the absolute maximum and absolute minimum of f (x, y, z) subject to g (x, y, z) = 54. The Lagrange
system to solve is:
2 = λ2x
4 = λ4y
6 = λ6z
x 2 + 2y 2 + 3z 2 = 54
The first equation gives us that λ 6= 0, so x = y = z = λ1 , plugging these into the fourth
equation gives λ = 31 or λ = −1
3 , which gives solution points of (3, 3, 3) and (−3, −3, −3),
respectively. f (3, 3, 3) = 36 and f (−3, −3, −3) = −36. So the answer is: (−3, −3, −3). The point is: ( , , ) 7. (a) (4 points) Let C be the curve of intersection of x 2 + y 2 = 1 and z = −x 2 . Find the
tangent line ~
l (t ) to C at the point (1, 0, −1).
Solution: C can be parametrized by ~
r (t ) = 〈cos t , sin t , − cos2 t 〉. The point (1, 0, −1)
0
corresponds to ~
r (t ) at t = 0. ~
r (t ) = 〈− sin t , cos t , 2 cos t sin t 〉. The tangent line of
interest is then: ~
l (t ) = 〈1, t , −1〉. (circle one):
(A) ~
l (t ) = 〈1 + t , t , −1 − t 〉; (B ) ~
l (t ) = 〈1 + 2t , t , −1 − 2t 〉; (C ) ~
l (t ) = 〈1, t , −1〉; (D) ~
l (t ) = 〈1 + t , t , −1〉; (b) (3 points) Find the length of the curve of ~
r (t ) = 〈e t , e t sin t , e t cos t 〉, for 0 ≤ t ≤ ln 2.
Solution: This length is given by R ln 2
0 (circle one): (A) p
p
e t 3 dt = 3 p
2; p
(B) 2 3; (C) p
3; p
(D) 2 2; 8. (a) (4 points) Find
Solution: R C 5x 2 y ds, where C is given by ~
r (t ) = 〈cos t , sin t , t 〉 for 0 ≤ t ≤ π. 2
C 5x y ds = R Rπ
0 p
p
5 cos2 (t ) sin(t ) 2 dt = 103 2 (circle one): (A) 0;
(b) (4 points) Find
(2, 3, 1). R C (B) p
5 2
;
3 (C) p
−5 2
;
3 (D) p
10 2
;
3 z 2 dx + x 2 dy + 2y 2 dz, where C is the line segment from (0, 0, 0) to Solution:
segment canR be parametrized by ~
r (t ) = 〈2t , 3t , t 〉 for 0 ≤ t ≤ 1,
R 2This line
1
2
2
2
2
2
.
giving C z dx + x dy + 2y dz = 0 2t + 12t + 18t dt = 32
3 (circle one): (A) 23
;
3 (B) 32
;
3 (C) 11; (D) 22
;
3 R
~ · d~
~ (x, y) = 〈x − y, x + y〉 and C is the path given by the
9. (8 points) Evaluate C F
r , where F
2
2
ellipse 4x + y = 4 transversed once and oriented clockwise.
Solution:
We can parameterized the ellipse by:
~
r (t ) = 〈cos(−t ), 2 sin(−t )〉 = 〈cos t , −2 sin t 〉, t ∈ [0, 2π] (Note the choice of −t to have a clockwise path.) This path has derivative:
~
r 0 (t ) = 〈− sin t , −2 cos t 〉.
Therefore, using the definition of the integral, we find:
Z
C ~ · d~
F
r= Z
0 Z
= 2π 0 Z
= 2π 0 Z
= 2π 0 2π ~ (~
F
r (t )) ·~
r 0 (t ) dt
〈cos t + 2 sin t , cos t − 2 sin t 〉 · 〈− sin t , −2 cos t 〉 dt
3 cos t sin t − 2(sin2 t + cos2 t ) dt
3 cos t sin t − 2 dt = −4π. R
C ~ · d~
F
r= T RIGONOMETRIC I DENTITIES
sin (θ + φ) = sin θ cos φ + cos θ sin φ
cos (θ + φ) = cos θ cos φ − sin θ sin φ
tan (θ + φ) = tan(θ) + tan(φ)
1 − tan θ tan φ sin (θ − φ) = sin θ cos φ − cos θ sin φ
cos (θ − φ) = cos θ cos φ + sin θ sin φ
tan(θ) − tan(φ)
1 + tan θ tan φ
cos(θ − φ) − cos(θ + φ)
sin(θ) sin(φ) =
2
cos(θ − φ) + cos(θ + φ)
cos(θ) cos(φ) =
2
sin(φ + θ) − sin(φ − θ)
sin(θ) cos(φ) =
2
sin (2θ) = 2 sin θ cos θ
tan (θ − φ) = cos (2θ) = cos2 θ − sin2 θ
2 tan θ
1 − tan2 θ
1 − cos (2θ)
sin2 θ =
2
1
+
cos
(2θ)
cos2 θ =
2
sin (2θ)
tan θ =
1 + cos (2θ) tan (2θ) = ...

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