classroom voting questions: multivariable...

Classroom Voting Questions:Multivariable Calculus

12.1 Functions of Two Variables

1. A function f(x, y) can be an increasing function of x with y held fixed, and be adecreasing function of y with x held fixed.

(a) True, and I am very confident

(b) True, but I am not very confident

(c) False, but I am not very confident

(d) False, and I am very confident

Answer: (a). True

by Carroll College MathQuest

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2. You awaken one morning to find that you have been transferred onto a grid which isset up like a standard right-hand coordinate system. You are at the point (-1, 3,-3),standing upright, and facing the xz-plane. You walk 2 units forward, turn left, andwalk for another 2 units. What is your final position?

(a) (-1,1,-1)

(b) (-3,1,-3)

(c) (-3,5,-3)

(d) (1,1,-3)

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Answer: (d). If you are facing the xz-plane, then you are standing on the y-axis,facing in the −y direction. Two steps forward puts you at (-1,1,-3). If you turn left,you will turn on to the x-axis and move in the +x direction. Two more steps puts youat (1,1,-3). So the correct answer is (d).


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3. Starting at the origin, if you move 3 units in the positive y-direction, 4 units in thenegative x-direction, and 2 units in the positive z-direction, you are at:

(a) (3,4,2)

(b) (3,-4,2)

(c) (4,3,2)

(d) (-4,3,2)

Answer: (d). If you move 4 units in the negative x-direction from the origin, yourx-coordinate must be -4.

by Mark Schlatter

MVC.12.01.015

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4. Which of the following points lies closest to the xy-plane?

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(a) (3,0,3)

(b) (0,4,2)

(c) (2,4,1)

(d) (2,3,4)

Answer: (c). (2,4,1) is closest to the xy-plane, since the distance to the xy-plane is|z|.

ConcepTests - to accompany Calculus 4th Edition, Hughes Hallet et al. John Wiley &Sons.

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5. Which of the following points lies closest to the origin?

(a) (3,0,3)

(b) (0,4,2)

(c) (2,4,1)

(d) (2,3,4)

Answer: (a). (3,0,3) is closest to the origin, since the distance to the origin is√

x2 + y2 + z2.


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6. Which of the following points lies closest to the y-axis?

(a) (3,0,3)

(b) (0,4,2)

(c) (2,4,1)

(d) (2,3,4)

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Answer: (b). (0,4,2) is closest to the y-axis, since the distance to the y-axis is√

x2 + z2.


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7. The point (2,1,3) is closest to:

(a) the xy plane

(b) the xz plane

(c) the yz plane

(d) the plane z=6

Answer: (b). The point is a distance of 2 from the yz plane, a distance of 1 from thexz plane, and a distance of 3 from both the xy plane and the plane z = 6.

by Mark Schlatter

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8. Which of the following points lies closest to the point (1,2,3)?

(a) (3,0,3)

(b) (0,4,2)

(c) (2,4,1)

(d) (2,3,4)

Answer: (d). (2,3,4) is closest to the point (1,2,3) since the distance to (1,2,3) is√

(x− 1)2 + (y − 2)2 + (z − 3)2.


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9. Sphere A is centered at the origin and the point (0,0,3) lies on it. Sphere B is givenby the equation x2 + y2 + z2 = 3. Which of the following is true?

(a) A encloses B

(b) A and B are equal

(c) B encloses A

(d) none of the above

Answer: (a). Both spheres are centered at the origin. Sphere A has radius 3, sphereB has radius

√3. Thus A encloses B.


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10. The points (1,0,1) and (0,-1,1) are the same distance from the origin.





Answer: (a). True

Carroll College

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11. The point (2, -1, 3) lies on the graph of the sphere (x− 2)2 + (y + 1)2 + (z− 3)2 = 25 .





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Answer: (False).


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12. In a table of values for a linear function, the columns must have the same slope as therows.





Answer: (False).


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13. The set of all points whose distance from the z-axis is 4 is the:

(a) sphere of radius 4 centered on the z-axis

(b) line parallel to the z-axis 4 units away from the origin

(c) cylinder of radius 4 centered on the z-axis

(d) plane z = 4

Answer: (c). A point (x, y, z) is a distance of 4 away from the z-axis if x2 + y2 = 42.Since this equation creates a circle of radius 4 for every plane parallel to the xy plane,the result is a cylinder.

by Mark Schlatter

MVC.12.01.100

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12.2 Graphs of Functions of Two Variables

14. What does a graph of the function f(x, y) = x look like?

(a) A line in the xy plane

(b) A line in three dimensions

(c) A horizontal plane

(d) A tilted plane

Answer: (d). For any (x, y) input, the output is equal to the x coordinate, and so thez coordinate is always equal to the x coordinate. This gives us a tilted plane, whichcrosses the xy plane on the y axis and has a slope of positive one in the x direction.


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15. Let h(x, t) = 3 + 3 sin(

π10

x)

cos (2πt) be the distance above the ground (in feet) of ajump rope x feet from one and after t seconds. The two people turning the rope stand10 feet apars. Then h(x, 1/4) is

(a) Concave up

(b) Concave down

(c) Flat

(d) Changes concavity in the middle

Answer: (c). At t = 1/4 the cosine is zero. Therefore, h(x, 1/4) = 3 is constant, so itsgraph is a straight line.


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16. The object in 3-space described by x = 2 is

(a) A point

(b) A line

(c) A plane

(d) Undefined

Answer: (c). a plane. While x is fixed at 2, y and z can vary freely.


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17. The set of points whose distance from the z-axis equals the distance from the xy-planedescribes a

(a) Plane

(b) Cylinder

(c) Sphere

(d) Cone

(e) Double cone (two cones joined at their vertices)

Answer: (e). Double cone. Fix a value for z, for example z = a, and draw the set ofpoints in the plane z = a that are equidistant from the z-axis and the xy-plane. It isthe set of points at a distance |a| from the z-axis, namely a circle of radius |a| in theplane z = a. Putting all these circles together, we get two cones, with vertex at theorigin, one above the xy-plane and one below it.


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18. The graph of f(x, y) = 2−x2−y2

will look most like

(a) a bowl opening up, but more shallow than x2 + y2

(b) a bowl opening up, but more steep than x2 + y2

(c) a bowl opening down

(d) a small hill in a large plane

Answer: (d). As a point in the xy plane moves away from the origin, the value of−x2− y2 decreases. Thus f(x, y) decreases as well. Since f(x, y) is always positive, itsgraph cannot look like a bowl opening down.

by Mark Schlatter

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CC KC MA233 F06: 18/18/32/32CC MP MA233 F06: 12/6/6/71CC KC MA233 F07: 14/17/30/39 time 4:00CC MP MA233 F07: 14/26/15/36 time 4:00CC KC MA233A F08: 8/0/58/33 time 3:30CC KC MA233B F08: 0/0/0/100 time 4:00CC KC MA233 F09: 28/0/28/44 time 5:00CC MP MA233 F09: 0/17/22/61 time 5:00CC HZ MA233 F10: 4/16/20/60 time 6:00CC HZ MA233 F11: 8/4/36/52 time 6:00CC TM MA233 F11: 0/45/18/36 time 5:15

19. The cross sections of g(x, y) = sin(x) + y + 1 with x fixed are

(a) lines

(b) parabolas

(c) sinusoidal curves


Answer: (a). When x is fixed (say with value c), the cross section has the equationg(c, y) = sin(c) + y + 1, which is a line with slope 1 and vertical intercept sin(c) + 1.

by Mark Schlatter

MVC.12.02.050

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20. The graph of the equation f(x, y) = 2 is a plane parallel to the xz-plane.





Answer: (False). This is a plane parallel to the xy-plane.


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21. The cross section of the function f(x, y) = x + y2 for y = 1 is a line.





Answer: (a). True

Carroll College

MVC.12.02.070

22. The graphs of f(x, y) = x2 + y2 and g(x, y) = 1− x2 − y2 intersect in a circle.





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Answer: (a). True

Carroll College

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23. The equation Ax + By + Cz + D = 0 represents a line in space.





Answer: (False). This represents a plane.

Carroll College

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24. In three space, x2 + y2 = 1 represents

(a) a circle

(b) a cylinder

(c) a sphere

Answer: (b). a cylinder, because any value of z will work.

Carroll College

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12.3 Contour Diagrams

25. Which of the following terms best describes the origin in the contour diagram in thefigure below?

(a) A mountain pass

(b) A dry river bed

(c) A mountain top

(d) A ridge

Answer: (b). A reasonable case can be made for (a) as well.


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26. Using the contour plot pictured, which path will result in the greatest change in alti-tude?

(a) From A to B

(b) From C to B

(c) From D to B

(d) All changes in altitude are approximately equal.

Answer: (d). All paths are crossing the same number of contour lines.

by Mark Schlatter

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27. Using the contour plot pictured, which path is the steepest?

(a) From A to B

(b) From C to B

(c) From D to B

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(d) All changes in altitude are approximately equal.

Answer: (b). The path from C to B is the one where the contour line are most closelyspaced.

by Mark Schlatter

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28. The contour lines of z = e(sin(x2+y2)) will be

(a) lines

(b) circles

(c) exponential curves


Answer: (b). When z is fixed, so is exp (sin (x2 + y2)), and thus so is x2 + y2.

by Mark Schlatter

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29. The contours of graph of f(x, y) = y2 + (x− 2)2 are either circles or a single point.



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Answer: (a). True


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30. If all the contours for f(x, y) are parallel lines, then the graph of f is a plane.





Answer: (False). The contours must not only be parallel lines, they must also beevenly spaced and either increasing or decreasing in order for the graph of f to be aplane.


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31. On a weather map, there can be two isotherms (contour lines) which represent thesame temperature but do not intersect.





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Answer: (a). True.


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12.4 Linear Functions

32. A plane has a z-intercept of 3, a slope of 2 in the x direction, and a slope of -4 in they direction. The height of the plane at (2,3) is

(a) -2

(b) -8

(c) -5

(d) not given by this information

Answer: (c). At (0,0), the plane is at height 3. When you move to (2,0), the plane isat height 3+2*2=7. When you move to (2,3), the plane is at height 7+3*(-4)=-5.

by Mark Schlatter

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33. Which of the following planes is parallel to the plane z = −2− 2x− 4y?

(a) z = −1− 2x− 2y

(b) (z − 1) = −2− 2(x− 1)− 4(y − 1)

(c) z = 2 + 2x + 4y

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Answer: (b).


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34. Any three points in 3 space determine a unique plane.





Answer: (False). If the three points lie on a line, there are many planes containingthem. A correct statement is that any three noncolinear points determine a uniqueplane.


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35. Any two distinct lines in 3-space determine a unique plane.





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Answer: (False). If two lines don’t intersect and are not parallel, then there is no planethat contains them both.


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36. If the graph of z = f(x, y) is a plane, then each cross section is a line.





Answer: (a). True


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12.5 Functions of Three Variables

37. Level surfaces of the function f(x, y, z) =√

x2 + y2 are

(a) Circles centered at the origin

(b) Spheres centered at the origin

(c) Cylinders centered around the z-axis

(d) Upper-hemispheres centered at the origin

Answer: (c).


MVC.12.05.010


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38. Any level surface of a function of 3 variables can be thought of as a surface in 3 space.





Answer: (False). A level surface may be a point or a line, not a surface.


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39. Any surface that is a graph of a 2-variable function z = f(x, y) can be thought of as alevel surface of a function of 3 variables.





Answer: (a). True


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40. Any level surface of a function of 3 variables can be thought of as the graph of afunction z = f(x, y).



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Answer: (False). Not every level surface is the graph of a function of 2 variables. Forexample the level surfaces might be spheres.


MVC.12.05.040

41. Suppose the temperature at time t at the point (x, y) is given by the function T (x, y, t) =5t− x2 − y2. Which of the following will not cause temperature to decrease?

(a) moving away from the origin in the positive x direction

(b) moving away from the origin in the positive y direction

(c) moving away from the origin in the direction of the line y = x

(d) standing still and letting time pass

Answer: (d). If x and y move away from the origin in any fashion, −x2− y2 decreases.

by Mark Schlatter

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CC KC MA233 F07: 0/0/22/78 time 2:00

42. If temperature at the point (x, y, z) is given by T (x, y, z) = cos(z − x2 − y2), the levelsurfaces look like:

(a) spheres

(b) Pringles brand potato chips

(c) planes

(d) bowls

Answer: (d). When the temperature is fixed, so is z − x2 − y2. Thus z = x2 + y2 + kfor some constant, and the level surface is a bowl.

by Mark Schlatter

MVC.12.05.060


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43. The level surfaces of the function f(x, y, z) = x2 +y2 + z2 are cylinders with axis alongthe y-axis.





Answer: (False). The level surfaces of this function are spheres.


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13.1 Displacement Vectors

44. The length of the sum of two vectors is always strictly larger than the sum of thelengths of the two vectors





Answer: (False). The geometry of vector addition shows that the sum of the lengthsis always at least as large as the lenth of the sum.


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45. ||~v|| = |v1|+ |v2|+ |v3|, where ~v = v1~i + v2

~j + v3~k.





Answer: (False). The length of a vector is not usually the sum of the magnitudes ofits components.


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46. ~v and ~w are parallel if ~v = λ~w for some scalar λ





Answer: (a). True


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47. Any two parallel vectors point in the same direction.




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Answer: (False). They could point in opposite directions.


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48. Any two points determine a unique displacement vector.





Answer: (False). The points must be distinct. If we have the same point twice, wedon’t have a unique displacement vector.


MVC.13.01.050

49. 2~v has twice the magnitude as ~v





Answer: (a). True


MVC.13.01.060

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50. In the picture, the unlabelled vector is closest to

(a) v + w

(b) v − w

(c) v + 2w

(d) 2v + w

Answer: (d). The vector looks the diagonal of a parallelogram with one side of w andanother of 2v.

by Mark Schlatter

MVC.13.01.070

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51. A “unit vector” is a vector with a magnitude of one. The vectors i = 〈1, 0, 0〉, j =〈0, 1, 0〉 and k = 〈0, 0, 1〉 are unit vectors that point in the x, y, and z directions,respectively.

True or False: The vector 〈 12, 1

2〉 is a unit vector.




24


Answer: (False). The magnitude of this vector is 1√2.

by Carroll College

MVC.13.01.075

CC KC MA233A F08: 8/92 time 1:40CC KC MA233B F08: 0/100 time 1:40CC KC MA233 F09: 15/85 time 2:20

52. True or False: The vector 1√3i− 1√

3j + 2√

3k is a unit vector.





Answer: (False). The magnitude of this vector is√

6/3.

Carroll College

MVC.13.01.080

CC KC MA233 F06: 40/60CC KC MA233 F06: 22/78HC AS MA203 F10: 10/10/20/60

53. Which of the following is a unit vector that is parallel to the vector 〈1,−2, 3〉?

(a) 〈16,−2

6, 3

6〉

(b) 〈 1√14

,− 2√14

, 3√14〉

(c) 〈 1

14,− 2

14, 3

14〉

(d) 〈− 1√14

, 2√14

,− 3√14〉

(e) More than one of the above

Answer: (e). Both (b) and (d) are unit vectors that are parallel to the given vector.


MVC.13.01.085

54. The vectors 2i− j + k and i− 2j + k are parallel.

25





Answer: (False). You can’t multiply one vector by a scalar and get the other vector.

Carroll College

MVC.13.01.090

CC KC MA233 F07: 13/87 time 2:00

55. Find a vector that points in the same direction as the vector 〈2, 1, 2〉, but has a mag-nitude of 5.

(a) 〈103, 5

3, 10

3〉

(b) 〈 10√3, 5√

3, 10√

3〉

(c) 〈 5√3, 5√

3, 5√

3〉

(d) 〈10, 5, 10〉(e) 〈30, 15, 30〉(f) More than one of the above

Answer: (a). First we divide this vector by its magnitude to produce a unit vector inthis direction, 〈2

3, 1

3, 2

3〉, and then we multiply it by 5.


MVC.13.01.100

CC KC MA233A F08: 73/8/12/4/4 time 4:00CC KC MA233B F08: 100/0/0/0/0 time 3:15CC KC MA233 F09: 60/15/0/15/5/5 time 4:30

13.2 Vectors in General

56. A plane is flying due south. There is a strong wind from the west. In what directiondoes the pilot have to point the plane to stay on course?

(a) South

(b) East

26

(c) West

(d) Southeast

(e) Southwest

Answer: (e). southwest


MVC.13.02.010

CC KC MA233 F06: 0/5/14/9/73CC KC MA233 F07: 0/0/4/21/75 time 1:30CC MP MA233 F07: 0/0/0/27/73 time 1:30CC KC MA233A F08: 0/0/0/4/96 time 1:15CC KC MA233B F08: 0/0/6/13/81 time 1:00CC MP MA233 F09: 0/0/0/10/90 time 3:00

57. A boat is traveling with a velocity of 30 mph due West relative to the water. Thecurrent is flowing 10 mph at an angle of 45◦ West of North. What is the boat’s netvelocity?

(a) 37.7 mph at 10.8◦ South of West

(b) 37.7 mph at 79.2◦ West of North

(c) 24.0 mph at 17.1◦ North of West

(d) 24.0 mph at 17.1◦ West of North

(e) None of the above

Answer: (b). We put the vectors into components and add: <-30 mph, 0 mph> +<-7.07 mph, 7.07 mph> = <-37.07 mph, 7.07 mph> = 37.7 mph at 10.8 degrees Northof West = 37.7 mph at 79.2 degrees West of North.

by Project MathVote

MVC.13.02.015

CC KC MA233 F09: 10/70/0/10/10 time 4:30

58. A car is traveling along the path from point A towards point D. When is the velocityvector closest to being parallel to j (assuming this path is in the xy plane)?

27

(a) A

(b) B

(c) C

(d) D

Answer: (c). At C, the velocity vector will be close to pointing straight down.

by Mark Schlatter

MVC.13.02.020

CC KC MA233 F06: 0/0/100/0CC KC MA233 F07: 13/0/79/8 time 2:00CC MP MA233 F07: 0/0/93/7 time 2:00CC MP MA233 F09: 0/0/90/10 time 3:00

13.3 The Dot Product

59. The only way that ~v · ~w = 0 is if ~v = 0 or ~w = 0.





28

Answer: (False). Try 〈1, 0〉 and 〈0, 1〉.


MVC.13.03.010

CC KC MA233 F07: 17/83 time 1:20CC MP MA233 F07: 14/86 time 2:00CC KC MA233 F09: 19/81 time 2:00CC MP MA233 F09: 14/86 time 1:00CC HZ MA233 F10: 5/0/19/76 ReviewHC AS MA203 F10: 15/5/10/70CC HZ MA233 F11: 4/4/4/87 time 1:15 ReviewCC TM MA233 F11: 15/0/42/37 time 2:00CC HZ MA233 F12: 0/0/16/84 Review

60. The zero vector ~0 (with magnitude ||~0|| = 0) is perpendicular to all other vectors.





Answer: (a). True: We define two vectors to be perpendicular if their dot product iszero. The zero vector dotted with anything is zero!


MVC.13.03.020

CC MP MA233 F06: 47/53CC MP MA233 F07: 36/64 time 2:00CC KC MA233 F07: 46/54 time 2:00

61. Any plane has only two distinct normal vectors





29

Answer: (False). There are an infinite number of vectors normal to any plane.


MVC.13.03.030

CC MP MA233 F06: 47/53CC MP MA233 F07: 33/67 time 2:00

62. Parallel planes share a same normal vector.





Answer: (a). True: A vector normal to one plane will be normal to any other parallelplane.


MVC.13.03.040

CC MP MA233 F06: 40/60

63. Perpendicular planes have perpendicular normal vectors.





Answer: (a). True


MVC.13.03.050

CC MP MA233 F06: 100/0

64. What is the angle between the vectors 〈√

3, 1〉 and 〈−√

3, 1〉?

30

(a) 30 degrees

(b) 60 degrees

(c) 90 degrees

(d) 120 degrees

(e) 150 degrees

(f) None of the above

Answer: (d). θ = cos−1(~x · ~y/|~x||~y|) = cos−1(−1

2) = 120◦.


MVC.13.03.055

CC KC MA233A F08: 0/0/4/83/0/13 time 3:15CC KC MA233B F08: 0/27/0/67/7/0 time 4:30CC KC MA233 F09: 0/19/5/57/0/19 time 5:00CC HZ MA233 F10: 5/10/0/57/0/24 time 3:45 ReviewCC HZ MA233 F11: 4/32/23/41/0/0 time 4:20 ReviewCC TM MA233 F11: 0/10/25/65/0/0CC HZ MA233 F12: 4/0/26/65/0/4 Review

65. The angle between the vectors −xi− j + k and xi + 2j − 3k:

(a) is 0 degrees

(b) is less than 90 degrees

(c) is greater than 90 degrees

(d) can be any of the above depending on the value of x.

Answer: (c). Since the dot product of the two vectors is −x2 − 5 < 0, the cosine ofthe angle is negative, and the angle between the two vectors must be more than 90degrees.

by Mark Schlatter

MVC.13.03.060

CC KC MA233 F07: 4/17/30/35 time 5:30CC KC MA233A F08: 4/0/92/4 time 4:30CC KC MA233B F08: 0/7/80/13 time 5:00CC KC MA233 F09: 5/29/43/24 time 5:00CC MP MA233 F09: 0/10/75/15

31

66. Two vectors have a dot product of 14. To guarantee the dot product is equal to 28,you could:

(a) double the angle between the vectors

(b) double the length of both vectors

(c) double the length of one vector


Answer: (c). Since the dot product is equal to the product of the magnitudes of thevector and the cosine of the angle between them, you only need to double the lengthof one variable.

by Mark Schlatter

MVC.13.03.070

CC MP MA233 F07: 0/25/75/0 time 2:00 good discussion, brought out misunder-standings of the definition of dot productCC KC MA233 F07: 13/0/87/0 time 2:30HC AS MA203 F10: 0/0/90/10

67. Which of the following is a point in the plane parallel to 3x + 4y − 2z = 6 containingthe origin?

(a) (1,1,1)

(b) (1,2,3)

(c) (3,2,1)


Answer: (d). A normal vector to the plane is 3i + 4j − 2k. Since the dot product ofall of the position vectors with this normal is not equal to zero, none of these points ison the plane.

by Mark Schlatter

MVC.13.03.080

CC KC MA233A F08: 4/13/4/78 time 6:00CC KC MA233B F08: 14/7/7/64 time 7:00CC KC MA233 F09: 24/6/12/58 time 6:00

68. In 2 space, consider the vector ~v = 5i + 7j. For which unit vector below will thecomponent of ~v perpendicular to that unit vector be largest?

32

(a) i

(b) (i− j)/√

2

(c) j

(d) (i + j)/√

2

Answer: (b). We want the unit vector which is most perpendicular to ~v.

by Mark Schlatter

MVC.13.03.090


69. A 100-meter dash is run on a track heading northeasterly. If the wind is blowing outof the south at a speed of 8 km/hr. The rules say that a legal wind speed measured inthe direction of the dash must not exceed 5 km/hr. Will the race results be disqualifieddue to an illegal wind?

(a) Yes, the race results will be disqualified because the wind exceeds 5 km/hr in thedirection of the race.

(b) No, the race results will not be disqualified because the wind does not exceed 5km/hr in the direction of the race.

(c) There is not enough information to answer this question.

Answer: (a). This is the first of 3 questions to lead students towards discovering vectorprojections. Students don’t need to use a dot product here - simple trig will do. Themagnitude in the direction of the race is simply the wind speed times the cosine of theangle between the race and the wind: (8)(cos π/4) = 4

√2 > 5.


MVC.13.03.092

CC KC MA233 F07: 43/52/5 time 5:00CC MP MA233 F09: 100/0/0 time 9:00 typo on B

70. A 100-meter dash is run on a track in the direction of the unit vector ~v = 1√2i + 1√

2j.

If the wind velocity ~w is ~w = 5i + 1j km/hr. The rules say that a legal wind speedmeasured in the direction of the dash must not exceed 5 km/hr. Will the race resultsbe disqualified due to an illegal wind?

33




Answer: (b). This is the second of 3 questions to lead students towards discoveringvector projections. ~v · ~w = 3

√2 < 5, so the race is legal.


MVC.13.03.095

CC KC MA233 F07: 17/83/0 time 5:30

71. A 100-meter dash is run on a track in the direction of the vector ~v = 2i+6j. The windvelocity is ~w = 5i + j km/hr. The rules say that a legal wind speed measured in thedirection of the dash must not exceed 5 km/hr. Will the race results be disqualifieddue to an illegal wind?




Answer: (b). This is the last of 3 questions to lead students towards discovering vectorprojections. The projection of ~w parallel to ~v is given by ~v·~w

|~v| = 4√

10/5 < 5.


MVC.13.03.097

CC KC MA233 F07: 13/87/0 time 5:00CC KC MA233A F08: 21/79/0 time 5:00CC KC MA233B F08: 25/75/0 time 6:00CC KC MA233 F09: 16/74/10 time 5:00CC HZ MA233 F10: 11/83/0 time 7:30 ReviewCC HZ MA233 F11: 9/91/0 time 9:30 ReviewCC TM MA233 F11: 37/63/0 time 9:00CC HZ MA233 F12: 5/91/5 Review

34

72. The picture shown is in 2 space. If the force vector is ~F = −4j, the total work towhich point will be the most positive?

(a) A

(b) B

(c) C

(d) D

Answer: (d). We want the dot product of the position vector with −4j to be the mostpositive.

by Mark Schlatter

MVC.13.03.100

CC MP MA233 F07: 0/0/0/100 time 2:00CC KC MA233 F07: 5/5/23/68 time 2:00CC KC MA233 F09: 10/10/16/64 time 3:00

73. An equation of the plane with normal vector i + j + k containing the point (1, 2, 3) isz = x + y .





35

Answer: (False). This plane does contain the point, but it has normal vector i+ j− k.

Carroll College

MVC.13.03.110

CC KC MA233 F07: 67/33 time 2:30CC HZ MA233 F12: 17/8/42/33

74. Which of the following vectors is normal to the plane z = −3x + 4y + 25?

(a) 〈3,−4, 1〉(b) 〈−3, 4, 1〉(c) 〈−3, 4, 25〉(d) 〈−3, 4,−1〉(e) More than one of the above

Answer: (e). Both (a) and (d) are normal to the given plane.


MVC.13.03.115

CC KC MA233A F08: 8/0/8/46/38 time 3:00CC KC MA233B F08: 31/0/0/50/19 time 4:00CC HZ MA233 F10: 17/0/28/33/22 time 5:00 ReviewCC HZ MA233 F11: 52/9/5/0/33 time 6:00 ReviewCC TM MA233 F11: 50/5/10/5/30 time 4:15

75. For any two vectors ~u and ~v, ~u · ~v = ~v · ~u.





Answer: (a). True.


MVC.13.03.120

CC KC MA233 F06: 100/0

36

76. For any two vectors ~u and ~v and any scalar k, it is true that k(~u · ~v) = (k~u) · ~v.





Answer: (a). True


MVC.13.03.130

CC KC MA233 F06: 68/32

13.4 The Cross Product

77. The cross product of 2i and 3j is

(a) 6k

(b) −6k

(c) 0

(d) 6ij

Answer: (a). The area of the parallelogram spanned by 2i and 3j is 6. The unit vectornormal to both i and j chosen by the right hand rule is k.

by Mark Schlatter

MVC.13.04.005

CC KC MA233 F06: 59/5/32/5CC KC MA233 F07: 96/4/0/0 time 2:30CC KC MA233 F09: 50/30/15/5 time 2:00

78. For the vectors ~a = 4i− j + 2k and ~b = −i + 5j + 3k, the cross product ~a×~b is

(a) −13i + 14j + 19k

(b) 13i + 14j − 19k

(c) −13i− 14j + 19k

(d) 13i− 14j − 19k

37

Answer: (c).


MVC.13.04.010

CC KC MA233 F07: 0/0/100/0 time 3:00CC KC MA233 F09: 5/5/85/5 time 2:30

79. A vector that is normal to the plane containing the vectors ~a = 4i − j + 2k and~b = −i + 5j + 3k is

(a) −13i + 14j + 19k

(b) 13i + 14j − 19k

(c) −13i− 14j + 19k

(d) 13i− 14j − 19k

Answer: (c). The result vector is always perpendicular to both vectors in the crossproduct.


MVC.13.04.020BJ BB MA305 F12: 7.69/23.08/57.69/11.54

80. If ~d = ~a×~b, then ~a · ~d =

(a) ~a× (~b ·~b)(b) 0

(c) ~a× ~a ·~b(d) (~a ·~b)×~b

Answer: (b). ~d is perpendicular to both ~a and ~b


MVC.13.04.030

CC KC MA233 F06: 0/100/0/0CC KC MA233 F07: 0/100/0/0 time 1:00CC KC MA233A F08: 0/79/13/8 time 3:15

38

CC KC MA233B F08: 0/81/13/6 time 3:30CC KC MA233 F09: 5/42/53/0 time 2:00BJ BB MA305 F12: 0/60/24/16

81. For any vectors ~u and ~v, ~u× ~v = ~v × ~u





Answer: (False).


MVC.13.04.040


82. For any vectors ~u and ~v, (~u× ~v)× (~v × ~u) = (~v × ~u)× (~u× ~v)





Answer: (a). True. The vectors (~u × ~v) and (~v × ~u) point in opposite directions, sothey have an angle of π between them, and so when we cross them, we get zero. Thismeans that both sides of this equation must be equal to zero.


MVC.13.04.050


39

14.1 The Partial Derivative

83. At point Q in the diagram below, which of the following is true?

(a) fx > 0, fy > 0

(b) fx > 0, fy < 0

(c) fx < 0, fy > 0

(d) fx < 0, fy < 0

Answer: (d). since both derivatives are negative.


MVC.14.01.010

CC KC MA233 F06: 0/32/23/46CC MP MA233 F06: 31/23/30/15CC MP MA233 F07: 0/31/31/37 time 3:00CC KC MA233 F07: 5/5/24/67 time 3:30CC KC MA233A F08: 0/4/46/50 time 3:00CC KC MA233B F08: 0/19/50/31 time 3:00CC KC MA233 F09: 5/37/32/26 time 2:00CC MP MA233 F09: 0/32/32/36 time 3:00CC HZ MA233 F10: 17/35/22/26 time 3:20CC MP MA233 F10: 0/12/47/41 time 5:00HC AS MA203 F10: 0/44/11/44CC HZ MA233 F11: 0/0/40/60 time 3:45CC TM MA233 F11: 0/10/90/0 time 2:15CC HZ MA233 F12: 4/25/21/50

40

BJ BB MA305 F12: 30.77/26.92/23.08/19.23

84. List the points P, Q, R in order of decreasing fx.

(a) P > Q > R

(b) P > R > Q

(c) R > P > Q

(d) R > Q > P

(e) Q > R > P

Answer: (b).


MVC.14.01.020

CC KC MA233 F06: 0/41/18/18/24 (typo in question when asked)CC MP MA233 F07: 0/75/6/0/19 time 1:30CC KC MA233A F08: 4/96/0/0 time 2:00CC KC MA233B F08 0/87/13/0/0 time 2:45CC KC MA233 F09: 10/90/0/0/0 time 1:30CC MP MA233 F09: 9/63/19/9/0 time 5:00CC HZ MA233 F10: 4/87/9/0/0 time 3:00CC HZ MA233 F11: 0/87/0/0/13 time 3:15CC TM MA233 F11: 15/65/20/0/0CC HZ MA233 F12: 8/80/4/0/8BJ BB MA305 F12: 19.23/53.85/15.38/0/11.54

41

85. Using the level curves of f(x, y) given in the figure below, which is larger, fx(2, 1) orfy(1, 2).

(a) fx(2, 1) > fy(1, 2).

(b) fx(2, 1) < fy(1, 2).

Answer: (a). Both derivatives are negative, but the contours in the x direction at (2,1)are more widely spaced than the contours in the y direction at (1,2). This means thatfy(1, 2) has a steeper negative slope, so fx(1, 2) > fx(2, 1).


MVC.14.01.030

86. Suppose that the price P (in dollars), to purchase a used car is a function of C, itsoriginal cost (also in dollars), and its age A (in years). So P = f(C, A). The sign of∂P∂C

is

(a) Positive

(b) Negative

(c) Zero

Answer: (a). Positive. If the age of a car is held constant, then as the original costincreases, so does the current purchase price. A great follow up question is to ask themwhat the contour diagram would look like for this function.


MVC.14.01.040

42

CC KC MA233 F06: 82/18CC MP MA233 F06: 80/20CC MP MA233 F07: 56/44 time 2:00CC KC MA233 F07: 86/14 time 3:00CC KC MA233A F08: 75/25 time 2:30CC KC MA233B F08: 44/56 time 2:30CC KC MA233 F09: 24/76 time 2:30CC MP MA233 F09: 91/9 time 2:00CC HZ MA233 F10: 52/39 time 3:15CC MP MA233 F10: 72/28 time 3:00CC HZ MA233 F11: 63/37 time 3:00CC TM MA233 F11: 33/67 time 4:00CC HZ MA233 F12: 48/52Option (c) Zero addedBJ BB MA305 F12: 48/40/12

87. Using the contour plot of f(x, y), which of the following is true at the point (4,2)?

(a) fx > 0 and fy > 0

(b) fx > 0 and fy < 0

(c) fx < 0 and fy > 0

(d) fx < 0 and fy < 0

43

Answer: (b). As you move in the positive x direction, the function is increasing. Asyou move in the positive y direction, the function is decreasing.

by Mark Schlatter

MVC.14.01.050

CC MP MA233 F07: 6/94/0/0 time 1:00CC HZ MA233 F10: 9/91/0/0 time 2:00 ReviewCC HZ MA233 F11: 7/89/0/4 time 2:30 ReviewCC TM MA233 F11: 9/86/5/0 time 2:30CC HZ MA233 F12: 0/96/4/0 ReviewBJ BB MA305 F12: 23.08/76.92/0/0

88. Using the contour plot of f(x, y), which of the following is closest to the partial deriva-tive of f with respect to x at (4,2)?

(a) 40

(b) 20

(c) 10

(d) 4

Answer: (c). As you move from x = 4 to x = 6 from the point (4, 2), the functionincreases by about 20 units, producing a slope of 10.

by Mark Schlatter

44

MVC.14.01.060

CC KC MA233 F06: 0/9/91/0CC MP MA233 F06: 7/36/57/0CC KC MA233 F07: 0/38/57/5 time 2:30CC MP MA233 F07: 6/13/69/13 time 1:00CC HZ MA233 F10: 4/22/70/4 time 2:45CC HZ MA233 F11: 0/16/80/4 time 4:00CC HZ MA233 F12: 0/25/67/8BJ BB MA305 F12: 0/50/46.15/3.85

89. At which point above the xy plane will both partial derivatives be positive?

(a) (-5,-5)

(b) (5,-5)

(c) (5,5)

(d) (-5,5)

Answer: (a). The point (-5,-5) is the only point where the surface is moving upwardin both the positive x and positive y directions.

by Mark Schlatter

MVC.14.01.070

45

CC KC MA233 F06: 100/0/0/0BJ BB MA305 F12: 79.17/0/12.5/8.33

14.2 Computing Partial Derivatives Algebraically

90. Which of the following functions satisfy Euler’s Equation, xfx + yfy = f?

(a) f = x2y3

(b) f = x + y + 1

(c) f = x2 + y2

(d) f = x0.4y0.6

Answer: (d).


MVC.14.02.010

CC KC MA233 F06: 0/23/36/41CC MP MA233 F06: 0/11/67/22CC KC MA233 F07: 0/29/29/43 time 3:30CC MP MA233 F07: 8/46/39/8 time 2:00 start class, no introCC KC MA233A F08: 0/58/17/25 time 5:00CC KC MA233B F08: 0/56/38/6 time 4:00CC KC MA233 F09: 0/76/24/0 time 4:30

91. If ∂f∂x

= ∂f∂y

everywhere, then f(x, y) is constant.





Answer: (False). As a counterexample consider f(x, y) = 4x + 4y.


MVC.14.02.020

46

CC KC MA233 F06: 18/77CC MP MA233 F06: 44/56CC HZ MA233 F11: 32/12/24/32 time 2:30BJ BB MA305 F12: 20/16/20/44

92. There exists a function f(x, y) with fx = 2y and fy = 2x.





Answer: (a). True. The function f(x, y) = 2xy works.


MVC.14.02.030

CC KC MA233 F06: 68/32CC MP MA233 F06: 82/18CC KC MA233 F07: 95/5 time 2:00CC KC MA233A F08: 48/52 time 2:00CC KC MA233 F09: 57/43 time 2:30CC HZ MA233 F10: 70/15/5/10CC MP MA233 F10: 73/0/13/13 ReviewCC HZ MA233 F11: 84/8/0/8 time 2:30CC TM MA233 F11: 25/40/20/15 time 3:00CC HZ MA233 F12: 67/13/8/4

14.3 Local Linearity and the Differential

93. Let f(2, 3) = 7, fx(2, 3) = −1, and fy(2, 3) = 4. Then the tangent plane to the surfacez = f(x, y) at the point (2, 3) is

(a) z = 7− x + 4y

(b) x− 4y + z + 3 = 0

(c) −x + 4y + z = 7

(d) −x + 4y + z + 3 = 0

(e) z = 17 + x− 4y

47

Answer: (b).


MVC.14.03.010

CC KC MA233 F06: 24/43/24/10CC KC MA233 F07: 18/82/0/0 time 3:30CC KC MA233A F08: 13/48/0/39/0 time 4:30CC KC MA233B F08: 69/13/13/6/0 time 3:30CC KC MA233 F09: 60/20/20/0/0 time 3:00CC MP MA233 F09: 9/73/0/14/4 time 5:00CC HZ MA233 F10: 14/45/18/9/14 time 5:00CC MP MA233 F10: 6/81/12/0/0 time 5:00CC HZ MA233 F11: 12/44/36/8/0 time 4:30CC TM MA233 F11: 64/0/13/5/13 time 4:00CC HZ MA233 F12: 18/55/9/9/0BJ BB MA305 F12: 16/12/0/0/72

94. The figure below shows level curves of the function f(x, y). The tangent plane approx-imation to f(x, y) at the point P = (x0, y0) is f(x, y) ≈ c + m(x − x0) + n(y − y0).What are the signs of c, m, and n?

(a) c > 0, m > 0, n > 0

(b) c < 0, m > 0, n < 0

(c) c > 0, m < 0, n > 0

(d) c < 0, m < 0, n < 0

(e) c > 0, m > 0, n < 0

Answer: (e). The slope in the x direction m is positive and the slope in the y directionn is negative. Note that in this case c is not the z-axis intercept: It is the value of thefunction at point P , which is positive.


MVC.14.03.020

48

CC KC MA233 F06: 10/52/5/5/29CC MP MA233 F06: 36/21/7/7/29CC KC MA233 F07: 0/86/0/0/14 time 5:00CC KC MA233A F08: 0/9/0/0/91 time 4:00CC KC MA233B F08: 0/31/13/6/50 time 3:40CC HZ MA233 F10: 9/23/4/0/64 time 4:15CC MP MA233 F10: 13/7/13/0/66 time 3:00 ReviewCC HZ MA233 F11: 12/0/4/8/76 time 4:45CC TM MA233 F11: 18/27/13/14/27 time 6:45CC HZ MA233 F12: 4/21/0/0/75BJ BB MA305 F12: 4/4/68/24/0

95. Suppose that f(x, y) = 2x2y. What is the tangent plane to this function at x = 2,y = 3?

(a) z = 4xy(x− 2) + 2x2(y − 3) + 24

(b) z = 4x(x− 2) + 2(y − 3) + 24

(c) z = 8(x− 2) + 2(y − 3) + 24

(d) z = 24(x− 2) + 8(y − 3) + 24

(e) z = 24x + 8y + 24

Answer: (d).


MVC.14.03.025

CC KC MA233 F09: 25/20/15/30/10 time 5:30CC HZ MA233 F10: 29/0/0/67/5 time 6:30CC MP MA233 F10: 0/7/0/93/0 time 8:00HC AS MA203 F10: 20/0/0/80/0CC HZ MA233 F11: 20/4/0/76/0 time 5:00CC HZ MA233 F12: 35/9/0/43/13

96. The differential of a function f(x, y) at the point (a, b) is given by the formula df =fx(a, b)dx + fy(a, b)dy. Doubling dx and dy doubles df .





49

Answer: (a). True


MVC.14.03.030

CC MP MA233 F06: 79/21

97. The differential of a function f(x, y) at the point (a, b) is given by the formula df =fx(a, b)dx + fy(a, b)dy. Moving to a different point (a, b) may change the formula fordf .





Answer: (a). True, the formula is always df = fx(a, b)dx + fy(a, b)dy, however we canalso argue that the statement is false because if we move to another point (c, d), wewill have different derivatives fx(c, d) and fy(c, d).


MVC.14.03.040

CC KC MA233 F06: 13/87

98. The differential of a function f(x, y) at the point (a, b) is given by the formula df =fx(a, b)dx+fy(a, b)dy. If dx and dy represent small changes in x and y in moving awayfrom the point (a, b), then df approximates the change in f .





Answer: (a). True


MVC.14.03.050

50

99. The differential of a function f(x, y) at the point (a, b) is given by the formula df =fx(a, b)dx + fy(a, b)dy. The equation of the tangent plane to z = f(x, y) at the point(a, b) can be used to calculate values of df from dx and dy.





Answer: (a). True


MVC.14.03.060

CC MP MA233 F06: 93/7

100. A small business has $300,000 worth of equipment and 100 workers. The total monthlyproduction, P (in thousands of dollars), is a function of the total value of the equipment,V (in thousands of dollars), and the total number of workers, N. The differential of Pis given by dP = 4.9dN + 0.5dV . If the business decides to lay off 3 workers and buyadditional equipment worth $20,000, then

(a) Monthly production increases.

(b) Monthly production decreases.

(c) Monthly production stays the same.

Answer: (b). since dN = -3 and dV = 20, dP = 4.9(-3) + 0.5(20) = -4.7. Thus, wesee that production, P, decreases (because dP is negative) by $4700.


MVC.14.03.070

CC KC MA233 F06: 86/14/0CC KC MA233 F07: 18/82/0 time 4:00CC KC MA233A F08: 61/35/4 time 4:00CC KC MA233B F08: 87/13/0 time 5:00CC KC MA233 F09: 75/25/0 time 4:00CC MP MA233 F09: 57/38/5 time 5:00CC HZ MA233 F10: 76/24/0 time 4:00CC HZ MA233 F11: 67/29/4 time 4:30CC TM MA233 F11: 90/10/0 time 4:20CC HZ MA233 F12: 17/83/0

51

101. Which of the following could be the equation of the tangent plane to the surfacez = x2 + y2 at a point (a, b) in the first quadrant?

(a) z = −3x + 4y + 7

(b) z = 2x− 4y + 5

(c) z = 6x + 6y − 18

(d) z = −4x− 4y + 24

Answer: (c). Since both partial derivatives will be positive at (a, b), so must thecoefficients of x and y.

by Mark Schlatter

MVC.14.03.080

CC KC MA233 F07: 14/0/87/0 time 4:00CC HZ MA233 F10: 4/22/57/17 time 4:30 ReviewCC HZ MA233 F11: 0/0/73/27 time 4:00 ReviewCC TM MA233 F11: 10/5/75/10CC HZ MA233 F12: 8/16/32/44BJ BB MA305 F12: 23.08/42.31/7.69/26.92

102. Suppose fx(3, 4) = 5, fy(3, 4) = −2, and f(3, 4) = 6. Assuming the function isdifferentiable, what is the best estimate for f(3.1, 3.9) using this information?

(a) 6.3

(b) 9

(c) 6

(d) 6.7

Answer: (d). Since we are moving .1 units in the x direction, the function increasesfrom 6 to approximately 6+.1*5 = 6.5. By similar reasoning, when we move in the ydirection, the height is approximately 6.5+(-.1)*-2 = 6.7

by Mark Schlatter

MVC.14.03.090

CC KC MA233 F06: 0/0/0/100CC MP MA233 F06: 50/0/7/43CC HZ MA233 F10: 32/0/0/68 time 4:00 ReviewCC MP MA233 F10: 36/0/7/57 time 4:00 ReviewCC HZ MA233 F11: 4/0/4/92 time 4:30CC TM MA233 F11: 4/4/9/83 time 4:15CC HZ MA233 F12: 10/10/5/76

52

103. We need to figure out the area of the floor of a large rectangular room, however ourmeasurements aren’t very precise. We find that the room is 52.3 ft by 44.1 ft so we getan area of 2,306.43 square feet, but our measurements are only good to within a coupleof inches, roughly an error of 0.2 feet in both directions, so our estimate of the area isprobably off by a bit. Use differentials to determine the likely error in our estimate ofthe floor’s area.

(a) 0.04 square feet

(b) 19.24 square feet

(c) 19.28 square feet

(d) 19.32 square feet

(e) 19.56 square feet

Answer: (c). Area is length times width so A = lw. We get the error from thedifferential dA = fldl + fwdw = wdl + ldw = 52.3 ∗ 0.2 +44.1 ∗ 0.2 = 19.28 square feet.


MVC.14.03.100

CC KC MA233A F08: 0/0/100/0/0 time 3:00CC KC MA233B F08: 0/0/100/0/0 time 2:00Note: Perhaps this would be a tougher question if the errors were different.

104. A giant stone cylinder suddenly appears on the campus lawn outside, and we of courseask ourselves: What could its volume be? We send out a student with a meter stick tomeasure the cylinder, who reports that the cylinder has a height of 3.34 meters (witha measurement error of 2 centimeters) and has a radius of 2.77 meters (give or take 3centimeters). We know that the volume of a cylinder is V = πr2h, so we estimate itsvolume as 80.5 m3. What is our measurement error on this volume?

(a) 5 cm

(b) 1.00 m3

(c) 1.89 m3

(d) 2.23 m3

Answer: (d). We get the error from the differential df = frdr + fhdh = 2πrhdr +πr2dh = 2π(2.77 m )(3.34 m )(0.03 m ) + π(2.77 m )2(0.02 m ) = 2.23 m3.


MVC.14.03.110

CC KC MA233A F08: 0/0/18/82 time 6:00CC KC MA233B F08: 0/0/8/92 time 9:00

53

14.4 Gradients and Directional Derivatives in the

Plane

105. The figure shows the temperature T ◦C in a heated room as a function of distance xin meters along a wall and time t in minutes. Which of the following is larger?

(a) ||∇T (15, 15)||(b) ||∇T (25, 25)||

Answer: (a). The contour lines are closer together at (15,15).


MVC.14.04.010

CC KC MA233 F06: 85/15CC HZ MA233 F10: 65/35 time 2:30 ReviewCC HZ MA233 F11: 96/4 time 3:00 ReviewCC TM MA233 F11: 86/13 time 5:00CC HZ MA233 F12: 68/32 Review

106. The table below gives values of the function f(x, y) which is smoothly varying aroundthe point (3, 5). Estimate the vector ∇(f(3, 5)). If the gradient vector is placed withits tail at the origin, into which quadrant does the vector point?

(a) I

54

(b) II

(c) III

(d) IV

(e) Can’t tell without more information

Answer: (d). because fx is positive and fy is negative.


MVC.14.04.020

CC KC MA233 F06: 5/10/25/60/0CC KC MA233 F06: 5/10/33/52/0 time 3:30CC KC MA233A F08: 5/0/5/90/0 time 5:00CC KC MA233B F08: 0/0/7/93/0 time 3:30CC KC MA233 F09: 5/35/20/40/0 time 3:30CC MP MA233 F09: 0/10/0/90 time 5:00CC HZ MA233 F10: 6/11/6/78/0 time 4:30CC TM MA233 F11: 23/23/7/46 time 4:30

107. Let ∇f(1, 1) = 3i − 5j . What is the sign of the directional derivative of f in thedirection of the vector ↖ and in the direction of the vector ↑?

(a) positive and positive

(b) positive and negative

(c) negative and positive

(d) negative and negative

Answer: (d). Both are negative.


MVC.14.04.030

CC KC MA233 F06: 0/18/24/59CC KC MA233 F07: 0/28/10/62 time 4:00CC MP MA233 F07: 0/7/7/83 time 2:00CC KC MA233A F08: 0/5/0/95 time 4:00CC KC MA233B F08: 0/7/27/67 time 4:00CC KC MA233 F09: 20/20/25/35 time 3:30CC MP MA233 F09: 0/35/0/65 time 3:00CC HZ MA233 F10: 10/10/15/65 time 5:00CC MP MA233 F10: 14/21/7/57

55

CC HZ MA233 F11: 0/4/8/88 time 8:00 the stats overstate their understandingCC TM MA233 F11: 6/31/6/56 time 5:30CC HZ MA233 F12: 5/32/14/50

108. Let ∇f(1, 1) = 3i − 5j . What is the sign of the directional derivative of f in thedirection of the vector ← and in the direction of the vector ↘?

(a) positive and positive

(b) positive and negative

(c) negative and positive

(d) negative and negative

Answer: (c).


MVC.14.04.040

CC HZ MA233 F10: 4/18/54/23 time 3:00 ReviewCC HZ MA233 F11: 12/0/76/12 time 5:45 ReviewCC TM MA233 F11: 10/26/42/21 time 3:30CC HZ MA233 F12: 20/8/72/0 Review

109. In which direction is the directional derivative of z = x2 + y2 at the point (2,3) mostpositive?

(a) i

(b) −i− j

(c) −i + j

(d) i + j

Answer: (d). From the point (2,3), you reach the next highest contour line mostquickly (picking from the four choices) if you move in the i+j direction.

by Mark Schlatter

MVC.14.04.045

CC MP MA233 F06: 0/0/0/100CC MP MA233 F07: 0/0/7/93 time 2:00

56

110. Which of the vectors shown on the contour diagram of f(x, y) in the figure below couldbe ∇f at the point at which the tail is attached?

(a) A

(b) B

(c) C

(d) D

Answer: (d). The vectors at A and C point in the opposite direction of grad f. Thevector at B is too short. The vector at D could be grad f at this point.


MVC.14.04.050

CC KC MA233 F06: 0/25/0/75CC KC MA233 F07: 0/38/0/62 time 3:00CC KC MA233A F08: 0/14/0/86 time 3:00CC KC MA233B F08: 0/40/0/60 time 3:30CC KC MA233 F09: 0/40/0/60 time 2:00BJ BB MA305 F12: 0/69.23/0/30.77

111. At which of the points P, Q,R, S in the figure below does the gradient have the largestmagnitude?

57

(a) P

(b) Q

(c) R

(d) S

Answer: (d). At point S the contour lines are closest together, so this is the point withthe largest gradient.


MVC.14.04.060

CC MP MA233 F06: 23/0/38/39CC MP MA233 F07: 0/0/20/80 time 2:00 good discussionCC KC MA233A F08: 0/0/0/100 time 1:30CC KC MA233B F08: 27/0/0/73 time 1:45CC KC MA233 F09: 5/0/5/90 time 1:45CC HZ MA233 F10: 10/10/10/70 time 2:15CC MP MA233 F10: 0/0/27/73 time 2:00HC AS MA203 F10: 5/0/0/95CC HZ MA233 F11: 4/12/4/80 time 3:00CC TM MA233 F11: 7/0/20/73 time 2:30CC HZ MA233 F12: 5/0/27/68

112. The surface of a hill is modeled by z = 25− 2x2− 4y2. When a hiker reaches the point(1, 1, 19), it begins to rain. She decides to descend the hill by the most rapid way.Which of the following vectors points in the direction in which she starts her descent?

(a) −4xi − 8yj

(b) 4xi + 8yj

58

(c) −4i− 8j

(d) 4i + 8j


Answer: (d). ∇f = −4xi−8yj, which evaluated at the point (1,1,19) is −4i−8j. Thisis the direction of most rapid increase, so to descent the hiker must go in the oppositedirection.


MVC.14.04.070

CC KC MA233 F07: 10/14/24/52 time 5:00CC KC MA233A F08: 0/5/33/45/19 time 4:00CC KC MA233 F09: 45/20/25/10/0 time 4:00CC HZ MA233 F10: 0/0/32/68/0 time 3:40HC AS MA203 F10: 47/11/26/16/0CC HZ MA233 F11: 28/4/56/12 time 4:00CC TM MA233 F11: 12/0/82/6 time 3:00CC HZ MA233 F12: 14/23/41/18/5

113. At which point will the gradient vector have the largest magnitude?

(a) (0,2)

(b) (-4,-4)

(c) (0,0)

59

(d) (6,-2)

Answer: (d). At the point (6,-2), the contour lines are most closely packed. A goodfollow up is to ask at what point will the gradient vector have the smallest magnitude.

by Mark Schlatter

MVC.14.04.080

CC MP MA233 F07: 7/0/0/93 time 3:00BJ BB MA305 F12: 0/3.85/7.69/88.46

114. At which point will the gradient vector be most parallel to j?

(a) (0,4)

(b) (-4,-4)

(c) (0,0)

(d) (6,-2)

Answer: (a). At the point (0,4), the contour line is close to being parallel to i. Sincethe gradient is perpendicular to the contour line at that point, the answer is a).

by Mark Schlatter

MVC.14.04.090

CC HZ MA233 F10: 76/5/9/9 time 3:00 ReviewCC HZ MA233 F11: 80/4/8/8 time 3:30 ReviewCC HZ MA233 F12: 52/4/16/28 Review

60

115. ∇f(1, 1) = 3i− 5j. What is the sign of the directional derivative of f in the direction~v = 4i + 2j?

(a) Positive

(b) Negative

Answer: (a). Positive


MVC.14.04.100

116. Suppose that the temperature at a point (x, y) on the floor of a room is given byT (x, y). Suppose heat is being radiated out from a hot spot at the origin. Which ofthe following could be ∇T (a, b), where a, b > 0?

(a) 2i + 2j

(b) −2i− 2j

(c) −2i + 2j

(d) 2i− 2j

Answer: (b). Larger values of T are closer to the origin, so both fx and fy must benegative.


MVC.14.04.110

HC AS MA203 F10: 53/35/6/6

14.5 Gradients and Directional Derivatives in Space

117. Suppose the temperature at a point (x, y, z) in a room is given by T (x, y, z). Supposeheat is being radiated out from a hot spot at the origin. Which of the following couldbe ∇T (a, b, c) where a, b, c are all positive?

(a) 2i + 2j − 4k

(b) −3i− 3j − 5k

(c) −2i + 2j + 5k

(d) 3i + 3j + 5k

61

Answer: (b). The gradient must point towards increasing temperature or, in this case,towards the origin. Since a, b, and c are all positive, grad T(a,b,c) must have negativecoordinates.

by Mark Schlatter

MVC.14.05.010

CC KC MA233 F06: 38/57/5/0CC MP MA233 F06: 0/8/0/92CC MP MA233 F07: 0/94/6/0 time 4:00CC KC MA233 F07: 0/57/0/43 time 3:30CC KC MA233A F08 17/75/0/8 time 4:00CC KC MA232B F08: 36/43/0/21 time 4:00CC KC MA233 F09: 17/50/0/33 time 3:30CC HZ MA233 F10: 14/41/0/46 time 4:00CC MP MA233 F10: 0/66/0/33 time 3:00CC HZ MA233 F11: 7/67/11/15 time 6:15CC TM MA233 F11: 10/5/0/85 time 3:00CC HZ MA233 F12: 8/48/12/32

118. Let f(x, y, z) = x2 + y2 + z2. Which statement best describes the vector ∇f(x, y, z)?It is always perpendicular to:

(a) vertical cylinder passing through (x, y, z).

(b) a horizontal plane passing through (x, y, z).

(c) a sphere centered on the origin passing through (x, y, z).

(d) None of the above

Answer: (c). The level surfaces of f(x, y, z) are spheres centered at the origin, and thegradient is always perpendicular to the level surface.

by Mark Schlatter

MVC.14.05.020

CC KC MA233 F06: 0/29/24/48CC KC MA233 F07: 19/5/71/5 time 3:00CC MP MA233 F07: 0/6/94/0 time 3:00CC KC MA233A F08: 4/12/84/0 time 2:00CC KC MA233B F08: 0/7/43/50 time 3:00CC KC MA233 F09: 11/6/78/6 time 3:00CC HZ MA233 F10: 0/32/59/9 time 4:00CC MP MA233 F10: 7/20/73/0CC HZ MA233 F11: 0/15/70/15 time 3:30CC TM MA233 F11: 6/23/65/6 time 3:45

62

CC HZ MA233 F12: 0/32/68/0

119. For f(x, y, z), suppose ∇f(a, b, c) · i > ∇f(a, b, c) · j > ∇f(a, b, c) · k > 0. The tangentplane to the surface f(x, y, z) = 0 through the point (a, b, c) is given by z = p+mx+ny.Which of the following is correct?

(a) m > n > 0

(b) n > m > 0

(c) m < n < 0

(d) n < m < 0

Answer: (c). The inequalities given tell us that fx(a, b, c) > fy(a, b, c) > fz(a, b, c) > 0.The equation of the tangent plane to the level surface can be written as fxx + fyy +

fzz + c = 0. Now if we solve for z we have z = − fx

fz

x − fy

fz

y − cfz

. Thus m = − fx

fz

<

n = − fy

fz

< 0.


MVC.14.05.030

CC KC MA233 F06: 91/9/0/0CC KC MA233 F07: 100/0/0/0CC KC MA233A F08: 80/12/8/0 time 3:30CC KC MA233B F08: 93/0/7/0 time 4:00CC KC MA233 F09: 72/0/17/11 time 4:00

120. The function f(x, y) has gradient ∇f at the point (a, b). The vector ∇f is perpendic-ular to the level curve f(x, y) = f(a, b).





Answer: (a). True


MVC.14.05.040

CC MP MA233 F06: 100/0CC HZ MA233 F10: 79/17/4/0 time 1:30

63

CC MP MA233 F10: 61/22/11/6 time 3:00CC HZ MA233 F11: 65/12/15/8 time 2:00CC TM MA233 F11: 84/16/0/0 time 2:00CC HZ MA233 F12: 68/24/8/0

121. The function f(x, y) has gradient ∇f at the point (a, b). The vector ∇f is perpendic-ular to the surface z = f(x, y) at the point (a, b, f(a, b)).





Answer: (False). The vector ∇f is a 2-vector; the vector perpendicular to the surfacehas a z-component.


MVC.14.05.050

CC MP MA233 F06: 21/79CC KC MA233A F08: 68/32 time 2:00CC HZ MA233 F10: 23/36/14/27 time 2:45CC MP MA233 F10: 53/23/12/12 time 3:00HC AS MA203 F10: 44/28/11/17CC HZ MA233 F11: 31/12/23/35 time 2:00CC TM MA233 F11: 42/32/16/10 time 2:15CC HZ MA233 F12: 44/28/8/20

122. The function f(x, y) has gradient ∇f at the point (a, b). The vector fx(a, b)i +fy(a, b)j + k is perpendicular to the surface z = f(x, y).





Answer: (False). The normal to the surface z = f(x, y) is obtained by writing it inthe form f(x, y)− z = 0 giving the normal as fxi + fy j − k.


64

MVC.14.05.060

CC KC MA233 F06: 33/67CC MP MA233 F06: 54/46CC HZ MA233 F12: 24/4/28/44

14.6 The Chain Rule

123. A company sells regular widgets for $4 apiece and premium widgets for $6 apiece. Ifthe demand for regular widgets is growing at a rate of 200 widgets per year, while thedemand for premium widgets is dropping at the rate of 80 per year, the company’srevenue from widget sales is:

(a) staying constant

(b) increasing

(c) decreasing

Answer: (b). Increasing: 4 ∗ 200− 6 ∗ 80 > 0

by Mark Schlatter

MVC.14.06.010

CC KC MA233 F06: 10/90/0CC KC MA233 F07: 9/91/0 time 3:00CC MP MA233 F07: 0/100/0 time 2:00 no introductionCC KC MA233A F08: 0/96/4 time 2:00CC KC MA233B F08: 0/100/0 time 2:30CC KC MA233 F09: 5/76/9 time 2:30CC HZ MA233 F10: 0/100/0CC HZ MA233 F11: 0/100/0 time 3:15CC TM MA233 F11: 0/100/0 time 3:15CC HZ MA233 F12: 0/100/0

124. Suppose R = R(u, v, w), u = u(x, y, z), v = v(x, y, z), w = w(x, y, z). In the chainrule, how many terms will you have to add up to find the partial derivative of R withrespect to x?

(a) 1

(b) 2

(c) 3

65

(d) 4

(e) 5

Answer: (c). 3: You need to follow the three paths from R to u to x, from R to v to x,and from R to w to x. Follow up: Ask students what these three terms in the partialderivative must be.

by Mark Schlatter

MVC.14.06.020

CC KC MA233 F06: 0/5/64/32/0CC KC MA233 F07: 0/0/57/43/0 time 3:00 (renumbered answers to avoid confusionof a = 2 etc.)CC KC MA233A F08: 0/4/96/0/0 time 3:00CC KC MA233B F08: 0/0/88/6/6 time 2:00CC KC MA233 F09: 0/0/80/15/5 time 4:00CC HZ MA233 F10: 0/4/96/0/0 time 3:00CC HZ MA233 F11: 0/0/100/0/0 time 2:15CC TM MA233 F11: 0/0/89/11/0 time 2:20CC HZ MA233 F12: 0/0/100/0/0

125. Let z = z(u, v) and u = u(x, y, t); v = v(x, y, t) and x = x(t); y = y(t). Then theexpression for dz

dthas

(a) Three terms

(b) Four terms

(c) Six terms

(d) Seven terms

(e) Nine terms


Answer: (c). Six terms. Follow up: Ask students what these six terms must be.


MVC.14.06.025

CC KC MA233 F06: 5/32/36/0/0/27CC KC MA233 F07: 0/4/87/4/0/4 time 5:00CC MP MA233 F07: 7/20/53/13/7/0 time 5:00 after an introduction and exampleCC KC MA233A F08: 0/17/46/0/4/33 time 5:00CC KC MA233B F08: 6/38/38/0/6/12 time 5:00CC KC MA233 F09: 0/30/35/15/0/20 time 5:00

66

126. The figures below show contours of z = z(x, y), x as a function of t, and y as a functionof t. Decide if dz

dt

∣

∣

t=2is

(a) Positive

(b) Negative

(c) Approximately zero

(d) Can’t tell without further information

Answer: (b). Negative. When t = 2, x = 2 and y = 1. x′(2) is large and negative,while y′(2) is small and positive. zx(2, 1) is large and positive., while zy(2, 1) is positivebut not as large. Thus the zxx

′ term dominates, telling us that the derivative z′(t) isnegative when t = 2.


MVC.14.06.030

CC KC MA233 F06: 27/73/0/0CC KC MA233 F07: 17/74/9/0 time 5:30CC MP MA233 F07: 53/33/13/0 time 5:00CC KC MA233A F08: 0/83/4/13 time 7:00CC KC MA233B F08: 0/75/19/6 time 6:00CC KC MA233 F09: 57/33/10/0 time 4:30CC HZ MA233 F10: 41/54/0/5 time 5:45CC HZ MA233 F11: 52/39/4/0 time 5:45CC TM MA233 F11: 44/44/12/0 time 7:30CC HZ MA233 F12: 4/61/22/13

127. Let s = f(x; y; z) and x = x(u; v; w); y = y(u; v; w); z = z(u; v; w). To calculate∂s∂u

(u = 1, v = 2, w = 3), which of the following pieces of information do you not need?

I f(1, 2, 3) = 5

II f(7, 8, 9) = 6

III x(1, 2, 3) = 7

67

IV y(1, 2, 3) = 8

V z(1, 2, 3) = 9

VI fx(1, 2, 3) = 20

VII fx(7, 8, 9) = 30

VIII xu(1, 2, 3) = −5

IX xu(7, 8, 9) = −7

(a) III, IV, VII, VIII

(b) I, IV, VI, VII

(c) II, V, VI, IX

(d) I, II, VI, IX

Answer: (d). We are working at u = 1, v = 2, w = 3 so we need the correspondingvalues of x, y, z which are x = 7, y = 8, z = 9. In the formula for the chain rule, thepartial derivatives of f are evaluated at (7, 8, 9), and the partial derivatives of x, y, zare evaluated at (1, 2, 3). Thus, we need (III), (IV), (V), (VII), and (VIII), but not(I), (II), (VI), (IX).


MVC.14.06.040

14.7 Second-Order Partial Derivatives

128. At the point (4,0), what is true of the second partial derivatives of f(x, y)?

68

(a) fxx > 0 and fyy > 0

(b) fxx < 0 and fyy < 0

(c) fxx > 0 and fyy = 0

(d) fxx < 0 and fyy = 0

Answer: (c). The slope in the positive x direction is positive and increasing as xincreases. Thus fxx > 0. The slope in the positive y direction is zero and remains soas y increases. Thus fyy = 0.

by Mark Schlatter

MVC.14.07.010

CC KC MA233 F06: 0/0/100/0CC MP MA233 F06: 0/0/77/23CC KC MA233 F07: 0/0/96/4 time 3:00CC KC MA233B F07: 0/0/100/0 time 2:00CC KC MA233A F08: 0/0/100/0 time 2:00CC KC MA233B F08: 0/0/100/0 time 2:40CC KC MA233A F09: 6/6/87/0 time 4:00CC KC MA233B F09: 6/0/94/0 time 2:30CC HZ MA233 F10: 0/0/100/0 time 2:30CC MP MA233 F10: 7/0/93/0 time 3:00HC AS MA203 F10: 6/0/89/6CC HZ MA233 F11: 0/0/87/13 time 3:15CC TM MA233 F11: 17/0/72/11 time 3:00CC HZ MA233 F12: 0/0/100/0

129. At the point (4,0), what is true of the second partial derivatives of f(x, y)?

69

(a) fxy > 0

(b) fxy < 0

(c) fxy = 0

Answer: (c). fxy = 0. As you move in the positive y direction, the slope in the xdirection does not change.

by Mark Schlatter

MVC.14.07.020

CC KC MA233 F06: 20/0/80CC MP MA233 F06: 23/0/77CC KC MA233 F07: 0/4/96 time 3:00CC KC MA233B F07: 15/0/85 time 1:00CC KC MA233A F09: 11/0/89 time 2:30CC KC MA233B F09: 12/6/82 time 2:00CC HZ MA233 F10: 38/0/62CC MP MA233 F10: 12/6/81 time 3:00HC AS MA203 F10: 31/6/63CC HZ MA233 F11: 13/0/87 time 2:00CC TM MA233 F11: 30/0/70 time 2:00CC HZ MA233 F12: 38/0/63

130. The figure below shows level curves of f(x, y). What are the signs of fxx(P ) andfyy(P )?

(a) fxx(P ) > 0, fyy(P ) ≈ 0

(b) fxx(P ) > 0, fyy(P ) < 0

(c) fxx(P ) ≈ 0, fyy(P ) ≈ 0

70

(d) fxx(P ) < 0, fyy(P ) > 0

Answer: (a). fxx(P ) > 0, fyy(P ) ≈ 0, because fx is positive and increasing, while fy isnegative and remaining fairly constant.


MVC.14.07.030

CC KC MA233 F06: 65/25/10/0CC MP MA233 F06: 39/61/0/0CC KC MA233 F07: 50/50/0/0 time 2:00CC KC MA233A F08: 96/4/0/0 time 3:00CC KC MA233B F08: 50/50/0/0 time 3:00CC KC MA233A F09: 68/26/0/5 time 3:00CC KC MA233A F09: 82/12/0/6 time 2:30CC HZ MA233 F10: 81/19/0/0 time 2:30CC HZ MA233 F11: 58/33/0/8 time 3:45CC TM MA233 F11: 58/37/5/0CC HZ MA233 F12: 48/33/14/5BJ BB MA305 F12: 58.33/29.17/8.33/4.17

131. The figure below shows level curves of f(x, y). What are the signs of fxx(Q) andfyy(Q)?

(a) fxx(Q) > 0, fyy(Q) < 0

(b) fxx(Q) < 0, fyy(Q) < 0

(c) fxx(Q) ≈ 0, fyy(Q) ≈ 0

(d) fxx(Q) < 0, fyy(Q) ≈ 0

71

Answer: (d). fxx(P ) < 0, fyy(P ) ≈ 0 because fx is positive and decreasing, while fy ispositive and remaining fairly constant.


MVC.14.07.040

CC KC MA233 F07: 0/7/0/93 time 2:00CC HZ MA233 F10: 5/0/0/95 time 2:15 ReviewCC HZ MA233 F11: 8/38/4/50 time 3:30

132. The figure below shows the surface z = f(x, y). What are the signs of fxx(A) andfyy(A)?

(a) fxx(A) > 0, fyy(A) < 0

(b) fxx(A) < 0, fyy(A) < 0

(c) fxx(A) ≈ 0, fyy(A) ≈ 0

(d) fxx(A) < 0, fyy(A) ≈ 0

Answer: (b). fxx(A) < 0, fyy(A) < 0 because fx is positive and decreasing, while fy isnegative and decreasing.


MVC.14.07.050

CC MP MA233 F06: 0/92/0/8CC KC MA233A F08: 0/100/0/0 time 2:30CC KC MA233B F08: 0/71/14/14 time 3:30CC HZ MA233 F10: 14/76/0/10 time 3:00CC HZ MA233 F11: 8/88/0/4 time 3:00CC TM MA233 F11: 33/50/11/5 time 6:00CC HZ MA233 F12: 5/82/0/14

72

BJ BB MA305 F12: 64/20/16/0

133. The figure below shows the surface z = f(x, y). What are the signs of fxx(P ) andfyy(P )?

(a) fxx(P ) > 0, fyy(P ) ≈ 0

(b) fxx(P ) > 0, fyy(P ) < 0

(c) fxx(P ) ≈ 0, fyy(P ) ≈ 0

(d) fxx(P ) < 0, fyy(P ) > 0

Answer: (d). fxx(P ) < 0, fyy(P ) > 0 because the surface is concave down in the xdirection and concave up in the y direction.


MVC.14.07.060

CC KC MA233 F06: 0/10/0/90CC MP MA233 F06: 0/31/0/69CC KC MA233 F07: 13/47/13/27 time 2:00CC KC MA233A F08: 0/0/0/100 time 2:00CC KC MA233A F09: 0/0/0/100 time 1:30CC KC MA233B F09: 0/18/6/76 time 1:30CC HZ MA233 F10: 0/24/0/76 time 1:30HC AS MA203 F10: 0/33/0/67CC HZ MA233 F11: 0/0/0/100 time 1:30CC TM MA233 F11: 0/16/0/84 time 2:00CC HZ MA233 F12: 4/20/4/72

134. The figure below shows the temperature T ◦C as a function of distance x in metersalong a wall and time t in minutes. Choose the correct statement and explain yourchoice without computing these partial derivatives.

73

(a) ∂T∂t

(t, 10) < 0 and ∂2T∂t2

(t, 10) < 0.

(b) ∂T∂t

(t, 10) > 0 and ∂2T∂t2

(t, 10) > 0.

(c) ∂T∂t

(t, 10) > 0 and ∂2T∂t2

(t, 10) < 0.

(d) ∂T∂t

(t, 10) < 0 and ∂2T∂t2

(t, 10) > 0.

Answer: (c). From the graph of the contour lines, we can see that along the line x = 10,the temperature T is increasing, thus ∂T

∂t(t, 10) > 0. Since the distance between the

contour lines is increasing as time increases along the line x = 10, the rate of changeof ∂T

∂t(t, 10) is decreasing so ∂2T

∂t2(t, 10) < 0.


MVC.14.07.070

CC KC MA233 F07: 5/0/95/0 time 3:00

135. The quadratic Taylor Polynomials (A)-(D) each approximate a function of two variablesnear the origin. Figures (I)-(IV) are contours near the origin. Match (A)-(D) to (I)-(IV).

74

A −x2 + y2

B x2 − y2

C −x2 − y2

D x2 + y2

(a) A - I, B - III, C - II, D - IV

(b) A - II, B - IV, C - I, D - III

(c) A - IV, B - II, C - III, D - I

(d) A - III, B - I, C - IV, D - II

(e) A - II, B - IV, C - III, D - I

Answer: (c). A - IV, B - II, C - III, D - I


MVC.14.07.080

75

136. In the contour plot below dark shades represent small values of the function and lightshades represent large values of the function. What is the sign of the mixed partialderivative?

(a) fxy > 0

(b) fxy < 0

(c) fxy ≈ 0

(d) This cannot be determined from the figure.

Answer: (b). If we move in the positive y direction fx decreases. If we move in thepositive x direction fy decreases. Thus fxy is negative.


MVC.14.07.090

CC KC MA233A F08: 0/52/48/0 time 3:00CC KC MA233B F08: 21/21/57/0 time 4:00CC KC MA233A F09: 16/74/5/5 time 4:00CC KC MA233B F09: 18/82/0/0 time 3:00CC HZ MA233 F10: 45/35/20/0 time 2:30CC HZ MA233 F11: 13/54/29/0 time 2:45CC HZ MA233 F12: 42/38/21/0

137. In the contour plot below dark shades represent small values of the function and lightshades represent large values of the function. What is the sign of the mixed partialderivative?

76

(a) fxy > 0

(b) fxy < 0

(c) fxy ≈ 0

(d) This cannot be determined from the figure.

Answer: (b). If we move in the positive y direction fx decreases. If we move in thepositive x direction fy decreases. Thus fxy is negative.

by Project MathVote

MVC.14.07.100

BJ BB MA305 F12: 4.17/62.5/8.33/25

15.1 Local Extrema

138. Which of these functions has a critical point at the origin?

(a) f(x, y) = x2 + 2y3

(b) f(x, y) = x2y + 4xy + 4y

(c) f(x, y) = x2y3 − x4 + 2y

(d) f(x, y) = x cos y

(e) All of the above

77

Answer: (a). Taking partial derivatives gues fx = 2x and fy = 6y2 so x = 0 and y = 0give a critical point. The other functions do not have a critical point at the origin.


MVC.15.01.010

CC KC MA233 F06: 91/9/0/0CC KC MA233 F07: 86/0/0/14 time 3:00(option (e) added)CC KC MA233A F08: 90/0/0/0/10 time 3:00CC KC MA233B F08: 44/0/0/0/56 time 3:00CC KC MA233 F09: 58/0/0/0/42 time 2:30

139. True or False? The function f(x, y) = x2y + 4xy + 4y has a local maximum at theorigin.





Answer: (False). Since fy(0, 0) = 4, the origin is not a critical point of f so it cannotbe a local maximum.


MVC.15.01.020

CC KC MA233 F06: 9/91CC KC MA233 F07: 0/100

140. Which of these functions does not have a critical point?

(a) f(x, y) = x2 + 2y3

(b) f(x, y) = x2y + 4xy + 4y

(c) f(x, y) = x2y3 − x4 + 2y

(d) f(x, y) = x cos y

(e) All have critical points

78

Answer: (c). Taking partial deriatives gives fx = 2xy3 − 4x3 and fy = 3x2y2 + 2.Because fy is always greater than or equal to 2, there are no critical points.


MVC.15.01.030

CC KC MA233 F07: 0/0/13/88 time 3:30CC MP MA233 F07: 0/50/38/12 time 5:00(option (e) added)CC KC MA233A F08: 0/0/20/24/56 time 3:40CC KC MA233B F08: 0/18/75/0/7 time 2:00CC KC MA233 F09: 0/10/5/10/75 time 3:00

141. Which of these functions has a critical point at the origin?

(a) f(x, y) = x2 + 2x + 2y3 − y2

(b) f(x, y) = x2y + xy

(c) f(x, y) = x2y2 − (1/2)x4 + 2y

(d) f(x, y) = x4y − 7y

Answer: (b). Taking partial derivatives gives fx = 2xy + y = (2x + 1)y and fy =x2 + x = (x + 1)x, so x = 0 and y = 0 give a critical point. The others do not have acritical point at the origin.


MVC.15.01.040


142. How would you classify the function f(x, y) = x2y + xy at the origin?

(a) This is a local maximum.

(b) This is a local minimum.

(c) This is a saddle point.

(d) We cannot tell.

(e) This is not a critical point.

79

Answer: (c). This is a saddle point. Taking partial derivatives gives fxx = 2y, fyy = 0,and fxy = 2x + 1, so at (0,0) the discriminant D = (2 · 0) · 0− (2 · 0 + 1)2 = −1. Thuswe have a saddle point at the origin.


MVC.15.01.050

CC KC MA233 F06: 0/0/33/67CC KC MA233 F07: 4/13/79/4 time 4:00(option(e) added)CC KC MA233A F08: 0/0/68/32/0 time 4:30CC KC MA233B F08: 0/6/75/19/0 time 4:30CC KC MA233 F09: 20/10/50/15/5 time 4:00

143. Which of these functions does not have a critical point with y = 0?

(a) f(x, y) = x2 + 2x + 2y3 − y2

(b) f(x, y) = x2y + xy

(c) f(x, y) = x2y2 − (1/2)x4 + 2y

(d) f(x, y) = x4y − 7y

Answer: (c). Taking partial derivatives gives fx = 2xy2 − 2x3 = 2x(y2 − x2) andfy = 2x2y + 2 = 2(x2y + 1). At critical point we need y2 = x2 and x2y = −1 soy3 = −1 or y = −1. Thus x may equal either 1 or -1, so the only two critical pointsare (1,-1) and (-1,-1).


MVC.15.01.060

144. Which of these functions does not have a critical point with x = −1?

(a) f(x, y) = x2 + 2x + 2y3 − y2

(b) f(x, y) = x2y + xy

(c) f(x, y) = x2y2 − (1/2)x4 + 2y

(d) f(x, y) = x4y − 7y

Answer: (d). Taking partial derivatives gives fx = 4x3y and fy = x4 − 7, so the twocritical points are x = ±71/4, y = 0.


MVC.15.01.070

80

145. Which of the following points are critical points?

(a) A and C

(b) A, C, and D

(c) A, B, and C

(d) A, B, C, and D

Answer: (c). At the points A, B, and C, there is no one direction to head to increaseas quickly as possible. Thus the gradient at all three points is the zero vector.

by Mark Schlatter

MVC.15.01.080

CC KC MA233 F06: 0/0/5/95CC KC MA233 F07: 4/21/17/58 time 3:30CC MP MA233 F07: 6/6/0/88 time 2:00 right idea, wrong reasons, need ABCD andAC as choicesOptions revised: Before the revision they were (a) B, C, and D; (b) A, C, and D; (c)A, B, and D; (d) A, B, and C. CC KC MA233A F08: 40/4/52/4 time 4:00CC KC MA233B F08: 26/0/37/37 time 3:00CC KC MA233 F09: 45/5/15/35 time 2:00CC MP MA233 F09: 38/0/9/52 time 3:00

146. Which of the following guarantees a saddle point of the function f(x, y) at (a, b)?

81

(a) fxx and fyy have the same sign at (a, b).

(b) fxx and fyy have opposite signs at (a, b).

(c) fxy is negative at (a, b).


Answer: (b). To make D = fxxfyy − (fxy)2 < 0, it suffices to make fxxfyy negative.

by Mark Schlatter

MVC.15.01.090

15.2 Optimization

147. Estimate the global maximum and minimum of the functions whose level curves aregiven below. How many times does each occur?

(a) Max ≈ 6, occurring once; min ≈ −6, occurring once

(b) Max ≈ 6, occurring once; min ≈ −6, occurring twice

(c) Max ≈ 6, occurring twice; min ≈ −6, occurring twice

(d) Max ≈ 6, occurring three times; min ≈ −6, occurring three times

82


Answer: (d). The global max is about 6, or slightly higher. It occurs three times,twice on the negative y-axis and once on the positive y-axis. The global min is -6,or slightly lower. It occurs three times, twice on the positive x-axis and once on thenegative x-axis.


MVC.15.02.010

CC KC MA233 F06: 19/5/5/62/10CC KC MA233 F07: 52/0/10/39/0 time 3:30CC KC MA233A F08: 4/0/0/92/4 time 2:30CC KC MA233B F08: 6/13/6/75/0 time 3:00CC KC MA233 F09: 5/10/45/40/0 time 2:30

148. What are the global maximum and minimum values of f(x, y) = x2 + y2 on thetriangular region in the first quadrant bounded by x + y = 2, x = 0, y = 0?

(a) Maximum = 2, Minimum = -2

(b) Maximum = 2, Minimum = 0

(c) Maximum = 4, Minimum = 2

(d) Maximum = 4, Minimum = 0

Answer: (d). The minimum value of x2 + y2 occurs at the origin, so the minimum is 0.The maximum occurs at the other two corners of the triangle, so at (2, 0) and (0, 2);the maximum value is 4.


MVC.15.02.020

CC KC MA233 F06: 15/15/0/70CC KC MA233 F07: 9/5/9/77 time 4:30CC KC MA233A F08: 0/13/4/83 time 4:40CC KC MA233B F08: 0/31/31/38 time 4:30CC KC MA233 F09: 10/33/5/52 time 4:30CC HZ MA233 F11: 17/33/0/45 time 5:00CC TM MA233 F11: 0/69/8/23 time 7:00

149. The function f(x, y) = x3 + 12xy + y4 has:

83

(a) no global maxes or mins

(b) a global max, but no global min

(c) a global min, but no global max

(d) both a global min and a global max

Answer: (a). As x→ −∞, f(x, y)→ −∞. As y →∞, f(x, y)→∞

by Mark Schlatter

MVC.15.02.030

CC KC MA233 F07: 26/22/13/39 time 5:00CC KC MA233A F08: 61/0/4/35 time 6:30CC KC MA233B F08: 31/13/37/19 time 9:00CC KC MA233 F09: 33/10/57/0 time 8:00CC HZ MA233 F11: 21/4/50/17 time 3:30CC TM MA233 F11: 26/20/20/33 time 5:00

150. Which of the following would be enough evidence to conclude that a smooth functionf(x, y) has a global min?

(a) D is always positive

(b) fxx > 0 and fyy > 0

(c) f(x, y) has no saddle points or local maxes


Answer: (d). (a) is not enough since it is true of −x2 − y2. (b) is not enough since itis true of z = x2 + 10xy + y2. (c) is not enough since it is true of any linear function.

by Mark Schlatter

MVC.15.02.040


15.3 Constrained Optimization: Lagrange Multipli-

ers

151. Find the maximum and minimum values of f on g = c.

84

(a) max = 5, min = 0

(b) max = 4, min = 0

(c) max = 3, min = 2

(d) max = 4, min = 2

Answer: (d). The maximum is f = 4 and occurs at (0, 4). The minimum is f = 2 andoccurs at about (4, 2).


MVC.15.03.010

CC KC MA233 F06: 0/5/0/95CC MP MA233 F06: 0/7/7/86CC KC MA233 F07: 4/0/0/96 time 2:00CC MP MA233 F07: 0/6/0/94 time 1:30(several options that never attracted votes were removed)CC MP MA233 F09: 7/0/0/93 time 2:00BJ BB MA305 F12: 4.17/0/4.17/91.67

152. Find the maximum and minimum values of f on the trapezoidal region below g = c inthe first quadrant.

85

(a) max = 5, min = 0

(b) max = 5, min = 2

(c) max = 4, min = 1

(d) max = 4, min = 0

(e) max = 3, min = 2

(f) max = 3, min = 0

(g) max = 4, min = 2

(h) max = 5, min = 2

(i) max = 2, min = 0

Answer: (d). The maximum is f = 4 and occurs at (0, 4). The minimum is f = 0 andoccurs at the origin.


MVC.15.03.020

CC MP MA233 F06: 0/0/0/60/0/7/26/0/7CC MP MA233 F07: 0/7/0/87/7/0/0/0/0 time 2:30

153. Find the maximum of the production function f(x, y) = xy in the first quadrant subjectto each of the three budget constraints. Arrange the x coordinates of the optimal pointin increasing order.

I x + y = 12

86

II 2x + 57y = 12

III 3x + y/2 = 12

(a) I < II, < III

(b) III < II < I

(c) II < III < I

(d) II < I < III

(e) III < I < II

Answer: (b). For I the solution is at (6,6). For II the solution is at (3, 1.2). For IIIthe solution is at (2,12).


MVC.15.03.030

154. This contour plot of f(x, y) also shows the circle of radius 2 centered at (0,0). If youare restricted to being on the circle, how many local maxes and mins does f(x, y) have?

(a) 1

(b) 2

(c) 3

(d) 4

Answer: (d). There appear to be four locations on the circle where the circle wouldbe tangent to a contour line (around (.2, -2), (.5, 1.8), (-1.8,1), and (-1.9, -.8)).

87

by Mark Schlatter

MVC.15.03.040

CC KC MA233 F06: 16/26/21/26CC MP MA233 F06: 40/27/20/13CC KC MA233 F07: 9/48/30/13 time 4:00CC MP MA233 F07: 0/33/33/33 time 5:00. No lecture, lots of discussion - leads intocontour and constraint being parallel, they came up with crossing the same contourtwice implies a maximum or minimum betweenCC KC MA233A F08: 9/50/27/14 time 4:00CC KC MA233B F08: 43/36/0/21CC KC MA233 F09: 10/19/14/57 time 4:00CC MP MA233 F09: 40/25/15/20 time 3:00CC MP MA233 F10: 0/18/18/64CC HZ MA233 F11: 23/10/10/57 time 3:00CC TM MA233 F11: 46/0/31/23 time 2:00BJ BB MA305 F12: 8/40/12/40

155. This plot shows the gradient vectors for a (hidden) function f(x, y) and a linear con-straint. Which point is closest to the global min of f(x, y) on this constraint?

(a) A

(b) B

(c) C

(d) D

Answer: (c). At C, if you move in either direction along the line, the gradient showsthat the function will be increasing.

88

by Mark Schlatter

MVC.15.03.050

CC MP MA233 F06: 0/0/100/0CC MP MA233 F07: 7/0/93/0 time 3:00CC KC MA233A F08: 9/0/91/0 time 2:00CC KC MA233B F08: 31/0/69/0 time 3:15

156. How many local maxs and mins does the function f(x, y, z) = ax + by + cz have onthe sphere x2 + y2 + z2 = 1?

(a) 1

(b) 2

(c) 3

(d) 4

(e) None

Answer: (b). Given a collection of parallel planes, there are only two which will betangent to a sphere.

by Mark Schlatter

MVC.15.03.060

CC KC MA233 F06: 32/68/0/0/0CC KC MA233 F07: 36/32/0/0/32 time 6:00(Answers reorganized to match clicker numbers a=1, b=2, etc.)CC KC MA233A F08: 0/64/0/36/0 time 4:00CC KC MA233B F08: 0/75/0/19/6 time 3:30CC HZ MA233 F11: 9/36/9/23/0 time 2:00

157. How many points will produce local max/min of f(x, y) = x2 − y2 over the regionx2 + y2 ≤ r2?

(a) 1

(b) 2

(c) 3

(d) 4

Answer: (d). Since the graph of f(x, y) looks like a saddle, there are no local max ormin on the interior, but four such points on the boundary (two local mins and twolocal maxs).

89

by Mark Schlatter

MVC.15.03.070

CC KC MA233 F07: 5/45/5/45 7:30CC KC MA233 F07: 71/29/0/0 5:00 (covering other section)CC KC MA233A F08: 18/14/0/68 time 5:00

158. The figure below shows the optimal point (marked with a dot) in three optimizationproblems with the same constraint. Arrange the corresponding values of λ in increasingorder. (Assume λ is positive.)

(a) I < II < III

(b) II < III < I

(c) III < II < I

(d) I < III < II

Answer: (a). Since λ is the additional f that is obtained by relaxing the constraintby 1 unit, λ is larger if the level curves of f are close together near the optimal point.The answer is I < II < III.


MVC.15.03.080

159. Minimize x2 + y2 subject to x2y2 = 4

(a) 1

(b) 2

(c) 2√

2

(d) 4

(e) 8

(f) 16

90

Answer: (d). 4. We solve 2x = 2xy2λ, 2y = 2x2yλ, and x2y2 = 4. So x2 = y2

and (x2)2 = 4, and since we are working in the first quadrant x = y =√

2. Thusf(√

2,√

2) = 4.


MVC.15.03.090

CC KC MA233A F08: 0/21/0/79/0/0 time 6:30CC KC MA233B F08: 0/7/7/79/7/0 time 6:30CC KC MA233 F09: 0/0/33/67/0/0 time 6:00

160. Maximize x2y2 subject to x2 + y2 = 4.

(a) 1

(b) 2

(c) 2√

2

(d) 4

(e) 8

(f) 16

Answer: (d). We solve 2xy2 = 2xλ, 2x2y = 2yλ and x2 +y2 = 4. This gives x2 = y2, so2x2 = 4, and since we are working in the first quadrant, x = y =

√2. f(

√2,√

2) = 4.


MVC.15.03.100

CC KC MA233 F07: 0/5/5/82/0/10 time 8:00CC KC MA233A F08: 0/0/4/88/0/8 time 7:00CC KC MA233 F09: 0/17/6/72/6/0

161. Maximize x2y2 subject to x + y = 4 with x, y ≥ 0.

(a) 1

(b) 2

(c) 2√

2

(d) 4

(e) 8

(f) 16

91

Answer: (f). 16. We solve 2xy2 = λ, 2x2y = λ, and x + y = 4. We get x = y = 2, sowe have f(2, 2) = 16.


MVC.15.03.110

CC KC MA233 F07: 0/0/0/9/4/86 time 4:30CC KC MA233A F08: 0/0/0/5/0/95 time 5:30CC KC MA233B F08: 0/6/0/6/0/88 time 5:00CC KC MA233 F09: 10/10/0/15/0/65 time 6:00

16.1 The Definite Integral of a Function of Two Vari-

ables

162. Suppose the contour plot shown shows the height of a pile of dirt in feet. Which of thefollowing is clearly a lower bound for the volume of dirt?

(a) 0*1+0*1+6*1+6*1

(b) 9*1+9*1+9*1+9*1

(c) 9

(d) 6*1+6*1+9*1+9*1

Answer: (a). The lower bounds for the height of dirt on the four squares are 0, 0, 6,and 6. We then use a Riemann sum.

92

by Mark Schlatter

MVC.16.01.010

CC MP MA233 F06: 60/0/30/10CC KC MA233 F06: 76/0/24/0CC KC MA233 F07: 57/15/0/33CC KC MA233B F08: 7/13/27/54 time 5:00CC KC MA233 F09: 50/10/20/20 time 3:30CC MP MA233 F09: 26/37/21/16 time 5:00CC MP MA233 F10: 68/12/12/6BJ BB MA305 F12: 41.67/25/16.67/16.67

163. Let R be the region 10 ≤ x ≤ 14 ; 20 ≤ y ≤ 30. The table below gives values off(x, y). Using upper and lower Riemann sums, what are the best possible upper andlower estimates for the integral

I =

∫

R

f(x, y)dxdy

(a) 23 < I < 990

(b) 92 < I < 300

(c) 160 < I < 396

(d) 160 < I < 300

(e) 92 < I < 396

Answer: (d). The function appears to be increasing in x and in y, so since ∆x∆y =2 · 5 = 10. Lower estimate = (2.3 + 3.7 + 4.2 + 5.8)∆x∆y = 16 · 10 = 160. Upperestimate = (5.8 + 6.2 + 8.1 + 9.9)∆x∆y = 30 · 10 = 300.


MVC.16.01.015

CC MP MA233 F07: 0/17/17/67/0CC KC MA233B F08: 20/7/7/0/67 time 7:00CC KC MA233 F09: 15/15/10/15/45 time 7:00CC HZ MA233 F10: 5/5/0/86/5 time 12:00

93

CC MP MA233 F10: 0/0/7/93/0 time 8:00CC HZ MA233 F11: 9/0/5/62/24 time 11:15CC TM MA233 F11: 33/0/33/0/33 time 10:00BJ BB MA305 F12: 16.67/12.5/20.83/25/25

164. Let R be the square defined by −1 ≤ x ≤ 1, −1 ≤ y ≤ 1. The sign of the definiteintegral of x4 over R is:

(a) positive

(b) negative

(c) zero

(d) cannot be determined

Answer: (a). Since over this entire region, x4 is nonnegative (and mostly positive),the integral is positive.

by Mark Schlatter

MVC.16.01.020

CC MP MA233 F06: 100/0/0CC KC MA233 F06: 100/0/0CC HZ MA233 F10: 71/5/19/0 time 3:15CC MP MA233 F10: 73/7/20/0 time 2:00CC HZ MA233 F11: 55/0/30/15 time 4:30CC TM MA233 F11: 40/0/60/0 time 3:00CC HZ MA233 F12: 48/4/48/0

165. The value of (1/π) times the integral of 1 + x over the unit circle R is:

(a) 0

(b) 1

(c) π

(d) π/2

Answer: (b). This integral gives the average value of 1 + x over the region R, becausethe area of the region is π. The average value is of the function is 1.

by Mark Schlatter

MVC.16.01.030

CC KC MA233 F07: 5/67/5/24 time 5:00CC KC MA233A F08: 4/75/21/0 time 4:30

94

BJ BB MA305 F12: 13.04/43.48/26.09/17.39

166. The integral∫

Rx dA over the region where R is the rectangle −1 ≤ x ≤ 1, −1 ≤ y ≤ 1

is

(a) positive

(b) negative

(c) zero

Answer: (c). Zero

ConcepTests - to accompany Calculus 4th Edition, Hughes-Hallet et al. John Wiley &Sons.

MVC.16.01.040

CC MP MA233 F06: 0/0/100CC KC MA233 F06: 0/6/94CC HZ MA233 F10: 0/5/91 time 1:15CC MP MA233 F10: 44/6/50CC HZ MA233 F11: 14/0/86 time 2:30CC TM MA233 F11: 10/0/89 time 3:00BJ BB MA305 F12: 16.67/4.17/79.17


TydA over the region where T is the rectangle −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero



MVC.16.01.050

CC MP MA233 F06: 79/14/7CC KC MA233 F07: 86/0/14 time 5:00CC KC MA233A F08: 77/5/18 time 3:45CC KC MA233B F08: 73/0/27 time 3:30CC KC MA233 F09: 70/5/25 time 2:45CC HZ MA233 F10: 91/0/0 time 1:30CC HZ MA233 F11: 90/0/10 time 2:15

95

CC TM MA233 F11: 80/10/10 time 3:15BJ BB MA305 F12: 65.22/8.7/26.09


R(x − x2)dA over the region where R is the rectangle −1 ≤ x ≤ 1,

−1 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero

Answer: (b). Negative


MVC.16.01.060

CC KC MA233 F06: 44/56/0CC HZ MA233 F10: 5/81/14 time 2:00


T(y − y2)dA over the region where T is the rectangle −1 ≤ x ≤ 1,

0 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero



MVC.16.01.070


L(x2 − x)dA over the region where L is the rectangle −1 ≤ x ≤ 0,

−1 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero

96



MVC.16.01.080

CC KC MA233 F07: 86/9/5 time 2:30CC KC MA233A F08: 52/30/17 time 3:00CC KC MA233B F08: 67/7/27 time 2:00CC KC MA233 F09: 55/40/5 time 2:30


L(y + y3)dA over the region where L is the rectangle −1 ≤ x ≤ 0,

−1 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero

Answer: (c). Zero


MVC.16.01.090


R(2x + 3y)dA over the region where R is the rectangle −1 ≤ x ≤ 1,

−1 ≤ y ≤ 1 is

(a) positive

(b) negative

(c) zero

Answer: (c). Zero


MVC.16.01.100

CC KC MA233A F08: 0/0/100 time 2:45CC KC MA233 F09: 0/0/100 time 1:30

97

16.2 Iterated Integrals


1

0

∫

1

0x2dxdy represents the

(a) Area under the curve y = x2 between x = 0 and x = 1.

(b) Volume under the surface z = x2 above the square 0 ≤ x, y ≤ 1 on the xy-plane.

(c) Area under the curve y = x2 above the square 0 ≤ x, y ≤ 1 on the xy-plane.

Answer: (b). The volume under the surface z = x2 above the square 0 ≤ x, y ≤ 1 onthe xy-plane.


MVC.16.02.010

CC MP MA233 F06: 0/88/12CC MP MA233 F07: 0/100/0 time 3:00 lecture and ”bread slicing” demo, then ques-tionsCC KC MA233 F07: 0/100/0 time 2:00CC KC MA233B F08: 0/93/7 time 3:00CC KC MA233 F09: 20/80/0 time 2:00BJ BB MA305 F12: 0/95.83/4.17


1

0

∫

1

xdydx represents the

(a) Area of a triangular region in the xy-plane.

(b) Volume under the plane z = 1 above a triangular region of the plane.

(c) Area of a square in the xy-plane.

Answer: both (a) and (b).


MVC.16.02.020

CC MP MA233 F06: 6/94/0CC MP MA233 F07: 0/100/0 time 1:30 asked why A wouldn’t also work...

175. Let f(x, y) be a positive function. Rank the following integrals from smallest to largest.

I1 =

∫

1

0

∫

1

x2

f(x, y)dydx I2 =

∫

1

0

∫

1

x3

f(x, y)dydx I3 =

∫

1

0

∫

1

0

f(x, y)dydx

98

(a) I1 < I2 < I3

(b) I1 < I3 < I2

(c) I2 < I1 < I3

(d) I2 < I3 < I1

(e) I3 < I2 < I1

(f) I3 < I1 < I2

Answer: (a). I1 < I2 < I3. Since the regions are nested and the integrands are thesame and positive, the only thing that matters is the size of the region we integrateover.


MVC.16.02.030

176.∫

1

0

∫

2−2x

0f(x, y)dydx is an integral over which region?

(a) The triangle with vertices (0,0), (2,0), (0,1).

(b) The triangle with vertices (0,0), (0,2), (1,0).

(c) The triangle with vertices (0,0), (2,0), (2,1).

(d) The triangle with vertices (0,0), (1,0), (1,2).

Answer: (b).


MVC.16.02.040

CC KC MA233 F07: 0/86/0/14 time 3:00CC KC MA233A F08: 0/91/0/9 time 2:30CC KC MA233B F08: 13/87/0/0 time 3:00CC KC MA233 F09: 5/85/0/10 time 2:30BJ BB MA305 F12: 8.7/69.57/8.7/13.04

177.∫

1

0

∫

2

2yf(x, y)dxdy is an integral over which region?

(a) The triangle with vertices (0,0), (2,0), (0,1).

(b) The triangle with vertices (0,0), (0,2), (1,0).

(c) The triangle with vertices (0,0), (2,0), (2,1).

(d) The triangle with vertices (0,0), (1,0), (1,2).

99

Answer: (c).


MVC.16.02.050

CC KC MA233 F07: 10/0/90/0 time 3:30BJ BB MA305 F12: 17.39/13.04/60.87/8.7

178. Which of the following integrals has the proper limits to integrate the shaded regionbelow?

(a)∫

1

−1

∫ −2x−1

−3f(x, y)dydx

(b)∫

1

−3

∫

1

− 1

2y− 1

2

f(x, y)dxdy

(c)∫

1

−1

∫

1

− 1

2x−1

f(x, y)dydx

(d)∫

1

−3

∫ − 1

2y− 1

2

−1f(x, y)dxdy


Answer: (b).

by Project MathVote

MVC.16.02.055

BJ BB MA305 F12: 20.83/54.17/12.5/8.33/4.17

179. Which of the following integrals is equal to∫

3

0

∫

4x

0f(x, y)dydx?

(a)∫

4x

0

∫

3

0f(x, y)dxdy

(b)∫

12

0

∫

3

y/4f(x, y)dxdy

100

(c)∫

12

0

∫ y/4

3f(x, y)dxdy

(d)∫

12

0

∫ y/4

0f(x, y)dxdy

(e)∫

3

0

∫

4x

0f(x, y)dxdy

Answer: (b).∫

12

0

∫

3

y/4f(x, y)dxdy


MVC.16.02.060

CC MP MA233 F07: 0/19/0/75/6 time: 4:30CC KC MA233 F07: 0/14/0/86/0 time 3:30CC KC MA233A F08: 4/92/4/0/0 time 4:00CC KC MA233B F08: 0/0/0/100/0 time 3:00CC KC MA233 F09: 15/0/0/85/0 time 3:00CC HZ MA233 F10: 0/91/0/4/4 time 2:15CC HZ MA233 F12: 13/74/0/13/0BJ BB MA305 F12: 0/60.87/0/39.13/0

180. The region of integration in the integral∫

2

0

∫

2x

0f(x, y)dydx is a

(a) rectangle

(b) triangle with width 2 and height 4

(c) triangle with width 4 and height 2


Answer: (b). As x varies from 0 to 2, y varies from 0 to 0 (when x = 0) to 0 to 4(when x = 2).

by Mark Schlatter

MVC.16.02.070

CC KC MA233 F07: 0/100/0/0CC KC MA233A F08: 0/100/0/0 time 1:40

181. The value of∫ r

−r

∫

√r2−x2

−√

r2−x2 xdydx is

(a) πr

(b) π/2

(c) πr2

101

(d) 0

Answer: (d). This is the integral of x over the circle of radius r centered over theorigin. Since the average value of x over this region is 0, the integral equals 0.

by Mark Schlatter

MVC.16.02.080

CC KC MA233 F09: 0/5/40/55 time 5:00

16.3 Triple Integrals

182. Which of the following is the mass of the solid defined by 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, and0 ≤ z ≤ 4 with density function δ(x, y, z) = x + y?

(a)∫

2

0

∫

4

0

∫

3

01dydzdx

(b)∫

2

0

∫

4

0

∫

3

0(x + y)dydxdz

(c)∫

2

0

∫

4

0

∫

3

0(x + y)dxdzdy

(d)∫

2

0

∫

4

0

∫

3

0(x + y)dydzdx

Answer: (d). The last integral is the only one with the density function as the integrandand with the parameters matched with their correct ranges.

by Mark Schlatter

MVC.16.03.010

CC MP MA233 F06: 0/0/36/64 review questionCC MP MA233 F07: 13/0/0/83 time: 2:00CC KC MA233 F07: 0/0/5/95 time 3:00CC KC MA233A F08: 4/0/8/88 time 2:00CC KC MA233B F08: 0/7/7/86 time 2:30CC KC MA233A F09: 0/0/19/81 time 1:30CC KC MA233B F09: 10/10/5/75 time 2:30CC HZ MA233 F10: 0/5/14/81 time 2:00CC HZ MA233 F11: 4/0/11/85 time 2:00CC TM MA233 F11: 0/0/15/84 time 5:00CC HZ MA233 F12: 17/0/4/79

183. The region of integration for the integral∫ r

−r

∫

√r2−x2

−√

r2−x2

∫

10

0f(x, y, z)dzdydx is a

102

(a) sphere

(b) cylinder

(c) cone


Answer: (b). Cylinder. The limits for x and y determine a circle of radius r centeredon the origin. Since z can take on any value regardless of x and y, the volume is acylinder.

by Mark Schlatter

MVC.16.03.020

CC KC MA233 F06: 32/41/9/18CC KC MA233 F07: 5/73/5/18 time 3:00CC KC MA233A F08: 12/75/0/12 time 3:20CC KC MA233B F08: 13/53/13/20 time 4:00CC KC MA233A F09: 5/48/19/29 time 3:00CC KC MA233A F09: 5/79/5/11 time 2:15CC HZ MA233 F10: 23/59/18/0 time 2:00CC HZ MA233 F11: 4/70/19/7 time 4:00CC TM MA233 F11: 5/77/9/9 time 3:30CC HZ MA233 F12: 13/67/17/4

184. What does the integral∫

1

0

∫

1

0

∫

1

0z dzdydx represent?

(a) The volume of a cube of side 1

(b) The volume of a sphere of radius 1

(c) The area of a square of side 1


Answer: (d). None of the above. The integral equals 1

2, while (a) = 1, (b) = 4

3, and (c)

= 1. If (x, y, z) are in meters, the result of this integration is in m4, which representsneither an area nor a volume.


MVC.16.03.030

CC KC MA233 F06: 18/0/14/68CC MP MA233 F07: 73/7/0/20 time: 2:00CC KC MA233 F07: 68/0/9/23 time 3:00CC KC MA233A F08: 37/0/0/63 time 3:30CC KC MA233B F08: 33/0/33/33 time 3:00

103


185. Which of the following integrals is equal to∫

3

0

∫

2

0

∫ y

0f(x, y, z)dzdydx?

(a)∫

2

0

∫

3

0

∫ y

0f(x, y, z)dzdxdy

(b)∫

2

0

∫

3

0

∫ y

0f(x, y, z)dzdydx

(c)∫

3

0

∫

2

0

∫ y

0f(x, y, z)dxdydz

(d)∫

3

0

∫

2

0

∫ z

0f(x, y, z)dydzdx

Answer: (a). As a follow-up question, ask “What’s wrong with (d)?”


MVC.16.03.040

CC KC MA233 F06: 96/0/0/4CC KC MA233A F08: 70/0/4/26 time 2:30CC KC MA233B F08: 80/0/0/20CC KC MA233A F09: 71/0/9/19 time 2:00CC KC MA233B F09: 95/0/0/5 time 2:00CC HZ MA233 F10: 83/0/0/17 time 2:30CC HZ MA233 F11: 64/0/0/36 time 4:30CC TM MA233 F11: 71/0/10/19 time 2:00CC HZ MA233 F12: 96/0/0/4

186.∫

1

−1

∫

√1−x2

−√

1−x2

∫

√1−x2−y2

−√

1−x2−y2(x2 + y2 + z2)dzdydx describes the mass of

(a) a cone that gets heavier toward the outside.

(b) a cone that gets lighter toward the outside.

(c) a ball that gets heavier toward the outside.

(d) a ball that gets lighter toward the outside.

Answer: (c).


MVC.16.03.050

CC KC MA233 F06: 0/4/23/73CC MP MA233 F07: 0/0/93/7 time: 2:00

104

CC KC MA233 F07: 9/41/46/5 time 5:00CC KC MA233A F09: 10/10/57/23 time 4:00CC KC MA233B F09: 33/0/67/0 time 3:00CC HZ MA233 F10: 0/17/35/48 time 5:30 Great pre & post-vote discussionsCC HZ MA233 F11: 4/4/58/35 time 4:45CC TM MA233 F11: 0/5/81/14 time 4:30CC HZ MA233 F12: 21/0/33/46

187. Which of the following integrals does not make sense?

(a)∫

3

1

∫

2

y−1

∫ y

0f(x, y, z)dzdxdy

(b)∫

3

1

∫ y

0

∫ y−1

2f(x, y, z)dxdydz

(c)∫

1

−1

∫

√1−x2

−√

1−x2

∫

√1−x2−y2

0f(x, y, z)dzdydx

(d)∫

1

−1

∫

√1−y2

−√

1−y2

∫

√1−x2−y2

0f(x, y, z)dzdxdy

Answer: (b). does not make sense because the limits of integration for y are 0 to y.


MVC.16.03.060

CC KC MA233 F07: 0/100/0/0CC KC MA233 F08A: 0/83/17/0 time 1:45CC HZ MA233 F10: 4/96/0/0 time 2:00CC HZ MA233 F11: 4/96/0/0 time 1:30CC HZ MA233 F12: 0/96/0/4

16.4 Double Integrals in Polar Coordinates

188. A point is at coordinates (r, θ) = (1, π). What are the rectangular coordinates of thispoint?

(a) (x, y) = (1, 0)

(b) (x, y) = (0, 1)

(c) (x, y) = (−1, 0)

(d) (x, y) = (0,−1)


105

Answer: (c). This angle puts the point on the negative x axis, at a distance of 1 unitfrom the origin.

by Project MathVote

MVC.16.04.004

CC KC MA233A F09: 0/5/86/9 time 2:30CC KC MA233B F09: 0/24/76/0 time 2:00CC HZ MA233 F10: 6/11/67/6/11 time 2:30HC AS MA203 F10: 11/6/61/17/6CC HZ MA233 F11: 5/0/91/0/0 time 2:00CC TM MA233 F11: 6/0/72/11/11 time 2:00CC HZ MA233 F12: 0/8/92/0/0

189. A point is at coordinates (x, y) = (0,−1). What are the polar coordinates of thispoint?

(a) (r, θ) = (1, 3π4

)

(b) (r, θ) = (1, 3π2

)

(c) (r, θ) = (1,−π)

(d) (r, θ) = (1,−π2)


Answer: (e). Both (b) and (d) are correct, demonstrating that any given point can beidentified by more than one set of polar coordinates.

by Project MathVote

MVC.16.04.006

CC KC MA233A F09: 0/41/5/23/32 time 2:00CC KC MA233B F09: 12/47/6/12/23 time 1:30CC HZ MA233 F10: 16/47/0/11/26 time 1:45HC AS MA203 F10: 11/50/22/0/17CC HZ MA233 F11: 14/57/0/5/24 time 1:45CC TM MA233 F11: 0/35/6/24/35 time 3:00CC HZ MA233 F12: 0/38/0/13/50

190. Which of the following regions resembles a quarter of a doughnut?

(a) 0 ≤ r ≤ 5, 0 ≤ θ ≤ π/2

(b) 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π

(c) 3 ≤ r ≤ 5, π ≤ θ ≤ 2π

106

(d) 3 ≤ r ≤ 5, π ≤ θ ≤ 3π/2

Answer: (d). To get a quarter, we need the range of the angles to be π/2. To ensurea doughnut, we need to make sure r does not reach 0.

by Mark Schlatter

MVC.16.04.010

CC MP MA233 F06: 19/6/0/75CC KC MA233 F06: 6/11/0/83CC KC MA233 F07: 0/0/0/100 time 2:00CC KC MA233P F07: 21/0/0/79 time 2:00CC KC MA233A F08: 0/0/4/96 time 1:40CC KC MA233B F08: 0/0/0/100 time 2:00CC KC MA233A F09: 4/0/0/96 time 1:30CC KC MA233B F09: 18/6/0/76 time 1:30CC HZ MA233 F10: 11/0/11/78 time 1:30CC HZ MA233 F11: 10/0/0/90 time 2:00CC TM MA233 F11: 0/0/6/94 time 2:00CC HZ MA233 F12: 16/4/0/80

191. Which of the following integrals is equivalent to∫

3

0

∫

2π

πrdθdr?

(a)∫

3

0

∫

0

−√

9−x2 1dydx

(b)∫

3

−3

∫

√9−x2

01dydx

(c)∫

3

−3

∫

√9−y2

01dxdy

(d)∫

0

−3

∫

√9−y2

−√

9−y21dxdy

Answer: (d). The area of integration is the bottom half of a circle of radius 3. Onlyd) produces this. Choice a) is the portion in quadrant IV, b) is the top half, and c) isthe right half).

by Mark Schlatter

MVC.16.04.020

CC MP MA233 F06: 6/31/19/44CC KC MA233A F07: 11/47/16/26 time 4:00CC KC MA233B F07: 21/36/7/36 time 4:00CC KC MA233A F08: 30/17/0/52 time 3:30CC KC MA233B F08: 8/75/17/0 time 3:30CC KC MA233A F09: 32/50/5/14 time 3:00CC KC MA233B F09: 0/59/0/41 time 2:30

107

CC HZ MA233 F10: 21/37/10/32 time 4:00CC HZ MA233 F11: 15/45/5/35 time 4:30CC TM MA233 F11: 39/33/17/11 time 3:30CC HZ MA233 F12: 4/13/0/83

192. What geometric shape is describe by the equation r = θ?

(a) line

(b) circle

(c) spiral


Answer: (c). spiral


MVC.16.04.030

CC MP MA233 F06: 12/12/76/0

193. What geometric shape is describe by the equation r = 4?

(a) line

(b) circle

(c) spiral


Answer: (b). a circle


MVC.16.04.040

CC MP MA233 F06: 0/94/0/6CC KC MA233A F07: 27/68/0/5 time 2:00CC KC MA233B F07: 50/21/0/29 time 2:00

194. What geometric shape is describe by the equation θ = 4?

(a) line

108

(b) circle

(c) spiral


Answer: (a). a line


MVC.16.04.050

CC MP MA233 F06: 94/0/0/6

195. What geometric shape is describe by the equation r = sin θ?

(a) line

(b) circle

(c) spiral


Answer: (b). circle


MVC.16.04.060

196. What geometric shape is describe by the equation r = 1/ sin θ?

(a) line

(b) circle

(c) spiral


Answer: (a). a line


MVC.16.04.070

197. Which of the following describes the upper half of the xy-plane?

(a) 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π

109

(b) 0 ≤ r ≤ ∞, 0 ≤ θ ≤ π

(c) 0 ≤ r ≤ θ, 0 ≤ θ ≤ π

(d) 2 ≤ r ≤ 4, π ≤ θ ≤ 3π/2

Answer: (b).


MVC.16.04.080

CC MP MA233 F06: 0/93/7/0 review questionCC KC MA233A F07: 0/100/0/0 time 2:00CC KC MA233B F07: 7/93/0/0 time 1:00

198. Which integral gives the area of the unit circle?

(a)∫

1

−1

∫

√1−x2

−√

1−x2 xdydx

(b)∫

2π

0

∫

1

0rdrdθ

(c)∫

2π

0

∫

1

0drdθ

(d)∫

1

0

∫

2π

0dθdr

Answer: (b).


MVC.16.04.090

CC KC MA233 F06: 5/76/5/14CC KC MA233A F07: 0/84/16/0 time 2:00CC KC MA233B F07: 14/64/14/7 time 2:00CC KC MA233A F08: 4/91/0/4 time 1:30CC KC MA233B F08: 0/83/0/17 time 1:30CC KC MA233A F09: 23/73/0/4 time 2:30CC KC MA233B F09: 6/71/0/23 time 2:30CC HZ MA233 F10: 0/95/5/0 time 1:30CC HZ MA233 F11: 0/100/0/0 time 1:30CC TM MA233 F11: 0/100/0/0 time 1:30CC HZ MA233 F12: 4/96/0/0

110

16.5 Integrals in Cylindrical and Spherical Coordi-

nates

199. What are the Cartesian coordinates of the point with cylindrical coordinates (r, θ, z) =(4, π, 6)?

(a) (x, y, z) = (0,−4, 4)

(b) (x, y, z) = (0, 4, 6)

(c) (x, y, z) = (−4, 4, 4)

(d) (x, y, z) = (4, 0, 4)

(e) (x, y, z) = (−4, 0, 6)

Answer: (e).

by Carroll College

MVC.16.05.003

CC MP MA233 F06: 0/0/0/6/94CC KC MA233 F06: 0/13/4/0/83CC KC MA233A F07: 0/0/0/5/95 time 2:00CC KC MA233B F07: 8/18/0/0/75CC KC MA233A F08: 0/17/0/0/83 time 1:30CC KC MA233B F08: 0/0/9/0/91 time 1:30CC KC MA233A F09: 0/0/0/5/95 time 1:30CC KC MA233B F09: 0/0/0/0/100 time 1:30HC AS MA203 F10: 0/6/0/12/82BJ BB MA305 F12: 4/20/0/0/76

200. What are the cylindrical coordinates of the point with Cartesian coordinates (x, y, z) =(3, 3, 7)?

(a) (r, θ, z) = (3, π, 7)

(b) (r, θ, z) = (3, π/4, 3)

(c) (r, θ, z) = (3√

2, π/4, 7)

(d) (r, θ, z) = (3√

2, π, 7)

(e) (r, θ, z) = (3√

2, π, 3)

Answer: (c).

by Carroll College

111

MVC.16.05.004

CC KC MA233A F08: 0/0/100/0/0 time 2:00CC KC MA233B F08: 0/0/92/8/0 time 2:00CC KC MA233 F09: 0/0/100/0/0 time 2:00BJ BB MA305 F12: 8/4/76/12/0

201. What are the Cartesian coordinates of the point with spherical coordinates (ρ, φ, θ) =(4, π, 0)?

(a) (x, y, z) = (0, 0,−4)

(b) (x, y, z) = (0, 0, 4)

(c) (x, y, z) = (4, 0, 0)

(d) (x, y, z) = (−4, 0, 0)

(e) (x, y, z) = (0, 4, 0)

Answer: (a).

by Carroll College

MVC.16.05.005

CC MP MA233 F06: 50/21/29/0/0CC KC MA233 F06: 77/9/14/0/0CC KC MA233A F07: 100/0/0/0/0 time 2:00CC KC MA233B F07: 64/0/7/27/0 time 2:30CC KC MA233A F08: 72/8/20/0/0 time 3:00CC KC MA233B F08: 23/0/15/46/15 time 3:00CC KC MA233 F09: 30/10/20/40/0 time 2:30HC AS MA203 F10: 53/18/12/18/0

202. What are the spherical coordinates of the point with Cartesian coordinates (x, y, z) =(0,−3, 0)?

(a) (ρ, φ, θ) = (3, π, π2)

(b) (ρ, φ, θ) = (3, π,−π2)

(c) (ρ, φ, θ) = (3, π2, π

2)

(d) (ρ, φ, θ) = (3, π2,−π

2)

(e) (ρ, φ, θ) = (3, π2, π)

112

Answer: (d).

by Carroll College

MVC.16.05.006

CC KC MA233 F06: 0/0/18/82/0CC KC MA233A F07: 5/5/27/63/0 time 2:30CC KC MA233B F07: 0/14/0/86/0 time 1:30CC KC MA233A F08: 4/0/8/76/12 time 3:00CC KC MA233B F08: 15/15/0/46/23 time 3:00CC KC MA233 F09: 0/5/0/90/5 time 3:00HC AS MA203 F10: 7/7/7/79/0

203. Which of the following regions represents the portion of a cylinder of height 4 andradius 3 above the 3rd quadrant of the xy plane?

(a) 1 ≤ r ≤ 3, 0 ≤ z ≤ 4, 0 ≤ θ ≤ π/2

(b) 0 ≤ r ≤ 3, 0 ≤ z ≤ 4, π ≤ θ ≤ 3π/2

(c) 0 ≤ r ≤ 4, 0 ≤ z ≤ 3, π ≤ θ ≤ 3π/2

(d) 0 ≤ r ≤ 3, 0 ≤ z ≤ 4, 0 ≤ θ ≤ π/2

Answer: (b). The 3rd quadrant implies π ≤ θ ≤ 3π/2. The answer must be b) insteadof c) since the height is 4 (and thus z ranges from 0 to 4).

by Mark Schlatter

MVC.16.05.010

CC KC MA233 F06: 0/100/0/0CC KC MA233A F08: 0/100/0/0 time 1:45CC KC MA233B F08: 0/92/8/0 time 1:45CC KC MA233A F09: 0/100/0/0 time 2:00

204. Which of the following is equivalent to

∫

5

−5

∫

3

0

∫

√25−x2

−√

25−x2

x dydzdx

(a)∫

3

0

∫

3

0

∫ π

0r2 cos θ dθdzdr

(b)∫

5

0

∫

3

0

∫ π

0r2 cos θ dθdzdr

(c)∫

3

0

∫

5

0

∫

2π

0r cos θ dθdzdr

(d)∫

5

0

∫

3

0

∫

2π

0r2 cos θ dθdzdr

113

Answer: (d). In cylindrical coordinates x = r cos(θ) and the extra r for cylindricalintegrals means the integrand is r2 cos(θ). The region of integration is a cylinder ofradius 3 and height 5, which corresponds to d).

by Mark Schlatter

MVC.16.05.020

CC KC MA233 F09: 5/72/18/5 time 4:30

205. Which of the following describes the bottom half of a sphere of radius 4 centered onthe origin?

(a) 0 ≤ ρ ≤ 4, π/2 ≤ φ ≤ π, 0 ≤ θ ≤ 2π

(b) 0 ≤ ρ ≤ 4, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π

(c) 0 ≤ ρ ≤ 4, 0 ≤ φ ≤ π, 0 ≤ θ ≤ π

(d) 0 ≤ ρ ≤ 4, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π

Answer: (a). Points below the xy plane have values of φ between π/2 and π.

by Mark Schlatter

MVC.16.05.030

CC KC MA233 F06: 67/24/0/10CC KC MA233A F07: 50/9/41/0CC KC MA233B F07: 93/0/7/0 time 2:00CC KC MA233A F08: 92/0/0/8 time 2:30CC KC MA233B F08: 100/0/0/0 time 2:00CC KC MA233 F09: 70/15/5/10 time 2:15

206. Which of the following describes the surface of the cylinder of radius 3 centered on thez-axis?

(a) 0 ≤ ρ <∞, θ = π, 0 ≤ φ ≤ π

(b) r = 3, θ = π2, −∞ < z <∞

(c) 1 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, −5 ≤ z ≤ 2

(d) r = 3, 0 ≤ θ ≤ 2π, −∞ < z <∞

Answer: (d). Follow up: What do the others describe?


MVC.16.05.040

114

CC KC MA233 F06: 0/0/0/100CC KC MA233A F08: 0/0/0/100 time 2:00CC KC MA233B F08: 0/0/0/100 time 2:00CC KC MA233 F09: 5/5/11/79 time 2:00

207. Which of the following describes the solid cylinder of radius 4, centered on the z-axis,with the central cylindrical core removed?

(a) 0 ≤ ρ <∞, θ = π, 0 ≤ φ ≤ π

(b) r = 3, θ = π2, −∞ < z <∞

(c) 1 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, −5 ≤ z ≤ 2

(d) r = 3, 0 ≤ θ ≤ 2π, −∞ < z <∞

Answer: (c).


MVC.16.05.050

CC KC MA233 F06: 0/0/90/10CC KC MA233A F07: 0/5/72/23 time 2:30CC KC MA233B F07: 7/0/57/36 time 2:30CC KC MA233A F08: 0/0/100/0 time 2:00CC KC MA233B F08: 0/0/100/0 time 2:00CC KC MA233 F09: 0/0/90/10 time 1:30

208. Which of the following integrals give the volume of the unit sphere?

(a)∫

2π

0

∫

2π

0

∫

1

0dρdθdφ

(b)∫ π

0

∫

2π

0

∫

1

0dρdθdφ

(c)∫ π

0

∫

2π

0

∫

1

0ρ2 sin φdρdθdφ

(d)∫ π

0

∫

2π

0

∫

1

0ρ2 sin φdρdφdθ

(e)∫ π

0

∫

2π

0

∫

1

0ρdρdφdθ

Answer: (c).


MVC.16.05.060

CC KC MA233 F06: 0/0/100/0/0CC KC MA233A F08: 0/4/88/8/0 time 1:30CC KC MA233 F09: 0/0/100/0/0 time 2:00

115

17.1 Parameterized Curves

209. Which of the following is an equation of a line in three dimensions (x, y, z)?

(a) x = 4

(b) y = 2x + 3

(c) z = 3x + 2y + 7

(d) All of the above


Answer: (e). All of these equations describe planes in three dimensions. The purposeof this question is to motivate parameterization.

by Project MathVote

MVC.17.01.005

CC KC MA233 F09: 0/14/36/29/21 time 2:30CC HZ MA233 F10: 5/5/45/23/23 time 2:00CC MP MA233 F10: 0/0/23/31/46 time 3:00CC HZ MA233 F11: 0/0/41/14/46 time 2:45CC TM MA233 F11: 0/0/62/24/15 time 3:00CC HZ MA233 F12: 0/0/33/21/46

210. Which of the following best describes the path of a particle defined by the parametricequations x(t) = cos(t2), y(t) = sin(t2)?

(a) a circle around which the particle moves faster and faster

(b) a parabola on which the particle travels at constant speed

(c) a parabola on which the particle travels faster and faster

(d) a circle on which the particle moves slower and slower

Answer: (a). Since the coordinates have the form (cos(a), sin(a)), the path must be acircle. Since the common input is t2, the particle is moving faster and faster aroundthe circle.

by Mark Schlatter

MVC.17.01.010

CC MP MA233 F06: 100/0/0/0CC KC MA233 F06: 83/0/0/17CC KC MA233 F07: 45/55/0/0 time 6:00

116

CC KC MA233A F08: 60/8/4/28 time 3:00CC KC MA233B F08: 40/20/27/13 time 5:00CC KC MA233 F09: 73/0/20/7 time 4:00CC MP MA233 F10: 61/0/0/38 time 2:00BJ BB MA305 F12: 40/24/28/8

211. Which of the following is not a parameterization of the entire curve y = x3?

(a) x(t) = t; y(t) = t3

(b) x(t) = t2; y(t) = t6

(c) x(t) = t3; y(t) = t9

(d) x(t) = 2t; y(t) = 8t3

Answer: (b). All parametrizations satisfy y(t) = (x(t))3, but the parametrization inb) only produces positive x values.

by Mark Schlatter

MVC.17.01.020

CC MP MA233 F06: 0/10/10/80CC KC MA233 F06: 22/9/9/61CC MP MA233 F07: 0/79/14/7 time: 6:00 review: the votes for (c) came from a typo(fixed)CC KC MA233 F07: 0/14/0/86 time 5:00CC KC MA233A F08: 0/36/4/60 time 2:00CC KC MA233B F08: 0/27/0/73 time 5:00CC KC MA233 F09: 0/47/6/47 time 4:30

212. What does the path of the particle described by x(t) = cos(t), y(t) = sin(t), z(t) = −tlook like?

(a) a circle in the xz plane

(b) a helix on which the particle is traveling up

(c) a helix on which the particle is traveling down

(d) a sine wave in the xz plane

Answer: (c). The path is a helix since, viewed from above, the particle is traveling onthe unit circle. The particle is moving down since, as t increases, z decreases.

by Mark Schlatter

MVC.17.01.030

117

CC KC MA233 F06: 5/0/95/0CC HZ MA233 F10: 9/14/77/0 time 1:30CC HZ MA233 F11: 0/4/96/0 time 3:30CC TM MA233 F11: 0/0/100/0 time 3:00CC HZ MA233 F12: 0/0/100/0

213. Which of the following parameterizations does not describe the quarter circle in thefigure below?

(a) (cos t, sin t), 0 ≤ t ≤ π/2

(b) (sin t, cos t), 0 ≤ t ≤ π/2

(c) (− cos t, sin t), π/2 ≤ t ≤ π

(d) (cos t,− sin t), 3π/2 ≤ t ≤ 2π

Answer: (a).


MVC.17.01.040

CC HZ MA233 F10: 60/15/15/10 time 2:45 ReviewBJ BB MA305 F12: 28/44/16/12

214. Let (cos at, sin at) be the position at time t seconds of a particle moving around a circle,where a > 0. If a is increased,

(a) The radius of the circle increases.

(b) The speed of the particle increases.

(c) The center of the circle changes.

(d) The path ceases to be a circle.

118

Answer: (b). The speed of the particle increases. As a increases, it takes less time forthe particle to complete one revolution, thus the speed increases.


MVC.17.01.050

CC MP MA233 F06: 44/56/0/0CC MP MA233 F07: 0/100/0/0 time: 3:00 review questionCC MP MA233 F09: 0/100/0/0 time 2:00CC HZ MA233 F10: 19/81/0/0 time 2:00CC HZ MA233 F11: 9/91/0/0 time 3:30CC TM MA233 F11: 24/71/0/5CC HZ MA233 F12: 29/58/0/13

215. Let (a cos t, a sin t) be the position at time t seconds of a particle moving around acircle, where a > 0. If a is increased,

(a) The radius of the circle increases.

(b) The speed of the particle increases.

(c) The center of the circle changes.

(d) The path ceases to be a circle.

Answer: (a). The radius of the circle increases.


MVC.17.01.060

CC KC MA233 F06: 88/4/0/8

216. Which of the following parametric curves does not trace out the unit circle?

(a) (cos t, sin t), 0 ≤ t ≤ 2π

(b) (sin2 t, cos2 t), 0 ≤ t ≤ 2π

(c) (sin(t2), cos(t2)), 0 ≤ t ≤ 2π

(d) (sin(2t), cos(2t)), 0 ≤ t ≤ 2π

Answer: (b).


119

MVC.17.01.070

CC HZ MA233 F10: 5/86/5/5 time 1:15CC HZ MA233 F12: 0/88/13/0BJ BB MA305 F12: 0/66.67/16.67/16.67

217. Which of the following parametric paths describe particles that are not traveling alonga straight line in 3-space?

(a) (1− t, 2 + 2t, 3− t)

(b) (1− t2, 2 + 2t2, 3− t2)

(c) (1, 2, 1− t)

(d) (1, t, 1− t2)


Answer: (d).


MVC.17.01.080

CC KC MA233 F07: 0/36/9/55 time 5:30 (added (e))CC KC MA233 F09: 13/0/6/37/44 time 4:00CC HZ MA233 F10: 5/0/5/23/68 time 2:30CC HZ MA233 F11: 0/4/0/14/82 time 4:00CC TM MA233 F11: 5/28/0/0/67 time 4:00CC HZ MA233 F12: 0/8/4/25/63

218. The value of c for which the lines l(t) = (c+4t, 2−t, 3+t) and m(t) = (4t, 1−8t, 4+4t)intersect is

(a) 4

(b) 0

(c) -4

(d) There is no such c.

Answer: (c). -4. The two parameters, t, in l(t) and m(t), represent different quantities,so we will call the second one s. Thus m(s) = (4s, 1− 8s, 4+4s). The lines intersect if

c + 4t = 4s 2− t = 1− 8s 3 + t = 4 + 4s

120

Solving the last two equations gives t = 1 and s = 0. Substituting into the firstequation gives c = −4. Thus (c) is the correct answer.


MVC.17.01.090

CC KC MA233A F08: 0/17/0/83 time 4:30CC KC MA233B F08: 0/40/13/47 time 4:00CC KC MA233 F09: 16/47/5/32 time 5:00CC MP MA233 F09: 0/50/0/50 time 5:00

17.2 Motion, Velocity, and Acceleration

219. A lighthouse at position L is in the middle of a lake. Its beam is turning counterclock-wise with constant angular velocity. At which point is the velocity vector of the beamlargest?

(a) A

(b) B

(c) C

121

(d) D

Answer: (c). Since the beam is moving with constant angular velocity, equal timesmean equal radian measure swept out. A small ∆t in time produces a greater differencein distance near C than at any other point.

by Mark Schlatter

MVC.17.02.010


220. A lighthouse at position L is in the middle of a lake. Its beam is turning counterclock-wise with constant angular velocity. At which point is the velocity vector of the beammost parallel to j?

(a) A

(b) B

(c) C

(d) D

Answer: (b). Since the velocity vector is always tangent to the curve, it is closest topointing straight up or down at B.

122

by Mark Schlatter

MVC.17.02.020

CC KC MA233 F07: 0/100/0/0 time 2:00CC KC MA233A F08: 10/76/14/0 time 3:00CC KC MA233B F08: 0/50/50/0 time 2:30CC MP MA233 F09: 5/45/50/0 time 3:00

221. A lighthouse at position L is in the middle of a lake. Its beam is turning counterclock-wise with constant angular velocity. At which point is the acceleration vector of thebeam most parallel to j?

(a) A

(b) B

(c) C

(d) D

Answer: (c). Velocity vectors on either side of C will be pointing up and down. Thedifference between those vectors will point almost straight up.

by Mark Schlatter

MVC.17.02.030

123

CC MP MA233 F09: 5/18/77/0 time 3:00

222. Which of the following describes the motion of a particle that is moving along a straightline and slowing down?

(a) ~a and ~v are parallel and point in the same direction.

(b) ~a and ~v are parallel and point in opposite directions.

(c) ~a and ~v are perpendicular.

(d) None of the above.

Answer: (b).


MVC.17.02.040

CC MP MA233 F06: 0/100/0/0CC KC MA233 F07: 0/96/4/0 time 2:30

223. True or False: If the speed of a particle is zero, its velocity must be zero.





Answer: (a). True


MVC.17.02.050

CC MP MA233 F07: 95/5 time 2:00 review question

224. A particle that is not accelerating must have zero velocity.





124

Answer: (False). A particle moving with constant speed along a straight line is notaccelerating.


MVC.17.02.060

CC MP MA233 F06: 0/100CC MP MA233 F07: 0/100 time 1:00 review question

225. A particle with constant speed must have zero acceleration.





Answer: (False). A particle moving with constant speed around a circle does not havezero acceleration.


MVC.17.02.070

CC MP MA233 F06: 44/56CC MP MA233 F07: 27/73 time 1:00 review questionCC KC MA233A F08: 33/67 time 1:00CC MP MA233 F09: 14/86 time 1:00

226. A particle with zero acceleration must have constant speed.





Answer: (a). True


MVC.17.02.080

125

CC MP MA233 F06: 100/0CC KC MA233 F08: 95/5

227. A particle with constant speed must have constant velocity.





Answer: (False). A particle moving around a circle with constant speed has changingvelocity since its direction changes.


MVC.17.02.090

228. The functions x(t) and y(t) describe the coordinates of a jeep in miles as it drivesaround the desert from noon (t = 0 hrs) until 2 pm (t = 2 hrs), when the jeep returnsto its starting location. We want to use these functions to predict how many miles willbe recorded on the odometer during this interval by doing an integral of some function∫

2

0f(t) dt. What units must the function f(t) have?

(a) miles

(b) miles/hr

(c) miles/hr2

(d) miles2/hr2


Answer: (b). If we integrate a function f(t) with respect to time, and the result is adistance in miles, then f(t) must have units of miles/hr. The purpose of this questionis to motivate the method for calculating arclength.

by Project MathVote

MVC.17.02.100

CC KC MA233 F09: 0/100/0/0/0 time 2:30

229. The functions x(t) and y(t) describe the coordinates of a jeep in miles as it drivesaround the desert from noon (t = 0 hrs) until 2 pm (t = 2 hrs), when the jeep returnsto its starting location. Which of the following must be true?

126

(a)∫

2

0x(t) dt = 0

(b)∫

2

0x′(t) dt = 0

(c)∫

2

0x′′(t) dt = 0

(d) More than one of the above


Answer: (b). If the jeep returns to its starting location, then any periods of positivevelocity x′(t) must be cancelled out by periods of negative velocity, so the overallintegral must be equal to zero. The purpose of this question is to demonstrate why weuse the integral of speed to calculate arclength, rather than the integral of velocity.

by Project MathVote

MVC.17.02.110

CC KC MA233 F09: 26/26/0/48/0 time 3:30

230. The functions x(t) = 2 sin πt and y(t) = 2 cos πt describe the coordinates of a jeep inmiles as it drives around the desert from noon (t = 0 hrs) until 2 pm (t = 2 hrs), whenthe jeep returns to its starting location. According to the jeep’s odometer, how far willit have traveled?

(a) 4 miles

(b) 2π miles

(c) 4π miles

(d) 8π2 miles


Answer: (c). We take derivatives of x(t) and y(t), then find that the speed is 2πmiles/hr. Thus if we integrate with this over a period of 2 hours, we find that the jeephas traveled a distance of 4π miles.

by Project MathVote

MVC.17.02.120

HC AS MA203 F10: 6/12/65/0/18

127

17.3 Vector Fields

231. Which of the following could be a formula for the vector field pictured?

(a) ~F (x, y) = xi

(b) ~F (x, y) = yi

(c) ~F (x, y) = xj

(d) ~F (x, y) = yj

Answer: (c). Since all the vectors point up or down, they are all parallel to j. Sincethe vectors point up for positive x and down for negative x, c) is the most plausiblechoice.

by Mark Schlatter

MVC.17.03.010

CC MP MA233 F06: 0/6/94/0CC KC MA233 F06: 0/0/100/0CC KC MA233A F08: 0/5/84/11 time 2:00CC KC MA233B F08: 0/0/87/13 time 2:30CC KC MA233 F09: 10/15/75/0CC MP MA233 F09: 0/5/77/18 time 2:00CC HZ MA233 F10: 0/0/100/0CC MP MA233 F10: 6/6/75/12 time 4:00HC AS MA203 F10: 5/5/65/25CC HZ MA233 F11: 4/11/78/7 time 2:15CC TM MA233 F11: 5/13/59/23 time 2:30CC HZ MA233 F12: 0/8/92/0

128

BJ BB MA305 F12: 8/8/44/40

232. Which of the following formulas will produce a vector field where all vectors point awayfrom the y axis and all vectors on a vertical line have the same length?

(a) ~F (x, y) = x3i

(b) ~F (x, y) = x2i

(c) ~F (x, y) = x3j

(d) ~F (x, y) = x2j

Answer: (a). To move away from the y axis, the vectors must point in the positive xdirection (or i direction) when x > 0 and point in the negative x direction when x < 0.

by Mark Schlatter

MVC.17.03.020

CC MP MA233 F06: 81/19/0/0CC KC MA233 F06: 88/12/0/0CC KC MA233A F08: 62/38/0/0 time 2:00CC KC MA233B F08: 35/45/5/15 time 2:30CC KC MA233 F09: 35/30/30/5 time 3:00CC MP MA233 F09: 68/32/0/0 time 2:00CC HZ MA233 F10: 52/30/13/4CC MP MA233 F10: 65/30/6/0 time 3:00CC HZ MA233 F11: 41/44/7/7 time 3:30CC TM MA233 F11: 73/23/3/0 time 2:45CC HZ MA233 F12: 96/4/0/0

233. Which of the following vector fields cannot be a gradient vector field?

(a) the one on the left

(b) the one in the middle

129

(c) the one on the right

Answer: (b). Gradient vectors are always perpendicular to contours. If we drawcontours perpendicular to the vectors for the middle field, the contours all meet at theorigin.

by Mark Schlatter

MVC.17.03.030

CC KC MA233 F06: 0/76/24CC KC MA233A F08: 0/100/0 time 1:45CC KC MA233B F08: 0/100/0 time 1:30CC KC MA233 F09: 0/75/25 time 2:00CC HZ MA233 F10: 0/78/22 time 1:30CC HZ MA233 F11: 8/84/8 time 3:00CC HZ MA233 F12: 4/88/8

234. Which formula below could produce the graph of the vector field:

(a) f(x)i

(b) g(x)j

(c) h(y)i

(d) k(y)j

Answer: (a).


MVC.17.03.040

CC MP MA233 F06: 100/0/0/0CC MP MA233 F07: 47/0/53/0 time 2:00 review question

130

CC HZ MA233 F10: 100/0/0/0 time 1:15BJ BB MA305 F12: 72/0/24/4

235. Which formula below could produce the graph of the vector field:

(a) f(x)i + f(x)j

(b) g(x)i− g(x)j

(c) h(y)i + h(y)j

(d) k(y)i− k(y)j

Answer: (b).


MVC.17.03.050

CC MP MA233 F06: 6/94/0/0CC KC MA233 F06: 0/100/0/0CC MP MA233 F07: 0/100/0/0 time 2:00 review questionCC KC MA233A F08: 0/95/0/5 time 3:00CC KC MA233B F08: 13/87/0/0 time 2:20CC KC MA233 F09: 5/85/5/5 time 4:00CC HZ MA233 F10: 0/87/4/4 time 2:00CC MP MA233 F10: 6/88/0/6 time 3:00CC HZ MA233 F11: 0/100/0/0 time 4:00CC TM MA233 F11: 0/100/0/0 time 3:15CC HZ MA233 F12: 0/96/0/4BJ BB MA305 F12: 8/88/4/0

131

236. Match the vector fields with the appropriate graphs.

1 ~F1 = ~r||~r||

2 ~F2 = ~r

3 ~F3 = yi− xj

4 ~F4 = xj

(a) 1 and III, 2 and I, 3 and IV, 4 and II

(b) 1 and IV, 2 and I, 3 and II, 4 and III

(c) 1 and II, 2 and I, 3 and IV, 4 and III

(d) 1 and II, 2 and IV, 3 and I, 4 and III

(e) 1 and I, 2 and II, 3 and IV, 4 and III

Answer: (c). 1 and II, 2 and I, 3 and IV, 4 and III


MVC.17.03.060

CC MP MA233 F06: 0/38/6/56CC KC MA233 F06: 0/0/100/0CC KC MA233A F08: 0/5/90/5 time 3:45CC KC MA233B F08: 0/12/69/19 time 4:30option (e) addedCC KC MA233 F09: 10/10/45/20/15 time 5:00CC MP MA233 F09: 23/54/5/18/0 time 4:00

132

CC HZ MA233 F10: 0/17/52/17/13 time 4:30CC HZ MA233 F11: 0/4/70/19/7 time 5:00CC TM MA233 F11: 0/18/50/23/9 time 5:00CC HZ MA233 F12: 0/0/58/42/0

237. The figure shows the vector field ~F = ∇f . Which of the following are possible choicesfor f(x, y)?

(a) x2

(b) −x2

(c) −2x

(d) −y2

Answer: (b). −x2, since ∇(x2) = −2xi , which is parallel to the x axis and pointstoward the y-axis.


MVC.17.03.070

CC MP MA233 F06: 0/83/0/17CC KC MA233 F06: 0/82/18/0CC KC MA233 F07: 4/70/26/0 time 3:30CC HZ MA233 F10: 5/55/41/0 time 2:30CC HZ MA233 F11: 0/59/37/4 time 3:00CC HZ MA233 F12: 0/54/46/0BJ BB MA305 F12: 8/56/32/4

238. Rank the length of the gradient vectors at the points marked on the contour plot below.

133

(a) 7 > 5 > 3 > 1

(b) 1 > 3 > 5 > 7

(c) 7 > 1 > 3 > 5

(d) 3 > 1 > 7 > 5

Answer: (a). The contours are closer together as we get farther from the origin.


MVC.17.03.080

17.4 The Flow of a Vector Field

239. The flow lines for the vector field pictured will be:

(a) straight lines

(b) circles

(c) ellipses

134

(d) parabolas

Answer: (d). As we follow the vectors from point to point, we trace out parabolas.

by Mark Schlatter

MVC.17.04.010

CC KC MA233 F07: 0/0/0/100 time 1:30CC KC MA233A F08: 0/0/0/100 time 1:45CC KC MA233B F08: 0/0/0/100 time 1:00CC HZ MA233 F10: 0/5/0/95 time 1:15CC HZ MA233 F11: 11/0/0/89 time 3:30CC TM MA233 F11: 5/11/6/78 time 2:30CC HZ MA233 F12: 0/4/29/67

240. The function that describes the distance of a particle from the x-axis as it follows aflow line is:

(a) linear

(b) exponential

(c) sinusoidal

(d) logarithmic

Answer: (b). Exponential. As a particle moves further and further from the x axis, thevectors get longer, meaning the velocity is increasing. Of the four possibilities, that isonly true for the exponential function.

by Mark Schlatter

MVC.17.04.020

135

CC MP MA233 F06: 0/67/33/0CC KC MA233 F06: 0/100/0/0CC MP MA233 F07: 40/40/13/7 time 2:00 review questionCC KC MA233A F08: 36/46/18/0 time 2:30CC KC MA233B F08: 79/21/0/0 time 2:40

241. Which parameterized curve is not a flow line of the vector field ~F = xi + yj?

(a) ~r(t) = et i + etj

(b) ~r(t) = et i + 2etj

(c) ~r(t) = 3et i + 3etj

(d) ~r(t) = 2et i + e2tj

Answer: (d). The flow lines ~r(t) = x(t)i+y(t)j solve the differential equation dx/dt =x, dy/dt = y. The solutions are all of the form x = aet and y = bet, so all but (d) areflow lines.


MVC.17.04.030

CC KC MA233 F06: 12/24/0/65CC KC MA233A F08: 20/25/0/55 time 3:30CC KC MA233A F08: 0/13/6/81 time 3:30HC AS MA203 F10: 0/17/17/67

242. Which parameterized curves are not flow lines of the vector field ~F = −yi + xj.

(a) ~r(t) = cos ti + sin tj

(b) ~r(t) = cos ti− sin tj

(c) ~r(t) = sin ti− cos tj

(d) ~r(t) = 2 cos ti + 2 sin tj

Answer: (b). The flow lines ~r(t) = x(t)i + y(t)j solves the system of differentialequations dx/dt = −y, dy/dt = x for all but (b).


MVC.17.04.040


136


243. The path x = t, y = et is a flow line of which vector field?

(a) i + yj

(b) i + x2j

(c) xi + xj

(d) yi + j

Answer: (a). We test these by substituting them into the equation ~r′(t) = ~F (~r(t)).


MVC.17.04.050

CC KC MA233 F06: 100/0/0/0

244. An object flowing in the vector field ~F = yi+xj is at the point (1, 2) at time t = 5.00.Estimate the approximate position of the object at time t = 5.01.

(a) (1.02, 2)

(b) (1.02, 2.01)

(c) (1.01, 2)

(d) (1.01, 2.02)

Answer: (b). In the x direction it is moving at speed y = 2, so in time ∆t = 0.01it will move ∆x = 0.02, so the new x position must be 1.02. In the y direction it ismoving at speed x = 1, so in time ∆t = 0.01 it will move ∆y = 0.01, so the new yposition must be 2.01.


MVC.17.04.060

CC KC MA233 F06: 0/83/0/17 time 4:00CC KC MA233A F08: 0/77/0/23 time 4:30

245. Two different curves can be flow lines for the same vector field.


137




Answer: (a). True. Every vector field has many flow lines. There is a flow line for agiven vector field through every point in the plane.


MVC.17.04.070

CC KC MA233 F07: 96/4 time 1:30

246. If one parameterization of a curve is a flow line for a vector field, then all its parame-terizations are flow lines for the vector field.





Answer: (False). A flow line of a velocity field represents the motion of a particle withvelocity specified by the vector field at every point. Changing the parameterizationcan change the velocity of the particle even though the particle moves through exactlythe same points.


MVC.17.04.080

CC KC MA233 F07: 62/38 time 2:00HC AS MA203 F10: 14/19/38/29

247. If ~r(t) is a flow line for a vector field ~F , then ~r1(t) = ~r(t− 5) is a flow line of the same

vector field ~F .





138

Answer: (a). True. These two lines move along exactly the same path, with the secondfollowing 5 time units behind. Their velocities at a given point at the same and so canbe given by the same velocity field ~F .


MVC.17.04.090

248. If ~r(t) is a flow line for a vector field ~F , then ~r1(t) = ~r(2t) is a flow line of the vector

field 2~F .





Answer: (a). True. Two particles with motion described by ~r(t) and ~r1(t) = ~r(2t)move along exactly the same path, with the second particle moving twice as fase asthe first at any given point. The velocity of the first at a point on its path is ~F , so thevelocity of the second is 2 ~F .


MVC.17.04.100

249. If ~r(t) is a flow line for a vector field ~F , then ~r1(t) = 2~r(t) is a flow line of the vector

field 2~F .





Answer: (False). A counterexample is given by ~F = xi and ~r(t) = et i. Then ~r1(t) =

2eti is not a flow line of 2~F = 2xi.


MVC.17.04.110

139

18.1 The Idea of a Line Integral

250. Suppose C is the path consisting of a straight line from (-1,0) to (1,0) followed by astraight line from (1,0) to (1,1). The line integral along this path is

(a) positive.

(b) zero.

(c) negative.

Answer: (a). Positive. The movement from (-1,0) to (1,0) yields a positive line integral(since the motion is in the direction of the field) which is greater in absolute value thanthe negative line integral from (1,0) to (1,1).

by Mark Schlatter

MVC.18.01.010

CC MP MA233 F06: 92/0/8CC KC MA233A F08: 56/22/22 time 3:00CC KC MA233B F08: 64/14/21 time 2:30CC KC MA233 F09: 61/33/6 time 2:30HC AS MA203 F10: 53/26/21BJ BB MA305 F12: 45.83/25/29.17

251. Given three curves, C1 (a straight line from (0,0) to (1,1)), C2 (a straight line from(1,-1) to (1,1)), and C3 (the portion of the circle of radius

√2 centered at the origin

moving from (1,-1) to (1,1)), rank the curves according to the value of the line integral

of ~F = −yi + xj on each curve.

140

(a) C1 < C2 < C3

(b) C2 < C1 < C3

(c) C3 < C1 < C2

Answer: (a). This vector field produces flow lines in the form of circles centered on theorigin flowing counterclockwise. Motion on C1 is always perpendicular to the vectorfield, so the line integral is 0. Motion on C2 is generally in the direction of the vectorfield (yielding a positive line integral), but motion on C3 is always parallel to the vectorfield, yielding the largest positive line integral.

by Mark Schlatter

MVC.18.01.020

CC KC MA233 F06: 88/6/6CC KC MA233A F08: 83/0/17 time 5:00CC KC MA233B F08: 50/36/14 time 7:00Note: Commas in answers changed to < signs, to clarify.CC KC MA233 F09: 47/18/35 time 4:00

252. The vector field ~F and several curves are shown below. For which of the paths is theline integral positive?

(a) C1

(b) C2

(c) C3

(d) C4

141

Answer: (b). C2. This is the only path that has a positive ~F · d~r throughout the path.


MVC.18.01.030

CC MP MA233 F06: 0/100/0/0BJ BB MA305 F12: 4.17/79.17/4.17/12.5

253. If the path C is a circle centered at the origin, oriented clockwise, which of the vectorfields below has a positive circulation?

(a) i

(b) ii

(c) iii

(d) iv

Answer: (c). The vector field in (i) circles counterclockwise around the origin, so itscirculation around C is negative. For the vector field in (ii), the path integral alongthe left half of C cancels out the path integral along the right half, so the circulation iszero. The vector field in (iii) has a component in the direction of C all the way around(except at the very top and bottom of C), so the circulation is positive. The vectorfield in (iv) is perpendicular to C at every point, so the circulation is zero.


MVC.18.01.040

142

254. True or false? Given two circles centered at the origin, oriented counterclockwise, andany vector field ~F , then the path integral of ~F is larger around the circle with largerradius.





Answer: (False). The path integral is determined by the values of the vector fieldalong the path, not simply by the size of the path.


MVC.18.01.050

CC KC MA233 F07: 40/60 time 2:30BJ BB MA305 F12: 12.5/25/33.33/29.17

255. True or false? If ~F is any vector field and C is a circle, then the integral of ~F aroundC traversed clockwise is the negative of the integral of ~F around C traversed counter-clockwise.





Answer: (a). True. The integral of a vector field along a path changes sign when thepath is reversed.


MVC.18.01.060

CC MP MA233 F06: 31/69CC KC MA233 F09: 69/31 time 3:00

256. The work done by the force field ~F = yi as an object moves along a straight line joining(1, 1) to (1,-1) is

(a) positive

143

(b) negative

(c) zero

Answer: (c). zero. Since the two points have the same x-coordinate, the straight linejoining them is vertical. On the other hand, the vector field points in the horizontaldirection, and so is always perpendicular to the path of the object. Thus it does nowork on the object as it moves.


MVC.18.01.070

CC KC MA233A F08: 0/0/100 time 2:30CC HZ MA233 F10: 0/0/100 time 2:00 Review

257. How much work does it take to move in a straight line from coordinates (1,3) to (5,3)

in the vector field ~F = −4i + 3j? Assume that coordinates are in meters and force isin Newtons.

(a) -25 Joules

(b) -16 Joules

(c) 7 Joules

(d) 16 Joules

(e) 25 Joules

Answer: (b). The object moves four meters in the positive x direction. The j compo-nent of the vector field is perpendicular to this, so it has no effect. The i componenthas magnitude 4, pointing opposite the direction of motion, so the object must do (4m)(4 N) = 16 J of positive work to oppose this and complete the motion.

by Project MathVote

MVC.18.01.080

CC KC MA233 F09: 0/94/0/6/0 time 2:30

18.2 Computing Line Integrals Over Parameterized

Curves

258. Which of the following is equivalent to the line integral of ~F (x, y) on the line segmentfrom (1,1) to (3,4)?

144

(a)∫

1

0~F (1 + 2t, 1 + 3t)dt

(b)∫

1

0~F (1 + 2t, 1 + 3t) · (2i + 3j)dt

(c)∫

1

0~F (3, 4) · (2i + 3j)dt

(d)∫

1

0~F (1 + t, 1 + t) · (2i + 3j)dt

Answer: (b). The proper form is∫ b

a~F (r(t)) · ~r′(t)dt. Here ~r(t) = (1 + 2t)i + (1 + 3t)j

and ~r′(t) = 2i + 3j.

by Mark Schlatter

MVC.18.02.010

CC KC MA233 F06: 37/63/0/0CC MP MA233 F07: 0/100/0/0 time 2:00CC KC MA233 F07: 0/100/0/0 time 3:30CC KC MA233A F08: 5/95/0/0 time 3:00CC KC MA233B F08: 0/93/7/0 time 3:00BJ BB MA305 F12: 24/56/8/12

259. Which of the following is equivalent to the line integral of ~F (x, y) on the line segmentfrom (1,1) to (3,4)?

(a)∫

2

0~F (1 + t, 1 + 1.5t) · (i + 1.5j)dt

(b)∫

2

0~F (1 + t, 1 + 1.5t) · (2i + 3j)dt

(c)∫

1

0~F (1 + t, 1 + 1.5t) · (i + 1.5j)dt

(d)∫

1

0~F (1 + t, 1 + 1.5t) · (2i + 3j)dt

Answer: (a). The proper form is∫ b

a~F (r(t)) · ~r′(t)dt. Here ~r(t) = (1 + t)i + (1 + 1.5t)j

(for 0 ≤ t ≤ 2) and ~r′(t) = i + 1.5j.

by Mark Schlatter

MVC.18.02.020


260. If C1 is the path parameterized by ~r1(t) = 〈t, t〉, 0 ≤ t ≤ 1, and if C2 is the path

parameterized by ~r2(t) = 〈t2, t2〉, 0 ≤ t ≤ 1, and if ~F = xi + yj, which of the followingis true?

(a)∫

C1

~F · d~r >∫

C2

~F · d~r

145

(b)∫

C1

~F · d~r <∫

C2

~F · d~r

(c)∫

C1

~F · d~r =∫

C2

~F · d~r

Answer: (c). The two parameterizations give the same path, in the same direction.Since a line integral does not depend on the way the path is parameterized, the twoline integrals are the same.


MVC.18.02.030

CC MP MA233 F06: 7/60/33CC KC MA233 F06: 37/58/5CC MP MA233 F07: 7/36/57 time 3:00CC KC MA233 F07: 12/48/40 time 2:40CC KC MA233A F08: 33/33/33 time 3:30CC KC MA233B F08: 10/45/45 time 3:00CC KC MA233 F09: 24/47/29 time 4:00HC AS MA203 F10: 10/45/45BJ BB MA305 F12: 50/12.5/37.5


parameterized by ~r2(t) = 〈1 − t, 1 − t〉, 0 ≤ t ≤ 1, and if ~F = xi + yj, which of thefollowing is true?

(a)∫

C1

~F · d~r >∫

C2

~F · d~r

(b)∫

C1

~F · d~r <∫

C2

~F · d~r

(c)∫

C1

~F · d~r =∫

C2

~F · d~r

Answer: (a). The two parameterizations give the same path, but one goes in the

opposite direction to the other. Since ~F points away from the origin, and C1 is orientedaway from the origin, the integral along C1 is positive, and along C2 is negative.


MVC.18.02.040

CC KC MA233 F06: 68/16/16CC KC MA233 F07: 84/4/12 time 3:00CC KC MA233A F08: 84/0/16 time 3:15CC KC MA233B F08: 69/0/31 time 3:30CC HZ MA233 F10: 100/0/0 ReviewHC AS MA203 F10: 40/5/55

146

BJ BB MA305 F12: 68/0/32


parameterized by ~r2(t) = 〈sin t, sin t〉, 0 ≤ t ≤ 1, and if ~F = xi + yj, which of thefollowing is true?

(a)∫

C1

~F · d~r >∫

C2

~F · d~r

(b)∫

C1

~F · d~r <∫

C2

~F · d~r

(c)∫

C1

~F · d~r =∫

C2

~F · d~r

Answer: (a). The two parameterizations move along the line y = x in the directionaway from the origin, but they have different endpoints. The path C1 goes from (0,

0) to (1, 1) and the path C2 goes from (0, 0) to (sin 1, sin 1) ≈ (0.84, 0.84). Since ~Fpoints away from the origin, and both paths are oriented so that the direction of travelis oriented away from the origin, the integral along the longer path is larger.


MVC.18.02.050

CC KC MA233 F07: 33/42/17 time 3:30CC KC MA233A F08: 95/0/5/ time 3:00CC KC MA233B F08: 61/15/23 time 3:00BJ BB MA305 F12: 22.73/27.27/50

263. Consider the path C1 parameterized by ~r1(t) = (cos t, sin t), 0 ≤ t ≤ 2π and the path

C2 paramterized by ~r2(t) = (2 cos t, 2 sin t), 0 ≤ t ≤ 2π. Let ~F be a vector field. Is it

always true that∫

C2

~F · d~r = 2∫

C1

~F · d~r?

(a) Yes

(b) No

Answer: (b). No. C1 is a circle of radius 1 centered at the origin and C2 is a circleof radius 2 centered at the origin. There is no reason why the integral around a circleof radius 2 should be twice the size of the integral around a circle of radius 1. Theintegral depends on the values of the vector field along the path, not just on the shapeof the path.


MVC.18.02.060

147

CC KC MA233A F08: 79/21 time 3:30

264. Consider the path C1 parameterized by ~r1(t) = (cos t, sin t), 0 ≤ t ≤ 2π and the path

C2 paramterized by ~r2(t) = (cos 2t, sin 2t), 0 ≤ t ≤ 2π. Let ~F be a vector field. Is it

always true that∫

C2

~F · d~r = 2∫

C1

~F · d~r?

(a) Yes

(b) No

Answer: (a). Yes. C1 is a circle of radius 1 centered at the origin, traversed once inthe counterclockwise direction, and C2 is the same circle traversed twice. Thus theintegral around C2 will be twice the integral around C1.


MVC.18.02.070

18.3 Gradient Fields and Path-Independent Fields

265. The vector field shown is the gradient vector field of f(x, y). Which of the followingare equal to f(1, 1)?

(a) f(1,−1)

(b) f(−1, 1)

(c) both of the above

148


Answer: (c). It is easy to see that the line integrals along the straight-line paths from(1,1) to (-1,1) and from (1,1) to (1,-1) are both equal to 0. Thus, by the fundamentaltheorem of line integrals, both f(1,−1) and f(−1, 1) are equal to f(1, 1).

by Mark Schlatter

MVC.18.03.010

CC KC MA233 F06: 62/10/28/0CC KC MA233 F07: 0/0/100/0 time 4:00CC KC MA233A F08: 92/0/8/0 time 2:20CC KC MA233B F08: 43/0/57/0 time 2:40CC KC MA233 F09: 44/22/28/6 time 2:30

266. Which of the vector fields below is not path independent?




Answer: (a). You can draw a closed curve around which the line integral will benon-zero, such as a circle around the origin.

by Mark Schlatter

MVC.18.03.020

CC MP MA233 F07: 90/0/10 time 2:00CC KC MA233 F07: 94/0/6 time 4:00CC KC MA233A F08: 83/0/17 time 3:00CC KC MA233B F08: 93/0/7 time 4:00CC KC MA233 F09: 84/5/11 time 2:00BJ BB MA305 F12: 20/15/65

149

267. Which of the following explains why this vector field is not a gradient vector field?

(a) The line integral from (-1,1) to (1,1) is negative.

(b) The circulation around a circle centered at the origin is zero.

(c) The circulation around a circle centered at the origin is not zero.


Answer: (c). Since the field is not path independent, it cannot be a gradient vectorfield.

by Mark Schlatter

MVC.18.03.030

268. The line integral of ~F = ∇f along one of the paths shown below is different from theintegral along the other two. Which is the odd one out?

150

(a) C1

(b) C2

(c) C3

Answer: (c). C3. The two paths C1 and C2 give the same line integral by the Funda-mental Theorem of Calculus for line integrals, since they have the same starting andending points. The integral over C3 could be different because it goes in the oppositedirection, so the starting and ending points are swapped.


MVC.18.03.040

CC MP MA233 F07: 0/0/100 time 1:00CC HZ MA233 F10: 11/0/91 time 1:45 Review

269. The figure below shows the vector field ∇f , where f is continuously differentiable inthe whole plane. The two ends of an oriented curve C from P to Q are shown, but themiddle portion of the path is outside the viewing window. The line integral

∫

C∇f · d~r

is

151

(a) Positive

(b) Negative

(c) Zero

(d) Can’t tell without further information

Answer: (a). The line integral of a gradient field is path independent, so this integralis the same as a straight line path from P to Q.


MVC.18.03.050

CC MP MA233 F07: 90/10/0/0 time 2:00CC KC MA233 F07: 37/19/43/0 time 4:00CC KC MA233 F09: 74/21/5/0 time 2:00CC HZ MA233 F10: 56/39/6/0 time 2:30 Review

270. Which of the diagrams contain all three of the following: a contour diagram of afunction f , the vector field ∇f of the same function, and an oriented path C from Pto Q with

∫

C∇F · d~r = 60?

152

(a) I

(b) II

(c) III

(d) IV

Answer: (b). II.


MVC.18.03.060

271. If f is a smooth function of two variables that is positive everywhere and ~F = ∇f ,which of the following can you conclude about

∫

C~F · d~r?

(a) It is positive for all smooth paths C.

(b) It is zero for all smooth paths C.

(c) It is positive for all closed smooth paths C.

(d) It is zero for all closed smooth paths C.

Answer: (d). (d) is true since gradient fields are path-independent, by the FundamentalTheorem of Calculus for Line Integrals. This contradicts (c), so that can’t be true. Asfor (a) and (b), the integral of f along any path is the difference between the value off at the ending point and the value of f at the starting point. Since the difference oftwo positive numbers could be any number, neither (a) nor (b) is true always.

153


MVC.18.03.070

272. What is the potential function for the vector field ~F = 2yi + 2xj?

(a) f(x, y) = 4xy

(b) f(x, y) = 2x2 + 2y2

(c) f(x, y) = 2xy

(d) This is not a conservative vector field.

Answer: (c).

by Project MathVote

MVC.18.03.080

18.4 Path-Dependent Vector Fields and Green’s The-

orem

273. What will guarantee that ~F (x, y) = yi + g(x, y)j is not a gradient vector field?

(a) g(x, y) is a function of y only

(b) g(x, y) is a function of x only

(c) g(x, y) is always larger than 1

(d) g(x, y) is a linear function

Answer: (a). Since the curl of ~F is (d/dx)g(x, y)− 1, we can guarantee that the curlis nonzero if g(x, y) is a function of y only.

by Mark Schlatter

MVC.18.04.010

CC KC MA233 F09: 90/5/5/0 time 2:30

274. The figure shows a curve C broken into two pieces C1 and C2. Which of the followingstatements is true for any smooth vector field ~F ?

154

(a)∫

C~F · d~r =

∫

C1

~F · d~r +∫

C2

~F · d~r(b)

∫

C~F · d~r =

∫

C1

~F · d~r −∫

C2

~F · d~r(c)

∫

C1

~F · d~r =∫

C2

~F · d~r(d)

∫

C~F · d~r = 0

(e) More than one of the above is true.

Answer: (b).


MVC.18.04.020

CC KC MA233A F08: 32/21/0/0/47 time 4:00CC KC MA233B F08: 26/47/16/11/0 time 3:30CC KC MA233 F09: 32/5/11/5/47 time 4:00HC AS MA203 F10: 11/11/28/0/50

275. A smooth two dimensional vector field ~F = F1i + F2j, with ~F 6= ~0 satisfies ∂F2

∂x= ∂F1

∂y

at every point in the plane. Which of the following statments is not true?

(a)∫

C~F · d~r = 0 for all smooth paths C.

(b)∫

C1

~F ·d~r =∫

C2

~F ·d~r for any two smooth paths C1 and C2 with the same startingand ending points.

(c) ~F = ∇f for some function f .

(d) If C1 is the straight line from -1 to 1 on the y-axis and if C2 is the right half of

the unit circle, traversed counter-clockwise, then∫

C1

~F · d~r =∫

C2

~F · d~r.

155

(e) More than one of the above is not true.

Answer: (a). Statements (b), (c), and (d) are true. Green’s theorem tells us that ~Fis path-independent, and path-independent fields are gradient fields, so statements (b)and (c) are true. Statement (d) is just a particular case of statement (b). Statement(a) would be true of it had the extra condition that C was a closed path, but it is nottrue for all paths.


MVC.18.04.030

19.1 The Idea of a Flux Integral

276. A river is flowing downstream at a constant rate of 5 ft/s. We take a rectangular netthat is 6 ft wide and 3 ft deep and place it in the river so that a vector perpendicularto the net (a normal vector) is parallel to the velocity of the water. What is the rateat which water flows through the net?

(a) 0 ft3/s

(b) 15 ft2/s

(c) 30 ft2/s

(d) 90 ft3/s


Answer: (d). We simply multiply length times width times velocity. The purpose ofthis question is to introduce the concept of flux.

by Project MathVote

MVC.19.01.005

CC KC MA233 F09: 6/12/0/76/6 time 3:00CC MP MA233 F09: 0/0/0/100/0 time 3:00

277. A river is flowing downstream at a constant rate of 5 ft/s. We take a rectangular netthat is 6 ft wide and 3 ft deep and place it in the river so that there is a 30 degreeangle between a vector perpendicular to the net (a normal vector) and the velocity ofthe water. What is the rate at which water flows through the net?

(a) 0 ft3/s

156

(b) 90 ft3/s

(c) 45 ft3/s

(d) ≈ 78 ft3/s


Answer: (d). We multiply length times width times velocity times the cosine of theangle. The purpose of this question is to introduce the concept of flux.

by Project MathVote

MVC.19.01.006

CC KC MA233 F09: 0/6/18/76/0 time 2:30CC MP MA233 F09: 0/0/5/95/0 time 4:00

278. Through which surface is the flux of ~F (x, y, z) = 2i negative?

(a) A square of side length 2 in the yz plane, oriented in the negative x direction

(b) A square of side length 2 in the xz plane, oriented in the positive y direction

(c) A square of side length 2 in the yz plane, oriented in the positive x direction

(d) A square of side length 2 in the xz plane, oriented up.

Answer: (a). We need the normal to the surface to be in the −i direction.

by Mark Schlatter

MVC.19.01.010

CC KC MA233 F06: 94/0/6/0CC MP MA233 F07: 73/0/27/0 time 3:30CC KC MA233 F07: 90/10/0/0 time 2:40CC KC MA233A F08: 95/0/5/0 time 2:00CC KC MA233B F08: 100/0/0/0 time 2:00CC MP MA233 F09: 100/0/0/0 time 4:00BJ BB MA305 F12: 60.87/8.7/21.74/8.7

279. Through which surface is the flux of ~F (x, y, z) = xi the most positive?

(a) A square of side length 2 in the yz plane, oriented in the positive x direction

(b) A square of side length 2 in the plane x = 4, oriented in the positive x direction

(c) A square of side length 4 in the plane x = 2, oriented in the positive x direction

(d) A square of side length 1 in the plane x = 8, oriented in the positive x direction

157

Answer: (c). In all of these cases, the flux equals the area of the square times the xcoordinate.

by Mark Schlatter

MVC.19.01.020

CC KC MA233 F06: 6/11/78/6CC KC MA233 F07: 0/0/100/0 time 2:00CC KC MA233 F09: 41/6/53/0 time 3:00CC MP MA233 F09: 0/6/94/0 time 3:00

280. Consider the flux of ~F = xi through a disk of radius 1 oriented as described below. Inwhich case is the flux positive?

(a) In the yz-plane, centered at the origin and oriented in the direction of increasingx.

(b) In the plane x = 2, centered on the x-axis and oriented away from the origin.

(c) In the plane y = 2, centered on the y-axis and oriented away from the origin.

(d) In the plane x + y = 2, centered on the x-axis and oriented away from the origin.

(e) More than one of the above has positive flux.

(f) None of the above.

Answer: (e). The flux through the disks described in (b) and (d) are both positive.

The flux in (a) is zero, since ~F = ~0 on the yz-plane. The flux in (c) is zero, since ~F isparallel to the disk in this case.


MVC.19.01.030

CC KC MA233A F08: 29/10/0/5/57/0 time 5:30CC KC MA233B F08: 13/20/7/0/60/0 time 5:00BJ BB MA305 F12: 33.33/8.33/4.17/4.17/45.83/4.17

281. Consider the flux of ~F = yi through a disk of radius 1 oriented as described below. Inwhich case is the flux positive?

(a) In the yz-plane, centered at the origin and oriented in the direction of increasingx.

(b) In the plane x = 2, centered on the x-axis and oriented away from the origin.

(c) In the plane y = 2, centered on the y-axis and oriented away from the origin.

158

(d) In the plane x + y = 2, centered on the x-axis and oriented away from the origin.

(e) More than one of the above has positive flux.

(f) None of the above.

Answer: (f). None. The flux through the surfaces in parts (a)-(d) is zero. In parts (a),(b), and (d), the flux is zero because contributions from positive and negative y-valuescancel. The flux in part (c) is zero because the vector field is parallel to the disk.


MVC.19.01.040

CC KC MA233 F07: 0/5/30/15/50/0 time 4:30CC KC MA233 F09: 0/23/35/18/6/18 time 4:00BJ BB MA305 F12: 0/0/16.67/12.5/41.67/29.17

282. Which vector field has a positive flux through the surface below?

(a) ~F = xj

(b) ~F = yj

(c) ~F = −zi

(d) ~F = (z + x)i

Answer: (d).


159

MVC.19.01.050

CC KC MA233 F06: 0/0/22/78CC MP MA233 F07: 0/0/0/100 time 3:00CC HZ MA233 F11: 12/8/8/71 time 3:45CC HZ MA233 F12: 52/4/4/39BJ BB MA305 F12: 21.74/13.04/4.35/60.87

283. Which vector field has a positive flux through the surface below?

(a) ~F = −yk

(b) ~F = yj

(c) ~F = −zi

(d) ~F = xk

Answer: (a).


MVC.19.01.060

CC KC MA233 F07: 100/0/0/0 time 2:30CC HZ MA233 F12: 85/0/15/0

284. Let ~F = xi + yj + zk. Which of the surfaces below has positive flux?

(a) Sphere of radius 1 centered at the origin, oriented outward.

(b) Unit disk in the xy-plane, oriented upward.

(c) Unit disk in the plane x = 2, oriented toward the origin.


160

Answer: (a). This vector field points away from the origin, so there is a positive fluxout of the sphere.


MVC.19.01.070

CC MP MA233 F07: 40/53/7/0CC KC MA233A F08: 80/0/10/10 time 2:00CC KC MA233B F08: 20/47/20/13 time 3:45

285. Choose the vector field with the largest flux through the surface below.

(a) ~F1 = 2i− 3j − 4k

(b) ~F2 = i− 2j + 7k

(c) ~F3 = −7i + 5j + 6k

(d) ~F4 = −11i + 4j − 5k

(e) ~F5 = −5i + 3j + 5k

Answer: (d). We want the vector field with the most negative k component.


MVC.19.01.080BJ BB MA305 F12: 4.17/12.5/4.17/79.17/0

161

286. Choose the vector field with the largest flux through the surface below.

(a) ~F1 = 2i− 3j − 4k

(b) ~F2 = i− 2j + 7k

(c) ~F3 = −7i + 5j + 6k

(d) ~F4 = −11i + 4j − 5k

(e) ~F5 = −5i + 3j + 5k

Answer: (c). We want the vector field with the most positive j component.


MVC.19.01.090

CC KC MA233A F08: 0/0/100/0/0 time 1:00CC KC MA233B F08: 0/0/100/0/0 time 1:00

287. Which of the following vector fields has the largest flux through the surface of a sphereof radius 2 centered at the origin?

(a) ~F1 = ~ρ||~ρ||

(b) ~F2 = ~ρ||~ρ||2

(c) ~F3 = xj

(d) ~F4 = ~ρ||~ρ||

Answer: (d). The flux of ~F3 is zero. The other fields are normal to the sphere andconstant over its surface, so it is a matter of which one has the largest magnitude.||~F4|| = 4, ||~F2|| = 1/2, and ||~F1|| = 1.

162


MVC.19.01.100

19.2 Flux Integrals For Graphs, Cylinders, and Spheres

288. The flux of the vector field ~F = 4ρ through a sphere of radius 2 centered on the originis:

(a) 0

(b) 8π

(c) 16π

(d) 32π

(e) 64π


Answer: (e). This vector field is normal to the surface, and constant across the surface,so we multiply the magnitude of the vector field by the area of the surface. The areaof the sphere is 4πr2 = 36π, and the magnitude of the vector field is || ~F || = 2, so theflux is 72π.

by Carroll College

MVC.19.02.010

CC MP MA233 F07: 0/0/31/0/46/23 time 3:00CC KC MA233 F07: 5/0/65/0/30/0 time 4:00CC KC MA233A F08: 0/0/0/5/95/0 time 3:00CC KC MA233B F08: 40/0/0/47/13/0 time 2:30Question rewritten to change radius from 3: Spheres of raduis 3 have equal surfacearea and volume.CC KC MA233 F09: 6/12/76/0/0/6 time 3:00CC MP MA233 F09: 0/0/0/7/93/0 time 5:00

289. The flux of the vector field ~F = 3ρ + 2θ + φ through a sphere of radius 1/2 centeredon the origin is:

(a) 0

(b) 3π

(c) 9π

163

(d) 27π

(e) 72π


Answer: (b). The only component of ~F that is normal to the sphere is the ρ component,which is constant over the surface of the sphere, so we multiply this magnitude (3) bythe sphere’s area π, to get 3π.

by Carroll College

MVC.19.02.020

CC MP MA233 F07: 0/100/0/0/0 time 2:00CC KC MA233 F07: 0/89/11/0/0 time 6:00CC KC MA233A F08: 0/100/0/0/0 time 3:00CC KC MA233B F08: 0/67/27/6/0 time 3:00(e) addedCC KC MA233 F09: 6/76/18/0/0/0 time 6:00CC MP MA233 F09: 0/93/0/7/0/0 time 4:00

290. The flux of the vector field ~F = 4θ + 2z through a cylinder of radius 1 centered on thez axis, between z = 0 and z = 2:

(a) 0

(b) π

(c) 2π

(d) 3π

(e) 4π

Answer: (a). Zero. A cylinder has three surfaces. There is no flux through the sidesbecause there is no r component to the vector field. There is a flux through the topand bottom because there is a z component to the vector field. However, the fluxesthrough the top and bottom are equal and opposite, because the vector field producesa negative flux through the bottom, and an equal positive flux through the top.

by Carroll College

MVC.19.02.030

CC MP MA233 F07: 46/0/9/18/27 time 3:00CC KC MA233 F07: 36/0/25/0/40 time 5:30CC KC MA233A F08: 24/10/5/10/51 time 5:00CC KC MA233B F08: 20/0/0/7/73 time 4:30CC KC MA233 F09: 0/6/23/6/65 time 3:30CC MP MA233 F09: 70/0/5/10/15 time 5:00

164

291. The flux of the vector field ~F = rr + rθ through a cylinder of radius 2 centered on thez axis, between z = 0 and z = 2:

(a) 0

(b) 2π

(c) 4π

(d) 8π

(e) 16π

Answer: (e). 16π. There is no flux through the top and bottom of the cylinder,because there is no z component. The r component gives us a flux through the sides,and this component is constant over the sides, with a magnitude of 2. We multiplythis by the area of the sides, 2πrz = 22π(2)(2) = 8π, so the flux is 16π.

by Carroll College

MVC.19.02.040

CC KC MA233 F07: 0/0/0/0/100 time 5:00CC KC MA233A F08: 0/0/0/23/77 time 3:00CC KC MA233B F08: 0/0/7/33/60 time 3:30CC KC MA233 F09: 0/6/6/35/53 time 4:00CC MP MA233 F09: 5/0/0/25/70 time 7:00

292. The flux of the vector field ~F = zr through a cylinder of radius 1/2 centered on the zaxis, between z = 0 and z = 3 is

(a) 0

(b) 2π

(c) 3π

(d) 9

2π

(e) 9π

Answer: (d). 9

2π. There is no flux out of the top and bottom of the cylinder. zr

does produce a flux out of the sides, but this vector changes in magnitude across thesurface, so we must do an integral. The circumference of the sides is 2πr = π, so weintegrate

∫

3

0πzdz = 9

2π.

by Carroll College

MVC.19.02.050

CC MP MA233 F07: 0/7/29/21/43 time 5:00 good discussion, no introductionCC KC MA233 F07: 0/0/0/0/100 time 6:30

165

CC KC MA233A F08: 0/5/5/38/53 time 5:30CC KC MA233 F09: 0/5/21/47/26 time 6:00

293. All but one of the flux calculations below can be done with just multiplication, butone requires an integral. Which one?

(a) ~F = 3ρ + 2φ through a sphere of radius 4.

(b) ~F = ρρ + θφ through a sphere of radius 3.

(c) ~F = rr + rz through a disk of radius 2, centered on the z axis, in the z = 2 plane.

(d) ~F = r2r + zθ through a cylinder of radius 1, between z = 1 and z = 3.

Answer: (c). The r component produces no flux, but the z component is normal to thesurface. It requires an integral because this component changes magnitude at differentpoints on the disk. At the origin, it is zero, while on the perimeter it is 2.

by Carroll College

MVC.19.02.060

CC KC MA233 F06: 0/6/6/89CC MP MA233 F07: 0/0/0/100 time 3:00 this crushed themCC KC MA233 F07: 0/30/9/53 time 5:30CC KC MA233A F08: 0/0/54/46 time 3:30CC KC MA233B F08: 0/7/93/0 time 4:00CC KC MA233 F09: 0/5/5/90 time 4:00CC MP MA233 F09: 0/5/65/30 time 5:00

20.1 The Divergence of a Vector Field

294. Moving from the picture on the left to the picture on the right, what are the signs of∇ · ~F ?

166

(a) positive, positive, negative

(b) zero, positive, negative

(c) positive, negative, zero

(d) zero, negative, positive

Answer: (b). On the left, the vector field swirls around, but has no expansion orcontraction. In the center the vector field is expanding away from the x = 0 line. Onthe right, the vector field is contracting towards the y = −1 line.

by Mark Schlatter

MVC.20.01.010

CC KC MA233 F06: 0/100/0/0CC KC MA233 F07: 0/71/19/10 time 4:30CC KC MA233A F08: 29/62/0/9 time 3:00CC KC MA233B F08: 0/100/0/0 time 3:00CC KC MA233 F09: 22/72/6/0 time 4:00CC MP MA233 F09: 10/76/10/4 time 3:00CC HZ MA233 F11: 8/92/0/0 time 2:30CC TM MA233 F11: 5/95/0/0CC HZ MA233 F12: 0/100/0/0

295. If ~F (x, y, z) is a vector field and f(x, y, z) is a scalar function, which of the followingis not defined?

(a) ∇f

(b) ∇ · ~F + f

(c) ~F +∇f

(d) ∇ · ~F +∇f



Answer: (d). ∇· ~F is a scalar and ∇f is a vector: You can’t add a vector and a scalar.


MVC.20.01.020

CC KC MA233 F07: 5/36/0/59 time 3:30 (before (e) and (f) added)CC MP MA233 F09: 0/0/30/30/30/10 time 5:00CC HZ MA233 F11: 0/0/16/68/16/0 time 5:30CC TM MA233 F11: 0/5/5/26/47/26 time 4:00CC HZ MA233 F12: 4/8/0/67/17/4

167

296. If ~F (x, y, z) is a vector field and f(x, y, z) is a scalar function, which of the followingquantities is a vector?

(a) ∇ · ~F(b) ∇f · ~u(c) ∇ · ∇f

(d) (∇ · ~F )~F

Answer: (d).


MVC.20.01.030

CC MP MA233 F07: 7/0/0/93 time 3:00 review questionCC MP MA233 F09: 0/38/0/62 time 5:00CC HZ MA233 F11: 0/4/4/92 time 2:00CC TM MA233 F11: 0/31/5/63 time 1:45CC HZ MA233 F12: 4/33/0/63

297. True or False? If all the flow lines of a vector field ~F are parallel straight lines, then∇ · ~F = 0.





Answer: (False). A vector field can have expansion or contraction in the direction ofthe flow.


MVC.20.01.040

CC KC MA233 F06: 0/100

298. True or False? If all the flow lines of a vector field ~F radiate outward along straightlines from the origin, then ∇ · ~F > 0.



168



Answer: (False). If the flow rapidly decreases in magnitude as it moves away from theorigin, the divergence can be zero, or negative.


MVC.20.01.050

CC KC MA233 F06: 90/10CC KC MA233 F07: 86/14 time 3:00CC MP MA233 F07: 0/100 time 3:00 review questionCC KC MA233 F09: 85/15 time 2:00

299. In Cartesian coordinates given the vector field ~F = F1 i + F2j + F3k,

~∇ · ~F =∂F1

∂x+

∂F2

∂y+

∂F3

∂z

.

Which of the following vector fields has zero divergence, so that it could represent theflow of a liquid which does not expand or contract?

(a) ~F = 2 sin(3z2)i + 5xyzj + 3e7xk

(b) ~F = 3 ln(yz)i + 2x3z7j + 4 cos(2x)k

(c) ~F = 6e2y j + 3 sin(4z)k


Answer: (b).

by Project MathVote

MVC.20.01.060

CC HZ MA233 F11: 0/100/0/0 time 1:30CC TM MA233 F11: 0/82/5/12CC HZ MA233 F12: 0/100/0/0

300. In cylindrical coordinates given the vector field ~F = F1r + F2θ + F3z,

~∇ · ~F =1

r

∂(rF1)

∂r+

1

r

∂F2

∂θ+

∂F3

∂z

.

What is the divergence of the vector field ~F = 2θr + 3zθ + 4rz?

169

(a) ~∇ · ~F = 2θ + 3z + 4r

(b) ~∇ · ~F = 9

(c) ~∇ · ~F = 2θr

(d) ~∇ · ~F = 0


Answer: (c).

by Project MathVote

MVC.20.01.070

CC HZ MA233 F11: 0/0/64/32/4 time 2:30CC HZ MA233 F12: 4/0/46/50/0BJ BB MA305 F12: 15.79/68.42/5.26/10.53/0

301. In spherical coordinates given the vector field ~F = F1ρ + F2θ + F3φ,

~∇ · ~F =1

ρ2

∂(ρ2F1)

∂ρ+

1

ρ sin φ

∂F2

∂θ+

1

ρ sin φ

∂(F3 sin φ)

∂φ

.

What is the divergence of the vector field ~F = 3

ρ2 ρ + 2rθ?

(a) ~∇ · ~F = 2rρ sin φ

(b) ~∇ · ~F = 3

ρ2

(c) ~∇ · ~F = cos φρ sin φ

(d) ~∇ · ~F = 0


Answer: (d).

by Project MathVote

MVC.20.01.080

20.2 The Divergence Theorem

302. Given a small cube resting on the xy plane with corners at (0, 0, 0), (a, 0, 0), (a, a, 0),and (0, a, 0), which vector field will produce positive flux through that cube?

(a) ~F = 3i

170

(b) ~F = xi− yj

(c) ~F = 2i + 3j + k

(d) ~F = zk

Answer: (d). Only ~F = zk has non-zero divergence.

by Mark Schlatter

MVC.20.02.010

CC KC MA233 F06: 0/0/10/90CC KC MA233 F07: 0/9/18/73 time 3:30CC MP MA233 F07: 0/14/14/71 time 4:00 review questionCC KC MA233A F08: 0/5/0/95 time 3:30CC KC MA233B F08: 7/13/0/80CC KC MA233 F09: 0/0/16/84 time 4:30CC MP MA233 F09: 0/0/10/90 time 5:00BJ BB MA305 F12: 26.09/4.35/34.78/34.78

303. Let ~F = (5x + 7y)i + (7y + 9z)j + (9z + 11x)k. This vector field produces the largestflux through which of the following surfaces?

(a) S1, a sphere of radius 2 centered at the origin.

(b) S2, a cube of side 2 centered at the origin, with sides parallel to the axes.

(c) S3, a sphere of radius 1 centered at the origin.

(d) S4, a pyramid contained inside S3.

Answer: (a). ∇ · ~F = 5 + 7 + 9 = 21 and thus is constant everywhere. So the fluxthrough any closed surface is simply the volume contained by the surface multipliedby 21. The surface containing the largest volume is S1.


MVC.20.02.020

CC KC MA233 F07: 95/5/0/0 time 3:30CC MP MA233 F07: 64/36/0/0 time 3:00 review questionCC KC MA233A F08: 76/24/0/0 time 4:00CC KC MA233B F08: 73/27/0/0 time 4:30CC KC MA233 F09: 47/37/5/10 time 4:00BJ BB MA305 F12: 25/33.33/12.5/29.17

171

304. True or False? The vector field ~F is defined everywhere in a region W bounded by asurface S. If ∇ · ~F > 0 at all points of W , then the vector field ~F points outward atall points of S.





Answer: (False). Although there must be a net flux out of the surface, there may beregions of the surface where the flux is inward.


MVC.20.02.030

CC KC MA233 F07: 55/45 time 3:30BJ BB MA305 F12: 25/50/16.67/8.33

305. True or False? The vector field ~F is defined everywhere in a region W bounded by asurface S. If ∇ · ~F > 0 at all points of W , then the vector field ~F points outward atsome points of S.





Answer: (a). True. In order to produce a positive net flux out of the surface S, thevector field must point outward at some points.


MVC.20.02.040BJ BB MA305 F12: 41.67/45.83/8.33/4.17

306. True or False? The vector field ~F is defined everywhere in a region W bounded by asurface S. If

∫

S~F · d ~A > 0, then ∇ · ~F > 0 at some points of W .



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Answer: (a). True. If there is a positive flux out of the surface, then the integral of thedivergence over the volume must be positive. This means that there must be points inthe volume with positive divergence.


MVC.20.02.050

307. Which of the following vector fields produces the largest flux out of the unit spherecentered at the origin?

(a) ~F1 = (ez + x3)i + exj + y3k

(b) ~F2 = (z2 + cos y)i− y3j + x3y3k

(c) ~F3 = z2 i− (x2 + z2)j + (z3 + zy2)k

(d) ~F4 = (x4 − y4)i− (z4 − 2x3y)j + (y4 − 2x3z)k

Answer: (c). ∇ · ~F1 = 3x2, ∇ · ~F2 = −3y2, ∇ · ~F3 = 3z2 + y2, ∇ · ~F4 = 4x3. The

divergence with the largest average value inside the unit sphere is ~F3.


MVC.20.02.060

CC KC MA233 F07: 5/5/60/30 time 6:00CC KC MA233A F08: 9/0/67/24 time 3:30CC KC MA233B F08: 7/0/40/53 time 3:30CC KC MA233 F09: 5/5/65/25 time 5:00

20.3 The Curl of a Vector Field

308. The pictures below show top views of three vector fields, all of which have no z com-ponent. Which one has the curl pointing in the positive k direction at the origin?

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(d) none of them

Answer: (b). At (0,0,0), a paddle wheel pointing in the positive k direction in thefield in the middle would rotate counterclockwise, since the vectors on the right wouldoverpower the vectors on the left.

by Mark Schlatter

MVC.20.03.010

CC KC MA233 F07: 0/29/0/71 time 4:30CC MP MA233 F07: 29/57/0/14 time 4:00 review questionCC KC MA233A F08: 0/74/26/0 time 4:00CC KC MA233B F08: 19/25/25/31 time 3:30CC KC MA233 F09: 5/10/30/55 time 3:30BJ BB MA305 F12: 20/50/15/15

309. Let ~F (x, y, z) be a vector field and let f(x, y, z) be a scalar function. If ~r = xi+yj+zk,which of the following is not defined?

(a) ∇× f

(b) ∇× ~F +∇f

(c) ∇× (~r ×∇f)

(d) f +∇ · ~F(e) More than one of the above

Answer: (a). We can only take the curl of a vector field. A scalar function has no curl.


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310. Which one of the following vector fields has a curl which points purely in the j?

(a) yi− xj + zk

(b) yi + zj + xk

(c) −zi + yj + xk

(d) xi + zj − yk

Answer: (c). The curls are −2k, −i− j − k, −2j, −2i.


MVC.20.03.030

CC KC MA233A F08: 4/17/79/0 time 3:00CC KC MA233B F08: 0/0/75/25 time 4:00CC KC MA233 F09: 0/15/75/10 time 3:30CC MP MA233 F09: 0/5/95/0 time 4:00

311. True or False? If all the flow lines of a vector field ~F are straight lines, then ∇× ~F = 0.





Answer: (False). Imagine a straight river where water moves downstream faster nearthe left bank than near the right bank. The difference in water speed causes an objectin the river to spin to the right as it floats straight downstream. The spinning is anindication of curl, even though the flow is straight.


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CC MP MA233 F07: 0/100 time 2:00 review questionCC KC MA233 F09: 6/94 time 2:30CC MP MA233 F09: 20/80 time 3:00

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312. True or False? If all the flow lines of a vector field ~F lie in planes parallel to thexy-plane, then the curl of ~F is a multiple of k at every point.





Answer: (False). A counterexample is ~F = zi. Because the vector field changes as wemove in the z direction, it can have non k curl.


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20.4 Stokes’ Theorem

313. Which of the following facts about ~F = ρρ is implied by Stokes’ Theorem?

(a) The line integral from (0,0,0) to (1,1,1) is equal to 3/2.

(b) ~F has positive divergence everywhere.

(c) The line integral on any closed curve is zero.

(d) The curl of ~F is non-zero.

Answer: (c). (a) and (b) are true, but only (c) is implied by Stokes’ Theorem. Since~F clearly has zero curl, by Stokes’ Theorem, the line integral on any closed curve iszero.

by Mark Schlatter

MVC.20.04.010

CC KC MA233A F08: 4/4/71/21 time 4:00CC KC MA233B F08: 0/0/100/0 time 3:40CC KC MA233 F09: 0/20/55/25 time 2:30

314. What can be said about the vector field ∇f in terms of curl?

(a) Its curl is negative.

(b) Its curl is zero.

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(c) Its curl is positive.

(d) Its curl depends on the function f .

Answer: (b). Since the line integral of grad f on any closed curve must be zero, Stokes’Theorem implies (under the right circumstances) that the curl of grad f must be zero.

by Mark Schlatter

MVC.20.04.020

CC KC MA233A F08: 0/50/14/36 time 2:00BJ BB MA305 F12: 0/8.7/4.35/86.96

315. The figure below shows the vector field ∇ × ~F . No formula for the vector field ~F isgiven. The oriented curve C is a circle, perpendicular to ∇× ~F . The sign of the lineintegral

∫

C~F · d~r

(a) is positive.

(b) is negative.

(c) is zero.

(d) can’t be determined without further information.

Answer: (b). Negative. The surface’s normal vector points downward, while the curlpoints up. Thus the flux of the curl through the surface is negative, and so Stokes’Theorem tells us that the circulation around C must be negative as well.


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CC KC MA233 F07: 20/20/40/15 time 4:00CC KC MA233A F08: 17/70/13/0 time 4:00CC KC MA233B F08: 0/21/79/0 time 5:00CC KC MA233 F09: 17/17/61/5 time 3:30CC HZ MA233 F11: 8/61/23/8 time 3:00CC TM MA233 F11: 10/74/10/5 time 3:00CC HZ MA233 F12: 5/86/5/5BJ BB MA305 F12: 18.18/50/27.27/4.55

316. The figure below shows the vector field ~F . The surface S is oriented upward andperpendicular to ~F at every point. The sign of the flux of ∇× ~F through the surface

(a) is positive.

(b) is negative.

(c) is zero.

(d) can’t be determined without further information.

Answer: (c). Zero. Because this vector field is perpendicular to the surface S atevery point, there must be no circulation around the perimeter of S. Thus, by Stokes’Theorem, the flux of ∇× ~F through the surface must be zero as well.


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CC KC MA233A F08: 30/0/39/30 time 3:40CC KC MA233B F08: 50/0/50/0 time 4:00CC KC MA233 F09: 21/5/68/5 time 3:00CC HZ MA233 F11: 20/4/68/8 time 2:45CC TM MA233 F11: 37/0/42/21CC HZ MA233 F12: 17/13/63/0/8

317. The vector field ~F has curl ∇ × ~F = 3i + 4j + 2k. What is the magnitude of thecirculation of ~F around the perimeter of the square with corners at coordinates (1,2,3),(4,2,3), (4,2,6), and (1,2,6)?

(a) 0

(b) 18

(c) 27

(d) 36

(e) 81


Answer: (d). 36. Stokes’ Theorem tells us that the circulation of a vector field aroundthe perimeter of a surface is equal to the flux of the curl of this vector field through thesurface. In this case, the surface is in the y = 2 plane, so only the j component of thecurl contributes to the flux, which is a constant 4 in magnitude. Because it is constant,we can simply multiply this by the area of the surface, and so we get 4× 9 = 36.


MVC.20.04.050

CC KC MA233A F08: 9/0/0/83/9/0 time 4:00CC KC MA233B F08: 29/0/0/57/7/7 time 4:00CC KC MA233 F09: 32/10/16/42/0/0 time 4:30After F10, the words “the magnitude of” were added to the question.CC HZ MA233 F11: 19/0/0/81/0/0 time 4:00CC TM MA233 F11: 15/0/0/80/5/0 time 4:00CC HZ MA233 F12: 18/0/0/82/0/0

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