Math 320/321 Exam

Math 320/321 Exam 1b – Spring 2021 – Solutions 1. (6 points) In the figure below is the contour diagram of f (x, y). Sketch the contour diagram for the function g(x, y) defined by g(x, y) = −f (x, −y). The contour diagram for g(x, y) will be the reflection of of the contour diagram of f , only reflected about the x-axis and having labels negated. y 12 11 10 x 9 8 2. (9 points) A train is traveling northwest at 10 miles per hour. A person in the train walks at 2 miles per hour from a window on the left side to a window directly across the train on the right side. (Left and right refer to the sides relative to a person facing the front of the train.) Assume that ~i points east, and ~j points north. Express your answers in terms of these unit vectors (i.e. resolve these vectors). Here we have |~v | = 10 and |w| ~ = 2 as pictured below. v represents the velocity vector for the train and w represents the velocity vector for the person walking. w v y x (a) What is the velocity vector of the train? √ √ ~v = −10 cos(45◦ )~i + 10 sin(45◦ )~j = −5 2~i + 5 2~j (b) What is the velocity vector of the person relative to the train? √ √ w ~ = 2 cos(45◦ )~i + 2 sin(45◦ )~j = 2~i + 2~j (c) What is the velocity vector of the person relative to the ground? √ √ ~v + w ~ = −4 2~i + 6 2~j 3. (6 points) Sketch a contour diagram for the function f (x, y) = y − e−x with at least four labeled contours. Describe in words how these contours are spaced. Answers will vary. The contours are not equally spaced. They are closer to each other as one moves left on the x-axis. The further one moves the right, the closer the functions are to being equally spaced (since e−x → 0 as x → ∞). y 3 2 f =1 1 −1 1 2 3 4x f =0 −1 f = −1 −2 f = −2 −3 4. (9 points) Describe completely in words the three different geometric objects represented by these sets of points in three space. An equation or formula is not sufficient. In a complete sentence, explain what these objects look like in words!! (a) The set of points whose distance from the origin is 5. This is a sphere of radius 5 centered at the origin. (b) The set of points whose y-coordinate is 7. This is a (vertical) plane 7 units above (which means ‘to the right of‘) the xz-plane. (c) The set of points whose x-coordinate is -2 and z-coordinate is 12. This is the line parallel to the y-axis, but at a height of 12 units above the xy-plane and 2 units behind/below the yz-plane. 5. (9 points) Company X sells widgets. Annual sales of widgets, S (in dollars), is a function of the number salespeople employed e and the number of damaging newspaper articles n published each year. That is, S = f (e, n). (a) Is the partial derivative Se (100, 50) positive, negative, or zero? Defend your answer. Se (100, 50) should be positive since adding a 101st salesperson should produce additional widget sales. (b) Is the partial derivative Sn (100, 50) positive, negative, or zero? Defend. Sn (100, 50) should be negative since having 51 (rather than 50) negative newspaper articles should harm annual widget sales. (c) Interpret the meaning of the statement Se (100, 50) = 300. If you were the CEO of this company, how might you respond to this piece of information? Se (100, 50) = 300 means that the change in sales company X expects to see by hiring a 101st salesperson is only $300. With such a tiny additional sales number, a CEO would be advised to not hire this additional salesperson since that person’s annual salary is likely to be much larger than this paltry increase in annual sales. 6. (6 points) Are the statements below true or false? Give reasons for your answers. (a) Any surface which is the level surface of a three-variable function g(x, y, z) can also be represented as the graph of a two-variable function. False. Take g(x, y, z) = x2 + y 2 + z 2 as an example. The level surface g = 1 is the unit sphere which cannot be the graph of a two-variable function. (b) If ~v · w ~ < 0 then the angle between ~v and w ~ is obtuse (i.e. greater than 90 degrees). True. If ~v · w ~ < 0 then cos θ < 0 since the lengths of ~v and w ~ are both positive. Since θ, the angle between the vectors, is between 0 and 180 degrees, we know θ is in either quadrant I or II. But cos θ is negative only in quadrant II. Math 320/321 Exam 2b – Spring 2021 – Solutions 1. (10 points) The path of a particle in space is given by the functions x(t) = 2t, y(t) = cos(t), z(t) = sin(t). Suppose the temperature in this space is given by a function T (x, y, z). , the rate of change of the temperature at the particle’s position. (Since the actual (a) Find dT dt function T (x, y, z) is not given, your answer will be in terms of derivatives of T ). ∂T dx ∂T dy ∂T dz ∂T ∂T ∂T dT = + + =2 + (− sin t) + (cos t) dt ∂x dt ∂y dt ∂z dt ∂x ∂y ∂z ∂T ∂T dT ∂T > 0, < 0, and > 0. At t = 0, is positive, (b) Suppose we know that at all points ∂x ∂y ∂z dt negative, or zero? dT ∂T ∂T ∂T ∂T ∂T We have |t=0 = 2 + (− sin 0) + (cos 0) =2 + > 0. dt ∂x ∂y ∂z ∂x ∂z 2. (8 points) Let g(x, y) = x2 + 2y 2 . The graph of g(x, y) is shown. (a) True or False? ∇g(1, 1) is normal to the graph of g. False. ∇g(1, 1) = 2x~i + 4y~j =⇒ ∇g(1, 1) = 2~i + 4~j. Notice there is no ~k component. g is a function of two variables. (b) If you answered True to part (a), determine the equation of the tangent plane to the graph of g at (1, 1, 3). If you answered False to part (a), determine a vector that is normal to the graph of g at (1, 1, 3). If z = x2 + 2y 2 , then x2 + 2y 2 − z = 0. We can then define F (x, y, z) = x2 + 2y 2 − z. Then, the surface is a level surface of F . ∇F (x, y, z) = 2x~i + 4y~j − ~k =⇒ ∇F (1, 1, 3) = 2~i + 4~j − ~k This vector, 2~i + 4~j − ~k, is normal to the graph at (1, 1, 3). 3 4 3. (8 points) Values of f (x, y) appear in the table below. Use it to estimate f~u (0, 0) if ~u = ~i − ~j. 5 5 HH H y x HH H 0 4 8 H H 0 3 6 100 90 81 85 79 68 65 61 55 f~u (0, 0) ≈ f~u (0, 0) = ∇f (0, 0) · ~u 3 = (fx (0, 0)~i + fy (0, 0)~j) · ( ~i − 5 10~ 15~ 3~ 4~ ≈ (− i − j) · ( i − j) 3 4 5 5 = −2 + 3 = 1 4~ j) 5 4. (12 points) For each of the following, find a vector normal to (a) the plane 2x − 3y + 4z = 0. ~n = 2~i − 3~j + 4~k (b) the vector ~v = ~i + 2~j. Answers will vary. ~n = 2~i − ~j will work. (c) the projection of ~v = 2~i + ~k onto ~u = ~i − ~k. Answers may vary. Any vector perpendicular to ~u will work. For example, ~n = ~i + ~k or ~n = ~j both work. (d) the cross product of ~v = 4~i + 6~j − ~k and w ~ = 2~i − ~j. Answers may vary. Two relatively obvious possible answers are ~v = 4~i+6~j−~k and w ~ = 2~i−~j. 5. (4 points) Measured in bushels per acre, the contour map below gives corn production C = f (R, T ) as a function of total rainfall, R, in inches, and average temperature, T , in degrees Fahrenheit, during the growing season. Estimate fR (12, 78). Be sure to include units in your response. 90 − 80 15 − 12 bushels per acre ≈ 3.33 inch of rainf all fR (12, 78) ≈ 6. (8 points) The surface z = g(x, y) is shown below. (a) Is −∇g(2, 5) i. a scalar a, ii. a vector in the form a~i + b~j, or iii. a vector in the form a~i + b~j + c~k for some real numbers a, b, and c? Choose one: (i) (ii) (iii) (ii) – notice that g is a function of two variables (b) Depending on your response to (a) above, discuss the sign(s) of a, b, and or c in (a). That is, determine whether these values are positive, negative, or zero. The opposite of the gradient vector will lie in the xy-plane. It will point in the direction of steepest decrease. In this case, that direction has a > 0 and b = 0. (c) Determine (i.e. resolve) a normal vector ~n for the tangent plane to this surface at (0, 12, 4). One normal vector would be simply ~k since the tangent plane is flat (parallel to the xyplane). In fact, any scalar multiple of that works. By opening the window, you are allowing cold outdoor air to mix with the warm indoor air, this will inevitably lower the temperature of the air inside the cabin and continue to lower the temperature as time goes on. The closer you are to the window the colder the temperature will be. So as the distance from the window increases, temperature will increase and as amount of time since the window was opened increases, the temperature will decrease. M(20, 0.07) = $396.02 The monthly payment on a 60-month loan of $20,000 with an annual rate of 7% is $396.02. M(30, 0.03) = $539.06 M(20, 0.07) + $150 = $396.02 + $150 = $546.02 ? $546.02 > $539.06  True It is true that M(20, 0.07) + $150 > M(30, 0.03) because $546.02 is greater than $539.06. Even though $30,000 is a larger loan to initially take out, since the $20,000 loan has a higher annual rate, along with the additional $150 charge a month, it has a larger monthly payment due. Math 320/321 Weekly Homework 2 – Solutions Due Friday, January 22 1. The diagram below shows the contour map for a circular island. Sketch the/any vertical cross-section of the island through the center. Your sketch should show concavity clearly. 2. For the graph of x2 + z 2 = 1 shown, draw three graphs – one for the trace x = −0.5, one for the trace x = 1, and a third for the trace y = 2. The trace x = −0.5: 2 z z= 1 −4 −3 −2 −1 q 3 4 1 2 q z = − 34 −1 y 3 4 −2 The trace x = 1: 2 z 1 z=0 −4 −3 −2 −1 1 −1 −2 2 3 y 4 The trace y = 2: 2 z 1 x2 + z 2 = 1 −2 −1 1 2 x −1 −2 3. For the graph of z = x2 shown, draw four graphs – one for the trace x = −0.5, one for the trace x = 1, a third for the trace y = 2, and a fourth for the trace y = 0. The trace x = −0.5: 2 z 1 z= −2 −1 1 1 4 2 y −1 −2 The trace x = 1: 2 z z=1 1 −2 −1 1 −1 −2 2 y The traces y = 0 and y = 2: 4 z 3 2 z = x2 1 −3 −2 −1 1 2 3 x −1 4. (a) You stand on the graph of the function at the point (-1, -1, 2) and start walking parallel to the x-axis (in the positive direction). Describe where you start and what you see when you look right and left. I am moving downhill, but it looks flat to my left and right as I do so. (b) You stand on the graph of the function at the point (0.75, 0, 1) and start walking parallel to the y-axis (in the positive direction). Describe where you start and what you see when you look right and left. It is flat as I move forward, but to my left it is a valley and to my right the ground elevates (a hill). 5. The figure at the right shows part of the contour diagram of P = f (N, V ). (a) Complete the diagram for N < 0 if f (−N, V ) = f (N, V ). Reflect the graph about the V -axis. The labels will stay the same. (b) Complete the diagram for N < 0, V < 0 if f (−N, −V ) = −f (N, V ). Reflect the graph about the origin. The labels will become negative in the third quadrant. Math 320/321 Weekly Homework 3 – Solutions Due Friday, January 29 1. GeoGebra is good at many things, but implicitly plotting equations in three variables such as x2 +y 2 = (2 + sin(z))2 is not one of them. (a) Try plotting this equation using GeoGebra3d. What happens? Submit a screen capture of something ‘easy’ such as an implicit plot of x2 + y 2 + 4z 2 = 1 to prove the program behaves badly with x2 + y 2 = (2 + sin(z))2 . (b) Describe as best you can the contours for the surface defined by x2 + y 2 = (2 + sin(z))2 . (c) Describe (shapes of) the cross-sections (i.e. when x = k and y = k) for the surface defined by x2 + y 2 = (2 + sin(z))2 . For which values of k do cross-sections even exist? When do they not exist? If you wish to submit screen captures of a few of these, that might be quite helpful in better understanding the surface. (d) One tool that does a decent job of plotting this implicitly described surface reasonably well is CalcPlot3D. By selecting Implicit Surface and de-selecting Function, produce a graph of the surface described by x2 + y 2 = (2 + sin(z))2 . View it on −4 ≤ x ≤ 4, −4 ≤ y ≤ 4, −6 ≤ z ≤ 6. Use more cubes/axis to get a finer picture. Post a screenshot or two of the image you construct. Does this image match the information you found in (b)? (a) Only the empty plot appears. (b) The contours will all be circles. They oscillate in radius between 1 and 4. (c) The cross-sections (for k small enough, k ≤ 9) will look like one of the two figures below. (d) The surface looks somewhat like a vase…lots of symmetry and periodicity! 2. The link https://www.geogebra.org/m/ee5nfsq6 is fun and educational. Prove that you know how to add vectors by submitting a screen capture showing that you are correct. 3. A particle moving with speed v hits a barrier at an angle of 30◦ and bounces off at an angle of 30◦ in the opposite direction with speed reduced by 30 percent. Find the velocity vector of the object after impact. y Before impact After impact 30◦ 30◦ x The velocity before impact is ~v = −v cos 30◦~i − v sin 30◦~j. This means the velocity after impact is √ given by w ~ = −0.7v cos 30◦~i + 0.7v sin 30◦~j = −0.35 3v~i + 0.35v~j ≈ −0.6v~i + 0.35v~j. • ·�—-· — – — – ., ., ‘ …. .. …….. -· – –· —–..___�, . >-.,, ·— – • –1– I� . – -·–·· — —- �-�”s::_C:= �-‘——– -. –·•1 —..�——-\–·-·—r.�–L———-�–+’– ·- -·· -· �- – -i_’�’ —-;————- ·–�-�yJ .. -� .J?L_f (�.�)–��–32r t. � 4________ -3 -2 – l O 2. ‘l —i-_;_——=–:,__�____:_-=—-=————-· ·•·-r————– -‘ . ______1,.__________ ·-·-··· —-1— – .. . .. -· ·-· — ·- –·—-·– -···——— ,–·-· –· —-&——- –·—···-·-·–·—· –.-‘,’f ·-· ————·· ‘·– — — –li– —·-· — . -·———� oP v . —– ;.:::,-·· ··.··- —-·· ____ Get�K 1¼, Mdj”.�� -�� -y,J, .i�_ 100 \bs.,… �:�.!”.”�i.� ·-· •” • ····-·–········—· … ·—�i�\ -�\�’? �-�-�.J.QO l’b, �” _______ .Jov1′(\_ � �e. o�� to � ::tN, -=- CIJff: �ro/Y\ tolling -··——–�$e ____ • �- J -�-L ..+ tlj_±_�–�——���–:: \\�l\\\-Jl\ _____ _ ____ t.” �1,1____�–· ·_____ 11 :::__ (�) u�lt ())ie_____ — -·-·-it- —�·-._::.,.—···· ·-· ··-······ ·-·�- II ;i \ I … 1 —4–�J1-; •.. —— ·, f, a ·ll\ _1·_. .,. _..;:.._ ::1!:.·””�””’�-‘(°t\-1″ ….. __ 1,� H.,,.. ,..-1 � s,10� “”�-\s 4 ._-__-_—+i� -����� �iii;��:�Mf1:__C-_�t”I–. -��j;-¥ !v��-�7-02_ ��-�t-�—�-_;_-C,D-�-.-,.- 1—–+-w� _ ar-t, givv-_ ..+� .___–4jf-�.1,wil$ f\.-.(; \I _u.11 •— _w�l .n1\\ “1 � ��- -lk � �”°�� . :;_� q� 61 _ -t-o �1 , a’)L __ P\141-\{f� � I I \J II �fH\P�- -•� f.,.. 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