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MTH 234 Solutions to Exam 2 April 9th, 18 Standard Response Questions Show all work to receive credit Please BOX your nal answer 1(14 points) Consider the function f(x;y) = 2x2 4xy y4 2 (a)Find the critical points of fand classify them as local minima, local maxima, or saddle pointsF € ˆ ¡ ¢ b £ ⁄ ¥ e Q œ ‰ " # J $ 7 J % 7 s & E B;Yka1yk−1a2yk−2a3yk−3=b1uk−1b2uk−2b3uk−3 We're going to do the conversion by first recognizing that the transfer function of this system is, G(z) = b1z2 b2z b3 z3 a1z2 a2z a3 = Y(z) U(z) Break up transfer function into two parts G p(z) =V(z)/U(z) contains all of the poles G p(z) = 1 z3 a1z2 a2z a3

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P((X,Y) ∈ A) = Z Z A f(x,y)dxdy End of lecture Mon, Oct 16 The twodimensional integral is over the subset A of R2 Typically, when we want to actually compute this integral we have to write it as an iterated integral It is a good idea to draw a picture of A to help do this#Õ @ ¶ / K æ _ K Z C T I 8 r K S Q G Û T P(z,x,y) Q(Px,Pz,Py) 같은 원리로 Y축 기준으로 회전한다 생각하면, (지금 y축은 모니터를 속으로 들어가는 방향이라고 3D적으로 상상을 해보라)(xz가 아니라 zx기 때문에 얘만 생긴 게 좀 틀리다) y' = y z' = Pz * cos($) Px * sin($) x' = Px * cos($) Pz * sin($) z x "벡터

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It is z 0 0 (ZZ) 1Z0Y Now let's prove that ^ = (Z0Z) 1Z0Y is in fact the minimizer, not just a point where the gradient of the sum of squares vanishes It seems obvious thatTitle Microsoft Word EVV FAQs as of 116docx Author Tim Created Date 11/6/ PMThe desired e ect P(Y z = y) by adjusting for X, as shown in the derivation of Eq (25) However, lacking graphs for guidance, it is not clear whether this condition follows from the assumptions embedded in the model;

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Protocolo los estados contratantes en el presente protocolo,Links with this icon indicate that you are leaving the CDC website The Centers for Disease Control and Prevention (CDC) cannot attest to the accuracy of a nonfederal website Linking to a nonfederal website does not constitute an endorsement by CDC or any of its employees of the sponsors or the information and products presented on the websiteThen, imagine a fluid with density ρ(x, y, z) and velocity field v ( x , y , z ) flowing through S Think of S as an imaginary surface that doesn't

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P h Y h P Z hh P Z P Z P Z §d d d d¨¸ ©¹ d d d d 19 Solution B Apply Bayes' Formula Let A Event of an accident B 1 Event the driver's age is in the range 16 B 2 Event the driver's age is in the range 2130 B 3 Event the driver's age is in the range 3065 B 4 Event the driver's age is in the range 6699 ThenSuppose x˜yand y˜z Then, x>y 1=2 and y>z 1=2 This implies that x>y 1=2 >(z 1=2) 1=2 >z 1=2 Therefore ˜is transitive We can show that ˘is not transitive by using the same counterexample used in (a) Suppose x= 0, y= 5, and z= 1 Since x= 0 = y 1 2, x yand y x, therefore x˘y Since y= 1 2 = z 1 2, y zand z y, therefore y˘zY(y) = P(Y y) = P(g(X) y) = Z A(y) p X(x)dx where A(y) = fx g(x) yg The density is p Y(y) = F0 Y (y) If gis strictly monotonic, then p Y(y) = p X(h(y)) dh(y) dy where h= g 1 Example 3 Let p X(x) = e x for x>0 Hence F X(x) = 1 e x Let Y = g(X) = logX Then F Y(y) = P(Y y) = P(log(X) y) = P(X e y) = F X(e) = 1 e e y and p Y(y) = eye e y for

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Solve for z w=p(yz) Rewrite the equation as Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Tap for more steps Cancel the common factor Divide by Subtract from both sides of the equation Multiply each term in by Tap for more stepsThis is also intuitive The probability that X and Y happen if we know that Z happens is the same as the probability that X happens when we know that Z happens and that then Y happens when we know that X and Z happen Let's note P Z the probability P ( Z) Then apply your formula P Z ( X, Y) = P Z ( X Y) P Z ( Y)Gral (Hint Express the region E = f(x;y;z)jx2y2z2 • 9 and x2y2 ‚ 1g (There is a typo in the original problem) in cylindrical coordinates and find R R R E dV) Solution In cylindrical coordinates, the sphere is given by the equation r2 z2 = 9 and the hole is given by r = 1 Hence, the bead is described by the inequalities ¡ p 9

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Y } z d o a s p g m f @ ^ g f v w t { \ f e ?A formal proof is therefore needed The assumptions explicit in the model take the form (a) X z = XCalculate the probability you entered from the ztable of p (z > 15) The ztable probability runs from 0 to z and z to 0, so we lookup our value From the table below, we find our value of Since that represents ½ of the graph, we subtract our value from 05 → 05 p (z > 15) =

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The CDC AZ Index is a navigational and informational tool that makes the CDCgov website easier to use It helps you quickly find and retrieve specific information Find links to key CDC topic areas in this alphabetical index Skip directly to site content Skip directly to AZ link EspañolWhere β0 is called the y–intercept and β1 is called the slope β0 is the value of y when x =0, and β1 is the change in y when x increases by 1 unit In many real–world situations, the response of interest (in this example it's profit) cannot be explained perfectly by a deterministic modelAnswer Let u x y z X x Y y Z z then X 00 X Y 00 Y Z 00 Z 0 X 0 X 1 Y 0 Y 1 Z 0 from MATH 42 at The University of Hong Kong

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Z 2ˇ 0 Zp 2 0 hx;y;z4i(r r ˆ)dˆd = Z 2ˇ 0 Z p 2 0 ˆ 2 p 2 ˆ5 8 dˆd = 2ˇ Z 2 0 ˆ2 2 p 2 ˆ5 8 dˆ = 1 3 1 6 = 1 6 1784 Use Stokes' Theorem to evaluate ZZ S curlFdS when F(x;y;z) = x2y3zi sin(xyz)j xyzk, and Sis the part of the cone y2 = x2 z2 that lies between the planes y= 0 and y= 3, oriented in the direction of the

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