7/28/2011 SECOND HOURLY PRACTICE V Maths 21a, O.Knill, Summer 2011Name:• Start by printing your name in the above box.• Try to answer each question on the same page as t h e question i s asked. If needed, usethe back or the next empty page for work. If you need additional paper, write your nameon it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. An swers which are illegible for the grader can not be given credit.• No notes, books, calculator s, computers, or other electronic ai d s can be allowed.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 10Total: 1101Problem 1) True/False questions (20 points)Mark for each of the 20 questions the correct letter. No justifications are needed.1)T F(0, 0) is a local minimum of t h e function f(x, y) = x6+ y6.2)T FIf f(x, y) has a local max at th e point (0, 0) with di scr i m i n ant D > 0, theng(x, y) = f(x, y) − x9+ y9has a local max at (0, 0) too.3)T FThe value of the function f(x, y) =√1 + 3x + 5y at (−0.002, 0.01) can bylinear app r oximation be estimated as 1 − (3/2) · 0.002 + (5/2) · 0. 01 .4)T FThe val u e of the double integralRπ/40R20x3cos(y) dxdy is the same as(R20x3dx)(Rπ/40cos(y) dy).5)T FThe chain rule can be written in the formddtf(~r(t)) = D~r′(t)f(~r(t))6)T FThe gradient of f at a point (x0, y0, z0) is ta n gent to the level surface of fwhich contains (x0, y0, z0).7)T FIf the Lagrange multiplier λ is positive, then the critical point under con-straint is a local minimum.8)T FIf the directional derivative D~vf(1, 1) = 0 for all vectors ~v, then (1, 1) is acritical point of f (x, y).9)T FFor any curve ~r(t), the vectors ~r′′(t) and ~r′(t) are always perpendicular toeach other.10)T FEvery criti cal point ( x , y) of a function f (x, y) for which the discriminantD is not zero is either a local maximum or a a local minimum.11)T FThe function f(x, y) = eyx2sin(y2) satisfies the partial d i ff er ential equationfxxyyyxyy= 0.12)T FIf (0, 0) is a critical point of f(x, y) and the discriminant D is zero butfxx(0, 0) < 0 th en (0, 0) can not be a local minimum.13)T FIn the second derivative test, one can replace the conditi on D > 0, fxx> 0with D > 0, fyy> 0 to check whether a point i s a local minimum.14)T FIf ~r(t) is a curve in space and f is a function of three variables, thenddtf(~r(t)) = 0 for t = 0 implies that ~r(0) is a critical point of f(x, y).15)T FThe function f(x, y) = (x4− y4) has neither a local maximum nor a localminimum at ( 0, 0).16)T FIf every point of t h e plane is a critical point for a function f t h en f is aconstant function.17)T FIf f(x, y) has a local max at th e point (0, 0) with di scr i m i n ant D > 0, theng(x, y) = f(x, y) − x9+ y9has a local max at (0, 0) too.18)T FThe value of the functi on f(x, y) = log(e + 3x + 5y) at (−0.002, 0.0 1) canby linear app r oximation be estimated as 1 − 0.006 + 0.05.19)T FThe chain rule tells thatddtf(~r(t)) = f(~r(t))r′(t)20)T FIf (0, 0) is a critical point of f(x, y) and the discriminant D is zero butfxx(0, 0) > 0 th en (0, 0) can not be a local maximum.2Problem 2) (10 points)a) (5 points) The pictur e below shows the contour map of a fu nction f (x, y) which hasmany critical points. Four of them are outlined for you on the y axes and a re lab e ledA, B, C, D and ordered in increasi n g y value. The picture shows also the gradient vectors.Determine from each of the 4 points whether i t is a loca l maximum, a local min i mum ora saddle point. No justificat i on is necessary in thi s problem.Point Max Min SaddleDCBACABDb) (5 points)Match the integrals with those obtained by changing th e order of integration. No justifi-cations a r e needed. Note that one of the Roman letters I)-V) will not be used, you haveto chose four out of five.Enter I,II,III,IV or V here. IntegralR10R11−yf(x, y) dxdyR10R1yf(x, y) dxdyR10R1−y0f(x, y) dxdyR10Ry0f(x, y) dxdyI)R10Rx0f(x, y) dydxII)R10R1−x0f(x, y) dydxIII)R10R1xf(x, y) dydxIV)R10Rx−10f(x, y) dydxV)R10R11−xf(x, y) dydx3Problem 3) (10 points)The Pac-Man region R is bounded by the lin es y = x, y = −x and the unit circle. Thenumbera =R RRx dxdyR RR1 dxdydefines the point C = (a, 0) called center of mass of the reg i on . Find it.Problem 4) (10 points)Find all the critical points o ff(x, y) = x3+ y3− 3x − 12yand indicat e whether they are local maxima, local minima or saddle points.Problem 5) (10 points)When Ramanujan, an amazing mathematician who wasborn i n India was sick in the hosp i t al and the Englishmathematician Hardy visited him, Ramanujan as ked”whats up?” Hardy answered. ”Nothing special. Eventhe number of the taxi cab was boring: 1729”. Ra-manujan answered: ”No, that is a remarkable number.It is the sm a l les t number, which can be wr i t t en in twodifferent ways as a sum of two perfect cubes. Indeed1729 = 13+ 123= 93+ 103.4a) (5 points) Find the linearization L(x, y, z) of th e function f(x, y, z) = x3+ y3− z3atthe point (9, 10, 12).b) (5 points) Use the technique of linear approximation to estimate 9.0013+10.023−12.0013.Problem 6) (10 points)a) (5 points) Find the equation ax + by + cz = d for the tangent plane to the level surfacef(x, y, z) = x3+ y3− z3= 1at the point (1, 1, 1). Note that this is the same Ramanujan function as in the previousproblem.b) (5 points) If we intersect the level surface f(x, y , z) = 1 with the plane z = 2, weobtain the equation for an impl i ci t curve x3+ y3= 9. It is a level curve for the functiong(x, y) = x3+ y3. Fi n d the tangent line to this curve at th e point (1, 2).Problem 7) (10 points)A function f(x, y) of two variables descr i bes the height of a mountain. You don’t knowthe function but you see its level curves. The mountain has its peak at the point A on thepicture.a) (3 points) At which point does f take its maximum under the constraint x = 0.b) (3 points) At which point does f take its maximum under the con st r ai nt y = 0.c) (4 points) At which of the points B or C is the length of the gradient vector la r ger ?Note: As always, you have to give explanations to get full credit. The points in a) andb) are not necessarily marked. Give it u p to an accu r a cy of 1/2. For example, …
View Full Document
Unlocking...