7/22/2010 SECOND HOURLY PRACTICE I Maths 21a, O.Knill, Summer 2010Name:• Start by printing your name in the above box.• Try to answer each question on the same page as the question is 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.• Provide details to all computations except for problems 1-3.• Please write neatly. Answers which are illegible for the grader can not be given credit.• No notes, books, calculators, computers, or other electronic aids 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), no justifications neededMark for each of the 20 questions the correct letter. No justifications are needed.1)T FEvery critical point of a smooth function f(x, y) of 2 variables is either amaximum or a minimum.Solution:The function f(x, y) = x + y does not have a critical point.2)T FIf f (x, y) = 10 and a region G in the plane is given, thenRRGf(x, y) dydx is10 times the area of the region.Solution:If f(x, y) = 1, then we get the area.3)T FIf a function f(x, y) has only one critical point (0, 0) in G = {x2+ y2≤ 1 }which is a local maximum and f(0, 0) = 1, thenRRGf(x, y) dxdy > 0.Solution:The critical point can be surrounded by a small region only, where f is positive.4)T FIf a curve ~r(t) cuts a level curve in a right angle and nonzero velocity at apoint which is not critical, then the d/dtf(r(t)) 6= 0 at that point.Solution:This is a consequence of the chain rule: the derivative is ∇f(r(t)) ·r′(t). The assumptionimplies that r′(t) is parallel to ∇f.5)T FThe linearization of a function f(x, y) at (0, 0) has a graph which is a planeax + by + cz = d tangent to the graph of f(x, y).Solution:The linearized function is linear and approximates the function at (0, 0). Its level curvesare planes and one of this plane is the tangent plane.6)T FThe surface area ~r(u, v) = hu2, v2, u2+v2i with 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 is equalto the surface area of ~r(u, v) = hu3, v3, u3+ v3i with 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.2Solution:Different parametrizations of the same surface h ave the same surface area. In this case,this is a parametrization of a plane.7)T FAssume ~r(u, v) is a parametrization of a surface g(x, y, z) = d and ~r(1, 2) =3, then ∇g(1, 2, 3) is parallel to ~ru(1, 2) ×~rv(1, 2).Solution:Both vectors are perpendicular to the surface at the p oint (1, 2, 3).8)T FAny region which is both type I and type II must be a rectangle.Solution:By definition of type I and type II.9)T FA given function f(x, y) defines two functions g(x, y) = fxx(x, y)fyy(x, y) −f2xyand h(x, y) = fxx(x, y). Assume b oth g and h are positive everywhere,then every critical point of f must be a local minimum.Solution:This is a consequence of the second derivative theorem.10)T FIf f(x, y) satisfies the Laplace equation fxx= −fyythen every critical pointof f with nonzero discriminant D is a saddle point.Solution:Yes, then D = −f2xx− f2xyis negative.11)T FThe Lagrange multiplier λ at a solution (x, y, λ) of the Lagrange equations∇f(x, y) = λ∇g(x, y), g(x, y) = 0 has the property that it is always positiveor zero.Solution:The sign of λ changes if g is replaced by −g but the critical points stay.312)T FThe partial differential equation ut= uxxis called the heat equation.Solution:This is an important partial differential equation.13)T FThe gradient ∇f(x, y, z) of a function of three variables is a vector tangentto the surface f(x, y, z) = 0 if (x, y, z) is on the surface.Solution:It is perpendicular, not tangent.14)T FThe value of log(2 + x) with natural log can be estimated by linear approx-imation as log(2) + x/2.Solution:Yes, because log′(x) = 1/x and log(2 + x) can be written as log(2) + log(1 + 1/x).15)T FThe tangent line to the curve f(x, y) = x3+ y3= 9 at (2, 1) can beparametrized as ~r(t) = h2, 1i + th8, 3i since h8, 3i is the gradient at (2, 1).Solution:This was the parametrization of a line perpendicular to the curve.16)T FAny function f(x, y) which has a local maximum also has a global maximum.Solution:There are functions which only have local maxima already in one dimensions. An exampleis f(x) = x4− x2. Now take f(x, y) = (x2+ y2)4+ x2+ y2.17)T FThe directional derivative of a function f(x, y) in the direction of the tangentvector to the level curve is zero.Solution:Yes, because it is then perpendicular to the curve.418)T FThe directional derivative of a function in the d irection ∇f/|∇f| of thegradient is always nonnegative at a point which is not a critical point.Solution:Indeed, an important fact19)T FThe chain rule tells d/dtf(t4, t3)|t=1is equal to the dot product of the gra-dient of f at (1, 1) and the velocity vector (4, 3) at (1, 1).Solution:Thats what the chain rule tells.20)T FFubini’s theorem assures thatR10R1xf(x, y) dxdy =R10R1yf(x, y) dydx.Solution:No this is a wrong identity.Problem 2) (10 points) No justifications are neededMatch the regions with the double integrals. Only 5 of the 6 choices match.a b c5d e fEnter a,b,c,d,e or f Integral of Function f(x, y)Rπ/20Rπ/20f(r, θ)r dθdrRπ/20Rπ/20f(x, y) dxdyRπ/20Rx0f(x, y) dydxRπ/20Ry0f(x, y) dxdyRπ/20Rx/20f(x, y) dydx6Solution:Enter a,b,c,d,e or f Integral of Function f(x, y)fRπ/20Rπ/20f(r, θ)r dθdrcRπ/20Rπ/20f(x, y) dxdybRπ/20Rx0f(x, y) dydxdRπ/20Ry0f(x, y) dxdyaRπ/20Rx/20f(x, y) dydxProblem 3) (10 points) (no justifications are needed)A function f(x, y) of two variables has level curves as shown in the picture.Enter A-E Descriptiona critical point of f(x, y).a point, where f is extremal under the constraint x = 2.a point, where f is extremal under the constraint y = 0.the point among points A-E, where the length of the gradient vector is largest.the point among points A-E, where the length of the gradient vector is smallest.a point, where Dh1,1if = 0 and Dh1,−1if 6= 0.a point, where Dh1,−1if = 0 and Dh1,1if 6= 0.a point, where Dh1,0if = 0 and Dh0,1if 6= 0.a point, where fx= 0 and fy6= 0.a point, where fy= 0 and fx6= 0.7ABCDESolution:Enter A-E DescriptionA a critical point of f(x, y).E a point, where f is extremal under the constraint x = 2.D a point, where f is extremal under the constraint y = 0.E the point among points A-E, where the length of the gradient vector is largest.A the point among points A-E, where the length of the gradient
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