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.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.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 > 0.4)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.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).6)T FThe surface area ~r(u, v) = h u2, 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 ≤ 1.7)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).8)T FAny region which is both type I and type II must be a rectangle.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.10)T FIf f(x, y) satisfies the Laplace equation fxx= −fyythen every critical pointof f with nonzero discriminant D is a saddle point.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.12)T FThe partial differential equation ut= uxxis called the heat 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.14)T FThe value of log(2 + x) with natural log can be estimated by linear approx-imation as log(2) + x/2.15)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).16)T FAny function f(x, y) which has a local maximum also has a global maximum.17)T FThe directional derivative of a function f(x, y) in the direction of the tangentvector to the level curve is zero.18)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.19)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).20)T FFubini’s theorem assures thatR10R1xf(x, y) dxdy =R10R1yf(x, y) dydx.2Problem 2) (10 points) No justifications are neededMatch the regions with the double integrals. Only 5 of the 6 choices match.a b cd 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) dydxProblem 3) (10 points) (no justifications are needed)3A 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.ABCDEProblem 4) (10 points)4A mass p oint with position (x, y) is attached by springsto the points A1= (0, 0), A2= (2, 0), A3= (0, 2), A4=(2, 3), A5= (3, 1). It has the potential energyf(x, y) = 31 − 14x + 5x2− 12y + 5y2which is the sum of the squares of the distances from(x, y) to the 5 points. Find all extrema of f using thesecond derivative test. The minimum of f is the posi-tion, where the mass point has the lowest energy.A1A2A3A4A5Hx,yLProblem 5) (10 points)The main building of a mill has a cone sh aped roof andcylindrical walls. If the cylinder has radius r, the heightof the side wall is h and the height of the roof is√3r,then the volume isV (h, r) = πr2h + hπr2/3 = (4π/3)hr2and area of the building isA(h, r) = πr2+ 2πrh + π2r2= π(3r2+ 2rh) .For fixed volume V (h, r) = 4π/3, minimize A(h, r) usingthe Lagrange multiplier method.Problem 6) (10 points)a) (5 points) Find the tangent plane to the surface x3y+yx3+z2x2= 6 at the point (1, 1, 2).b) (5 points) Find the tangent line to the curve x4− y4= 15 at the point (2, 1).Problem 7) (10 points)5a) (4 points) Compute the moment of inertiaI =Z ZG(x2+ y2) dydxof the half disc D = {x2+ y2≤ 1, x ≥ 0 }.b) (6 points) Evaluate the following double integralZe1Z1log(x)yey− 1dydx ,where log is the natural log as usual.Problem 8) (10 points)a) (4 points) Find the linearization L(x, y) of the function f(x, y) = x5· y3.b) (6 points) Estimate 10.015· 4.9993using the linear approximation found in part a).Problem 9) (5 points)Find the surface area of the surface parametrized by~r(t, s) = hs cos(t), s sin(t), ti ,where 0 ≤ t ≤ 5π and 0 ≤ s ≤ 2.Hint. You can use the anti derivative formulaZ√1 + s2ds = s√1 + s2/2 + arcsinh(s)/2computed in class without having to derive it again.Problem 10) (10 points)We minimize the surface of a roof of height x an d width 2x and length L =√2y if thevolume V (x, y) = x2√2y of the roof is fixed and equal to√2. In other words, you have to6minimize f(x, y) = 2x2+ 4xy under the constraint g(x, y) = x2y = 1. Solve the problemwith
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