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HARVARD MATH 21A - Second Hourly Practice III

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7/22/2010 SECOND HOURLY PRACTICE III 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 th e 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. 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) (20 points)1)T F(1, 1) is a local maximum of the function f(x, y) = x2y − x + cos(y).Solution:(1, 1) is not even a critical point.2)T FIf f is a smooth function of two variables, then the number of critical pointsof f inside the unit disc is finite.Solution:Take f(x, y) = x2for example. Every point on the y axis {x = 0 } is a critical point.3)T FThe value of the function f(x, y) = sin(−x + 2y) at (0.001, −0.002) can bylinear approximation be estimated as −0.003.Solution:The correct approximation would be f(0, 0) + 0.001(−1) − 0.002(2) = −0.005.4)T FIf (1, 1) is a critical point for the function f(x, y) then (1, 1) is also a criticalpoint for the function g(x, y) = f(x2, y2).Solution:If ∇f(1, 1) = (fx(1, 1), fy(1, 1)) = (0, 0) then also ∇g(1, 1) = (fx(1, 1)2x, fy(1, 1)2y) =(0, 0). Note that the statement would not be true, if we would replace (1, 1) say with(2, 2) (as in the practice exam).5)T FIf the velocity vector ~r′(t) of the planar curve ~r(t) is orthogonal to thevector ~r(t) for all times t, then the curve is a circle.Solution:d/dt(~r (t) ·~r(t)) = ~r(t) ·~r′(t) = 0 by assumption. This shows that |~r(t)| is constant.6)T FThe gradient of f(x, y) is normal to the level curves of f.2Solution:This is a basic and important fact.7)T FIf (x0, y0) is a maximum of f(x, y) under the constraint g(x, y) = g(x0, y0),then (x0, y0) is a maximum of g(x, y) u nder the constraint f(x, y) =f(x0, y0).Solution:Assume you have a situation f, g, where this is true and where the constraint is g = 0,produce a new situation f, h = −g, where the first statement is still true but where theextrema of h under the constraint of f is a minimum.8)T FIf ~u is a unit vector tangent at (x, y, z) to the level surface of f(x, y, z) thenthe directional derivative satisfies Duf(x, y, z) = 0.Solution:The directional derivative measures the rate of change of f in the direction of u. On alevel surface, in the direction of the surface, the function does not change (because f isconstant by definition on the surface).9)T FIf ~r(t) = hx(t), y(t)i and x(t), y(t) are polynomials, then the tangent line isdefined at all points.Solution:Take the example ~r(t) = ht2, t3i. At t = 0, we have a cusp and the gradient is zero. Wedo not have a tangent line there.10)T FThe vector ~ru(u, v) is tangent to the surface parameterized by ~r(u, v) =hx(u, v), y(u, v), z(u, v)i.Solution:The vector ~ruis tangent to a grid curve and so tangent to the surface.11)T FThe second derivative test allows to check whether an extremum found withthe Lagrange multiplier method is a maximum.Solution:No, the second derivative test applies for function f(x, y) without constraint.312)T FIf (0, 0) is a critical point of f(x, y) and the discriminant D is zero butfxx(0, 0) > 0 then (0, 0) can not be a local maximum.Solution:If fxx(0, 0) > 0 then on the x-axis the function g(x) = f(x, 0) has a local minimum. Thismeans that there are points close to (0, 0) where the value of f is larger.13)T FLet (x0, y0) be a saddle point of f (x, y). For any unit vector ~u, there arepoints arbitrarily close to (x0, y0) for which ∇f is parallel to ~u.Solution:Just look at the level curves n ear a saddle point. The gradient vectors are orthogonalto the level curves which are hyperbola. You see that they point in any direction except4 directions. To see this better, take a pen and draw a circle around the saddle pointbetween two of your knuckles on your fist. At each p oint of the circle, now draw thedirection of steepest increase (this is the gradient direction).14)T FIf f(x, y) has two local maxima on the plane, then f must have a localminimum on the plane.Solution:Look at a camel type surface. There is no local minimum.15)T FGiven a unit vector v, define g(x) = Dvf(x). If at a critical point, for allvectors v we have Dvg(x) > 0, then f is a local maximum.Solution:On every line through the critical point, we have a local minumum. So, it is a localminimum, not a local maximum16)T FIf x4y + sin(y) = 0 then y′= 4x3/(x4+ cos(y)).Solution:The sign is wrong.17)T FThe critical points of F (x, y, λ) = f(x, y) −λg(x, y) are solutions to the La-grange equations when extremizing the funct ion f(x, y) under the constraintg(x, y) = 0.4Solution:The critical points of F are p oints where fx= λgx, fy= λgy, g = 0 which is exactly theLagrange equations.18)T FThe volume under the graph of f (x, y) = x2+y2inside th e cylinder x2+y2≤1 isR10R2π0r3dθdr.Solution:x2+ y2= r2.19)T FThe surface area of the unit sphere is 4π.Solution:We computed that in class.20)T FThe area of a disc of radius 2r is 4 times larger than a disc of radius r.Solution:We know the area to be πr2.5Problem 2) (10 points)Match the regions with the corresponding double integrals.a0.00.20.40.60.81.00.20.40.60.81.0b0.00.20.40.60.81.00.20.40.60.81.0c0.00.20.40.60.81.00.20.40.60.81.0d0.00.20.40.60.81.00.20.40.60.81.0e0.00.20.40.60.81.00.20.40.60.81.0f0.00.20.40.60.81.00.20.40.60.81.0Enter a,b,c,d,e or f Integral of f(x, y)R10R√xx2f(x, y) dydxR10R√y0f(x, y) dxdyR10R1y2f(x, y) dxdyEnter a,b,c,d,e or f Integral of f(x, y)R10R√1−x20f(x, y) dydxR10R1(1−x)2f(x, y) dydxR10R√1−x2(1−x)2f(x, y) dydxSolution:a),b),c),f),d),e).Problem 3) (10 points)a) Use the technique of linear approximation to estimate f(log(2) + 0.001, 0.006) for f(x, y) =e2x−y. (Here, log means the natural logarithm).b) Find the equation ax + by = d for th e tangent line which goes through the point (log(2), 0).6Solution:a) L(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0)f(x0, y0) = e2 log 2= 4fx(x0, y0) = 8fy(x0, y0) = −4L(x, y) = 4 + 0.001 · 8 − 4 · 0.006 =3.984 .b) We have a = 8 and b = −4 and get d = 8 log(2) so that the line has the equation8x − 4y = 8 log(2) .Problem 4) (10 points)Find a point on the surface g(x, y, z) =1x+1y+8z= 1


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