Name: ID Number:TA: Section Time:MTH 234Exam 2: PracticeOctober 19, 201050 minutesSects: 13.1, 13.3,14.1-14.7.No calculators or any other devices allowed.If any question is not clear, ask for clarification.No credit will be given for illegible solutions.If you present different answers for the same problem,the worst answer will be graded.Show a ll your work.Box your answers.1. (a) (15 points) Find the position r and velocity vector functions v of a particle thatmoves with an acceleration function a(t) = h0, 0, −10i m/sec2, knowing that theinitial velocity and position are given by, respectively, v(0) = h0, 1, 2i m/sec andr(0) = h0, 0, 3i m.(b) (5 points) Draw an approximate picture of the graph of r(t) for t ≥ 0.Solution:(a)a(t) = h0, 0, −10i,v(t) = hv0x, v0y, −10 t + v0zi, v(0) = h0, 1, 2i ⇒v0x= 0,v0y= 1,v0z= 2.v(t) = h0, 1, −10 t + 2i.r(t) = hr0x, t + r0y, −5t2+ 2t + r0zi, r(0) = h0, 0, 3i ⇒r0x= 0,r0y= 0,r0z= 3.r(t) = h0, t, −5t2+ 2t + 3i.(b)zy3parabola2. (a) (10 points) Find and sketch the domain of the function f(x, t) = ln(3x + 2t).(b) (10 points) Find all possible constants c such that the function f(x, t) above issolution of the wave equation, ftt− c2fxx= 0.Solution:(a) The argument in the ln function must be positive. then, the domain isD =(x, t) ∈ R2: 3x + 2t > 0.tx2−3t = −(3/2)x(b)ft=23x + 2t, fx=33x + 2t,ftt= −4(3x + 2t)2, fxx= −9(3x + 2t)2,0 = ftt− cfxx= −4(3x + 2t)2+ c29(3x + 2t)2=1(3x + 2t)2(−4 + 9c2) ⇒⇒ 9c2= 4, ⇒ c = ±23.3. (a) (10 points) Find the direction in which f(x, y) increases the most rapidly, and thedirections in which f(x, y) decreases the most rapidly at P0, and also find the valueof the directional derivative of f(x, y) at P0along these directions, wheref(x, y) = x3e−2y, and P0= (1, 0).(b) (10 points) Find the directional derivative of f(x, y) above at the point P0in thedirection given by v = h1, −1i.Solution:(a) The direction in which f increases the most rapidly is given by ∇f, and the one inwhich decreases the most rapidly is −∇f. So,∇f(x, y) = h3x2e−2y, −2x3e−2yi, ⇒ ∇f(1, 0) = h3, −2i , −∇f(1, 0) = h−3, 2i .The value of the directional derivative along these directions is, respectively, |∇f (1, 0)|and −|∇f(1, 0)|, where|∇f(1, 0)| =√9 + 4 =√13 .(b) A unit vector along h1, −1i is u =1√2h1, −1i, then,Duf(1, 0) = ∇f(1, 0) · u = h3, −2i·1√2h1, −1i =5√2,Duf(1, 0) =5√2.4. (a) (10 points) Find the tangent plane approximation of f(x, y) = x cos(πy/2) −y2e−xat the point (0, 1).(b) (10 points) Use the linear approximation computed above to approximate the valueof f(−0.1, 0.9).Solution:(a)f(x, y) = x cos(πy/2) − y2e−xf(0, 1) = −1,fx(x, y) = cos(πy/2) + y2e−xfx(0, 1) = cos(π/2) + 1 = 1,fy(x, y) = −x sin(πy/2)π2− 2ye−xfy(0, 1) = −2,Then, the linear approximation L(x, y) is given byL(x, y) = (x − 0) − 2(y − 1) − 1, ⇒L(x, y) = x − 2y + 1 .(b) The linear approximation of f(−0.1, 0.9) is L(−0.1, 0.9), which is given byL(−0.1, 0.9) = −0.1 − 2(−0.1) − 1 = −0.1 − 1 = −1.1, ⇒L(−0.1, 0.9) = −1.1 .5. (20 points) Find every local and absolute extrema of f(x, y) = x2+ 3y2+ 2y on the unitdisk x2+ y2≤ 1, and indicate which ones are the absolute extrema. In the case of theinterior stationary points, decide whether they are local maximum, minimum of saddlepoints.Solution:We first compute the interior stationary points, which are (x, y) solutions of∇f = h2x, 6y + 2i = h0, 0i ⇒ x = 0, y = −13.The point (0, −1/3) belongs to the disk x2+ y2≤ 1 so we have to decide whether it isa local maximum, minimum or saddle point:fxx= 2, fyy= 6, fxy= 0,D = fxxfyy− (fxy)2= 12 > 0, fxx> 0 ⇒0, −13is a local minimum.This point is also a candidate for absolute minimum, so we record the value of f,0, −13⇒ f0, −13= 0 +39−23= −13.We now look for extreme point on the boundary x2+ y2= 1. We evaluate f(x, y) alongthe boundary. From the equation x2+ y2= 1 we compute x = ±p1 − y2. This functionis differentiable for y ∈ (−1, 1), but is not differentiable at y = ±1. Since we need to usethe chain rule to find the extrema of g(y) = f(x(y), y) and the chain rule does not holdat y = ±1, we need to consider these points, (0, ±1) separately:(0, 1) ⇒ f(0, 1) = 5, (0, −1) ⇒ f(0, −1) = 1.Now we find local extrema on g(y) = f(x(y), y) in the interval y ∈ (−1, 1). The functiong is given byg(y) = (1 − y2) + 3y2+ 2y ⇒ g(y) = 1 + 2y2+ 2y.The local extrema for g are the points y solutions of g0(y) = 0, that is, 4y + 2 = 0, sowe conclude y = −1/2 and x = ±p1 − 1/4 = ±√3/2, that is,±√32, −12⇒ f ±√32, −12!=34+34− 212=12.Therefore, the absolute extrema are(0, 1) absolute maximum ,0, −13absolute minimum