Name                                     MATH 107 PRACTICE MIDTERM III Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work. PROBLEM 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why. A) Let f(x,y) be a function of two variables and P be a point.  If the value of the function tends towards 5 for every line segment that ends at P, then     Solution False, the limit has to be the same for all curves not just lines.  For example, if P is the origin (0,0) and                                xy2 + x2y             f(x,y)  =                                                               x4 + y2 Then it is easy to check that the limit is zero along all line segments ending at the origin, but along y  =  x2 the limit is 1/2.   B)  If  z  =  f(x - y) and f is a continuous function then     Solution True.  Use the chain rule with u  =  x - y:         zx  =  f '(u)(ux)  =  f '(u)(1)  =  f '(u)         zy  =  f '(u)(uy)  =  f '(u)(-1)  =  -f '(u) and         f '(u) + (-f '(u))  =  0 PROBLEM 2 Find the dimensions of the rectangular box with the largest volume in the first octant such that one of the vertices is at the origin and the opposite vertex lies on the ellipsoid          x2 + 2y2 + 3z2  =  6 The volume of such a box is given by          V  =  xyz We use LaGrange Multipliers         grad F  =  <2x,4y,6z>  =  l  =  l grad V This gives us four equations          2x  =  lyz        4y  =  lxz        6z  =  lxy        x2 + 2y2 + 3z2  =  6   Solving the first three equations for l gives        l  =  2x/yz        l  =  4y/xz        l  =  6z/xy Setting the equations equal to each other and multiplying by xyz gives         2x2  =  4y2  =  6z2  or         x2  =  2y2  =  3z2  Substituting into the fourth equation gives         x2 + x2 + x2  =  6         x2  =  2         x  =  Resubstituting gives         2  =  2y2           y  =  1 and         2  =  3z2         z  =  (2/3)0.5    PROBLEM 3 Consider the function                    5         z  =                              xy A.     Use a calculator to sketch the level curves corresponding to z = (-2,-1,0,0.5,1,2,5,10).   Draw them on a whole piece of paper.   Solution We solve for y                   5         y  =                              xz   The contour map is pictured below           B.    Suppose that z represents the altitude function, and you are to travel on this surface above the unit circle in a counter-clockwise direction.  Discuss your travels.   Solution      Beginning at the point (1,0) we are infinitely high.  Then we slide steeply down, reaching a local minimum elevation at the point (/2, /2) of 5/2.  Then steeply rise heading towards an infinite elevation at the point (0,1).  Then suddenly appear at an infinitely low chasm and begin climbing to an elevation of -5/2 at the point (-/2, /2).  Then we again slide down to an infinitely low chasm at the point (-1,0)  We appear on the other side of this point at an infinitely high cliff and again slide down to the altitude of 5/2 at the point (-/2, -/2) .  Then we rise up again to the infinitely high cliff at the point (0,-1).  We instantly appear on the other side in another infinitely low chasm and rise up to a local high point at (/2, -/2) .  Then slide again to the infinitely low chasm at (1,0)   PROBLEM 4  Find the following limits if they exist.  A.     Solution We consider the path to the origin from the positive x axis (y  =  0).  We get                         x4          lim                    =    1         x -> 0      x4  Now approach the origin via y  =  x.  We have                        x4  + 2x4 + x4                 4x4         lim                                        =                  =  2                  x -> 0            x4 + x4                      2x4  Since the two limits approach two different values, we can conclude that the limit does not exist. B.     For this limit we use polar coordinates                       r5cos5q - r4sin4q         lim                                            = lim      r3cos5q - r2sin4q  =  0         r -> 0                r2                       r -> 0          Therefore, the limit exists and is equal to 0   PROBLEM 5  Let           f(x,y)  =  x2 + xy,     x(u,v)  = u2v,  and     y(u,v)  =  u - v Find A)       Solution       fx  =  2x + y  B)    Without calculating any derivatives, write the appropriate chain rule for          Solution         fuu  =  (fu)u  =  (fxxu + fyyu)u  =  fxuxu + fxxuu + fyuyu + fyyuu          =  (fxxxu + fxyyu)xu + fxxuu + (fxyxu + fyyyu)yu + fyyuu  C)   (7 Points)Find      Now calculate:         x(1,2)  =  12(2)  =  2                    y(1,2)  =  1 - 2  =  -1         xu  =  2uv  =  2(1)(2)  =  4            xuu  =  2v  =  2(2)  =  4         yu  =  1            yuu  =  0            fxx  =  2            fxy  =  1            fyy  =  0         fx  =  2x + y  =  2(2) + (-1)  =  3            fy  =  x  =  2 Plugging in, we get         [(2)(4) + (1)(1)](4) + (3)(4) + [(1)(4) + (0)(1)](1) + (2)(0)         =  (9)(4) + 12 + (4)(1)  =  36 + 12 + 4  =  52   PROBLEM 6  If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, and R3, then             1                1              1               1                     =               +              +                             R              R1             R2             R3                         If the resistances are measured as R1 = 25 ohms, R2 = 40 ohms, and R3 = 50 ohms, with possible errors of 0.5% in each case, use differentials to estimate the maximum error in the calculated value of R.  Solution We use the differential formula         -R-2 dr  =  -R1-2 dR1 - R2-2 dR2 - R3-2 dR3           R-2 dr  =  .005(R1-2 R1 + R2-2 R2 + R3-2 R3)          R-2 dr  =  .005(R1-1  + R2-1  + R3-1 )  =  .005R Hence         dr  =  .005R3 =  .005(1/25 + 1/40 + 1/50)3  =  .000425     PROBLEM 7  Find parametric equations for the tangent line of the curve of intersection of the paraboloid z  =  x2 + y2 and the ellipsoid  4x2 + y2 + z2  =  9 at the point (-1,1,2). This line is on the tangent plane of both surfaces.  Thus is is perpendicular to both normal vectors.   We compute the two normal vectors.  The normal vector for the parabola is          <2x, 2y, -1>  =  <-2, 2, -1> The normal vector for the ellipsoid is         <8x, 2y, 2z>  =  <-8, 2, 4> Now cross the two vectors to find a vector perpendicular to both normal vectors.                     10i + 16j + 12k Now use the formula for parametric equations of a line given a point and a parallel vector         x(t)  =  -1 + 10t        y(t)  =  1 + 16t        z(t)  =  2 + 12t