Math32a 1 R Kozhan Midterm 2 Nov 21 2011 Name UID Circle your TA and discussion session 1A Tues Mike O Brien 1C Tues Jeff Lin 1E Tues Jordy Greenblatt 1B Thur Mike O Brien 1D Thur Jeff Lin 1F Thur Jordy Greenblatt Instructions If you get stuck move on to the next question You don t have a lot of time Show all work if you want to get full credit I reserve the right to take off points if I cannot see how you arrived at your answer even if your final answer is correct No books notes electronics incl calculators and cell phones are allowed Good luck Question Max Your score 1 6 2 14 3 10 4 10 5 10 Total 50 1 Problem 2 If you don t know how to solve part of this problem you may skip it and solve the other parts a 3 points Classify and sketch very roughly the surface given by the equation x2 2y 2 z 2 2x 6z 12 0 b 2 points Find the parametric equation of its axis of symmetry c 3 points Describe all possible traces of this surface in planes parallel to xz plane d 4 points Viewing z as a function of independent variables x and y find the derivative z z and y at the point 1 3 1 x e 2 points Find the tangent plane at the point 1 3 1 Solution a Let us complete the squares x2 2y 2 z 2 2x 6z 12 x 1 2 1 2y 2 z 3 2 9 12 x 1 2 2y 2 z 3 2 2 so our surface is the hyperboloid x 1 2 2y 2 z 3 2 2 of two sheets which is obtained from the hyperboloid x2 2y 2 z 2 2 by shifting by 1 in x direction and by 3 in z direction b Since x2 2y 2 z 2 2 has y axis as its axis of symmetry the axis of symmetry of x 1 2 2y 2 z 3 2 2 is the line through 1 0 3 parallel to y axis So its parametric equation is x 1 y 0 t z 3 c Traces in planes parallel to xz plane are obtained when we take y k for some constant k Thus traces are x 1 2 z 3 2 2k 2 2 which are ellipses in fact circles if 2k 2 2 0 a single point if 2k 2 2 0 and empty if 2k 2 2 0 3 d Denote F x y z x2 2y 2 z 2 2x 6z 12 Then z Fx 2x 2 x Fz 2z 6 Fy 4y z y Fz 2z 6 z z so at the point 1 3 1 x 0 y 23 Note another approach is to differentiate both sides of the equality x2 2y 2 z 2 2x 6z 12 0 keeping in mind that z depends on x and y while x and y are independent variables E g differentiating with respect to x would give 2x 2z z z 2 6 0 x x which gives the same result z 2x 2 x 2z 6 and so on e Note the equation of the tangent plane to the surface given by z f x y at x0 y0 z0 is z z0 fx x0 y0 x x0 fy x0 y0 y y0 The equation of the tangent plane to the surface given by F x y z 0 at x0 y0 z0 is Fx x0 y0 z0 x x0 Fy x0 y0 z0 y y0 Fz x0 y0 z0 z z0 0 In either case one obtains that the tangent plane at the point 1 3 1 is z 1 23 y 3 4 Problem 3 10 points If you weigh a solid body in air m1 and when submerged in water m2 then basic Archimedean principle tells you that the volume of the body is m1 m2 and therefore its density is 1 s m1m m 2 Suppose you obtained measurements m1 100 gram and m2 90 gram and you know that the maximal error is no more than 0 01 in absolute value when measuring m1 and no more than 0 02 in absolute value when measuring m2 Estimate the maximal error of s Solution 1 Note that the model is the following m1 100 and a Denote f m1 m2 m1m m 2 m2 90 are the results of our mesurements while the actual true values are m1 m1 and m2 m2 where m1 and m2 are unknown errors satisfying m1 0 01 m2 0 02 Our measurements give us that s is f m1 m2 while in fact the true value of the density is f m1 m1 m2 m2 Therefore we made the error equal to f m1 m1 m2 m2 f m1 m2 Denote this s the increment of s We know from our math that this increment s can be approximated by the differential ds Now let us find the differential of s ds m1 m2 dm1 dm2 2 m1 m2 m1 m2 2 Here dm1 and dm2 is just a different notation for variables m1 m2 We know the values m1 100 and m2 90 we don t know the exact values of dm1 and dm2 though so ds 0 9 dm1 dm2 0 1 Note this is just a function of two independent variables dm1 and dm2 When is this error ds maximal under the restrictions that dm1 0 01 dm2 0 02 It s maximal if dm1 0 01 and dm2 0 02 or if dm1 0 01 dm2 0 02 so ds 0 9 0 01 0 02 0 029 One may also use the triangle inequality on equation 0 1 and also get ds 0 9 dm2 dm2 0 9 0 01 0 02 0 029 5 Problem 4 a 5 points Does there exist a number c that makes the following function continuous everywhere on R2 Justify your answer 2 x2 y if x y 6 0 0 2 x 1 y 2 1 f x y c if x y 0 0 b 5 points Find the limit if it exists or show that the limit does not exist x2 y 1006 x y 0 0 x4 y 2012 lim Hint does this function remind you xy x2 y 2 Solution a Note that the numerator of that expression is a continuous function everywhere and the denominator is a continuous function that doesn t vanish so the function f x y is continuous everywhere except possibly at 0 0 To check this point we just need to find the limit lim x y 0 0 f x y and see when it is equal to f 0 0 c Well 02 02 0 f x y x y 0 0 02 1 02 1 lim So the function is continuous at 0 0 if and only if c 0 2 1006 y uv 2 1006 b Note that xx4 y Recall that 2012 is exactly the function u2 v 2 where …