Vectors Calaulus 1.36

Practice Final 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 Q be a three dimensional solid and let          F(x,y,z)  =  (x2y + sin z)i+ (cos x - xy2)j + (3xy + z)k and let S be the boundary of Q with outwardly pointing normal.  The the volume so Q is given by         Solution True, We have         divF  =  2xy - 2xy + 1  =  1 Using the divergence theorem, we see that          which is just the volume of the solid. B.  Let F  =  3xy i + cosx j and let C1 and C2 be as shown below.  Then                 Solution C.  A new particle, the fluxon, has been discovered to be emitted from the sun.  The particle emits  a force field          F(x,y,z)  =  (y2 - z) i + (x2 - z) j + (x2 + y2) k where the origin represents the center of the sun.  If the total flux through the earth's northern hemisphere has been calculated as 10,000, then the total flux through the earth's southern hemisphere  must also be 10,000. Solution False,  since divF  =  0, the total flux must be zero.  If the flux through the northern hemisphere is   10,000, then the flux through the southern hemisphere must be -10,000. Problem 2    You are the captain of the spaceship Potential that you have programmed to follow the vector-valued function          r(t) =  (t2 + 5) i + (t - 3) j + t3 k  where t is measured in hours.  However, at time t  =  2, your engines fail and your ship begins drifting in deep space.  There is a deep space station located at (6,2,38).   A.  Find the vector-valued function that describes the Potential's flight after the engines failed.  Use t  =  2 to represent the time at which your engines first shut down. (Hint:  This should be a linear vector valued function.) Solution The flight will go in a linear path in the direction of the unit tangent vector with speed equal to the speed when the engines fail.  We have         r'(t)  =  2t i + j + 3t2 k         r'(2)  =  4i + j + 12k When t  =  2, the spacecraft is at          r(2)  =  9i - j + 8k  The flight can be described by         rf(t)  =  r(2) + (t - 2)r'(2)  =  9i - j + 8k + (t - 2)(4i + j + 12k)         =  (1 + 4t)i + (-3 + t)j + (-16 + 12t)k  B.  Will your ship make it to the station, or will you float helplessly for eternity? Solution We are looking for a time t with          rf(t)  = 6i + 2j + 38j Setting the j components equal we get         -3 + t  =  3        t  =  5 However         rf(5)  =  9i + 2j + 44k since the i and k components of rf and the station are different, we can conclude that our  spaceship will drift away to eternity. Problem 3  Show that the helix         r(t)  =  (R cos t) i + (R sin t) j + t k  where R is a positive constant, has the property that N(t) . r(t) is a constant.  Find this constant. Solution We first calculate T(t)         r'(t)  =  (-R sin t) i + (R cos t) j + k          || r'(t)||  =  (R2 sin2 t + R2 cos2 t + 1)1/2  =  (R2 + 1)1/2 Hence         T(t)  =  (R2 + 1)-1/2 [(-R sin t) i + (R cos t) j + k] Next, we have         T'(t)  =  (R2 + 1)-1/2 [(-R cos t) i + (-R sin t) j] and         ||T'(t)||  =  (R2 + 1)-1/2 [R2 cos2 t + R2 sin2 t]1/2  =   (R2 + 1)-1/2 [R]  Dividing gives         N(t)  =  -cos t i - sin t j Finally, we take the dot product         N(t) . r(t)  =  [(R cos t) i + (R sin t) j + t k] . [-cos t i - sin t j]         =  R cos2 t + R sin2 t  =  R Problem 4    The probability density function for an event is given by         where R is the square with vertices (4,0), (6,2), (4,4), and (2,2). A.  Use the appropriate change of variables (Jacobians) to find k that is solve          Solution We sketch the picture and find the equation of the four lines that border the square.         We can also write         0  <  x - y  < 4        and        4  <  x + y  <  8 We let          u  =  x - y        v  =  x + y Adding the two equations gives         u + v  =  2x        x  =  1/2 (u + v) Subtracting the two equations gives         v - u  =  2y        y  =  1/2 (v - u) We can compute the Jacobian         We have                 Setting this equal to 1 gives                     3         k  =                                      896          B.  Find the probability that  0  <  x - y  <  1 Solution Since         u  =  x - 1 we just adjust the limits appropriately         Problem 5   Switch the order of integration and write as one double integral         Solution The key to this problem is to sketch the picture which is shown below         Now we can realize the region as being bounded from below by y  =  x2 and above by y  =  x + 2. We have                 Problem 6  Set up the integrals that give the following.  Use the most appropriate coordinate system. A.  The mass of the solid that lies inside the sphere         x2 + y2 + z2  =  9 and outside the cone         z2  =  x2 + y2  that has density function         Solution We use spherical coordinates.  The sphere becomes         r  =  3 and to find the equation of the cone, we add z2 to both sides to get         2z2  =  x2 + y2 + z2          2r2cos2 f  =  r2         cos2 f  = 1/2        f  =  p/4 Now we can write         B.  The surface area of the part of the paraboloid          z  =  x2 + y2 that lies inside the cylinder         x2 + y2  =  4 Solution We calculate the partials         zx  =  2x        zy  =  2y        (1 + zz + zy)1/2  =  (1 + 4x2 + 4y2)1/2 Since the region is a circle of radius 2, we convert to polar coordinates to get         Problem 7   Find the work done by the force field          F(x,y)  =  (3x2 - y) i + (x2 - y3) j as a particle moves counterclockwise around the rectangle with vertices (2,1), (5,1), (5,5), and (2,5). Solution We use Green's Theorem.  We have         Nx - My  =  2x + 1 We have         Problem 8 Verify Stokes Theorem where          F(x,y,z)  =  (2x) i + (2y) j + (z sin z3) k and S is part of the paraboloid          z  =  9 - x2 - y2  that lies above the plane z  =  8 oriented upward. Solution We first compute the line integral          We notice that the intersection of the paraboloid and the plane is given by          9 - x2 - y2  =  8        x2 + y2  =  1 This is the circle of radius 1 raised 8 units above the xy-plane.  Its parameterization is given by         r(t)  =  (cos t) i + (sin t) j + 8k          r'(t)  =  (-sin t) i + (cos t) j   so that          F . dr  =  [(2cos t) i + (2sin t) j + (8 sin 83) k] . [(-sin t) i + (cos t) j]         =  -2 sin t cos t + 2 sin t cos t  =  0 Since the integrand is zero, so is the integral. Now we use Stokes Theorem.  We have          So that the surface integral is zero.