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.

Vectors Calaulus 1.35

Practice 3 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,z) be a function with continuous second order partial derivatives and let F(x,y,z) be the gradient of f(x,y,z).  If S is the ellipsoid              x2                      +  y2 + z2  =  1             4 oriented outward, then          Solution False Using the divergence theorem, we have         If gradf(x,y,z)  =  F, then          divF  =  fxx + fyy + fzz  There theorem would be true if the function was harmonic, however if it not harmonic.  All bets are off.  For example, if         f(x,y,z)  =  1/6 (x2 + y2 + z2) and          divF  =  1/6 (2 + 2 + 2)  =  1 Hence the integral represents the volume of the ellipse which is certainly not zero.   B.  Let F(x,y) be a conservative vector field, then           Solution   True,   The first line integral traces out the line segment from (0,1) to (1,0) and the second traces out the quarter-circle from (0,1) to (1,0).  Notice that in the first integral         r1(t)  =  (1 - t)i + tj         r1'(t)  =  -i + j   and in the second integral          r2(q)  =  (cos q)i + (sin q)j         r1'(t)  =  (-sin q)i +(cos q) j   By the fundamental theorem of line integral, the integral is independence of path, hence the two integrals are equal. Problem 2 Show that  for any closed surface S.   Solution We use Stokes' Theorem.  We have         Where C is the boundary of the surface S.  But since S is a closed surface, it has not boundary.  Hence C is a curve of zero length and the right hand integral is zero. Problem 3   A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2).  The current can by represented by the vector field         F(x,y,z)  =  (2x + 2z)i + (1 - 3z)j + (2x - 3y + 5)k Find the total work done by the current. Solution The important thing to not here is that         Since F is conservative, we can use the fundamental theorem of line integrals.  We seek a potential function f.  We have         fx  =  2x + 2z Integrating with respect to x gives         f  =  x2 + 2xz + C(y,z) Now taking the derivative with respect to y gives         fy  =  Cy(y,z)  =  1 - 3z Integrating with respect to y gives         C(y,z)  =  y - 3yz + C(z) so that         f  =  x2 + 2xz + y - 3yz + C(z) Now we take the derivative with respect to z to get         2z - 3y + C'(z)  =  2x - 3y + 5 so that          C'(z)  =  5 Integrate with respect to z to get         C(z)  =  5z Hence          f(x,y,z)  =   x2 + 2xz + y - 3yz + 5z The fundamental theorem of line integrals gives that the integral is         f(1.5,1,2) - f(0.2,0.1,0.3)  =           [(1.5)2 + 2(1.5)(2) + 1 - 3(1)(2) + 5(2)] - [(0.2)2 + 2(0.2)(0.3) + 0.1 - 3(0.1)(0.3) + 5(0.3)]         =  11.58 Problem 4 Evaluate where F is the vector field          and S is the rectangular solid with vertices (0,0,0), (1,0,0), (1,2,0), (0,2,0), (0,0,3), (1,0,3), (1,2,3), (0,2,3). Solution We use the divergence theorem.  We have         divF  =  1 - 1 + 2  =  2 We have         Since the integrand of the right hand side is just a constant, its value is equal to the constant times the volume of the solid.  Since the solid is a rectangular solid with side lengths 1, 2, and 3, we have         2(Volume E)  =  2(1)(2)(3)  =  12 Problem 5 Find the work done by sailing a ship from the point (2,3) the the point (-1,2) against the wind with velocity field         F(x,y)  =  yi + (3x + 2y)j Solution First notice that F is not a conservative vector field.  We need to parameterize the curve and perform the line integral.  The curve can be parameterized by         r(t)  =  (2 + (-1 - 2)t)i + (3 + (2 - 3))j  =  (2 - 3t)i + (3 - t)j We have         dr  =  -3i - j                 F(t)  =  (3 - t)i + (3(2 - 3t) + 2(3 - t))j  =  (3 - t)i + (12 - 11t)j  The integrand becomes         F . dr  =  -3(3 - t) + (-1)(12 - 11t)  =  -21 + 14t Now we integrate         Problem 6 Find the flux of F through the surface S where          F(x,y,z)  =  3zi - 4j + yk and S is the part of the plane         x + y + z  =  1 in the first octant with upwardly pointing unit normal. Solution We use Stokes' theorem.  We have         We can write the surface as         z  =  1 - x - y Using Stokes theorem we get

Vectors Calaulus 1.34

Stokes' Theorem Stokes' Theorem The divergence theorem is used to find a surface integral over a closed surface and Green's theorem is use to find a line integral that encloses a surface (region) in the xy-plane. The theorem of the day, Stokes' theorem relates the surface integral to a line integral. Since we will be working in three dimensions, we need to discus what it means for a curve to be oriented positively.  Let S be a oriented surface with unit normal vector N and let C be the  boundary of S.  Then C is positively oriented if its orientation follows the right hand rule, that is if you right hand curls around N in the direction of C's  orientation, then your thumb will be pointing in the direction of N.                           Now we are ready to state Stokes' Theorem.  The proof will be left for a more advanced course. Stokes' Theorem Let S be an oriented surface with unit normal vector N and C be the positively oriented boundary of S.  If F is a vector field with continuous first order partial derivatives then                Example Let S be the part of the plane         z  =  4 - x - 2y with upwardly pointing unit normal vector.  Use Stokes' theorem to find         Where         F  =  yi + zj - xyk Solution First notice that without Stokes' theorem, we would have to parameterize  three different line segments.  Instead we can find this with just one double  integral. We have         and          N dS  =  i + 2j + k So that          Curl F . N dS  =  1 + x + 2y - 1  =  x  2y We integrate         Curl and Circulation Just as the divergence theorem assisted us in understanding the divergence of a function at a point, Stokes' theorem helps us understand what the Curl of a vector field is.  Let P be a point on the surface and Ce be a tiny circle around P on the surface.   The          measures the amount of circulation around P.  You can see this by noticing that if F flows in the direction of the tangent vector, then F . dr will be positive.  If it flows in the opposite direction, then it will be negative.  The stronger the force field in the direction of the tangent  vector, the greater the circulation. Since the region enclosed by Ce is tiny, the surface integral can be approximated by          or         Curl F . N  =  Circulation per unit area So the curl tell us how much the force field rotates around the point.               We can see that if this is a small piece of the surface containing  P, then          Curl F . N  >  0