Solutions to Sample Exam

1. The partial derivatives are found by holding one variable fixed. We see:

   

   The interpretation of is to give the rate of change of the amount in the account (B) over time. Hopefully, this is an increasing function. The other partial derivative, tells you how much the amount of money in the bank will increase as you adjust the principal amount invested.

2. 1. To find a normal vector to the surface S we compute the gradient of the function g(x,y,z) = z - f(x,y) at the point indicated. So,

      

      At the point we have the normal vector as .

   2. To find the tangent plane to S at this point, we use the normal vector to find that ax + by + cz = d where a, b, and c are the , , and components of the normal vector and d is a constant that is obtained by plugging in the point . Thus, the tangent plane is z = 1.
3. We know that critical points occur where the gradient is equal to . We also know that the gradient is always perpendicular to the contours. Thus, from the contour diagram, we see that B and C are possible critical points, while A is not. Near B, the contours are circular and increasing as one approaches B, thus, the graients near B point toward B and B is a local maximum. Point C is a saddle point, since moving in the y direction puts one on a contour with a lower z value, while moving in the x direction increases the value of z.
4. 1. For this integral, we are at values of z that are less than 1, since we are inside a the upper hemisphere of a sphere of radius 1. This means that the quantity is always negative. Since we are integrating a negative function over a volume, we expect the result to be negative.
   2. The function (-xz) is an odd function of the variable x. Thus, when x is positive (the right half of the hemisphere) the integral gives a negative value (since -x is negative and z is always positive), and when x is negative, we get a positive value. Thus, the integral over the right half of the hemisphere will cancel with the integral over the left half, giving a net result of zero.
5. To show that the two equations are both satisfied by this parameterization, simply plug x(t), y(t), and z(t) into each equation. This gives

   

   Thus, the parameterization satisfies both equations and must therefore represent a curve on each surface. Since the equations represent two different planes, and the curve that is parameterized is a straight line, this line must represent the intersection of the two planes.

6. The path along which the line integral is to be evaluated is the straight line y = x from the point (0,0) to the point (3,3). Thus, a proper parametrization would be x = t, y = t, . This gives and . The vector field along the curve is , so the line integral is

   

   This is a difficult integral to do, but the problem is simplified by using the Fundamental Theorem of Calculus for Line Integrals. Note that the vector field above is a gradient field ( ) with a potential function given by . This means that

   

7. For this disk, a unit normal vector would be . Since the disk is parallel to the xy plane, then . The disk is in the xy plane, so this means that the equation of the surface we want to find the flux through is z = 0, and the vector field here is . The portion of the surface we are interested in lies above the region R, which is the disk . So,

   

8. 1. To check whether is a conservative vector field (ie. has a potential function), we calculate the curl. A simple calculation reveals that , so is a gradient field. We now try to find a potential function by setting . This means that

      

      we now take partial derivatives with respect to y and compare with the second component of the vector field.

      

      This means that we now have . Taking the partial derivative of this with respect to z and comparing with the thrid component of the vector field we find

      

      And so we find that the potential function is constant.

   2. Since the vector field is only a two-dimensional vector field, we can check the simpler relations to determine whether it is conservative. We see that, letting ,

      

      Since these partial derivatives are not equal, we know that is not a gradient field and therefore has no potential.

9. 1. To find the divergence of , we first write the vector field using x, y and z and then compute div( ) . So

      

      Now we compute the partial derivatives. For simplicity, I show only the first term, .

      

      Similar terms arise from the other derivatives. In the end, you will find that div , as long as .

   2. Because the divergence of this vector field is undefined at the origin, we cannot use the divergence theorem directly to compute the flux through a cube centered at the origin. However, we can break the region of integration up into two pieces.

      Since the vector field is pointing radially outward, it will be easy to calculate the flux through a sphere centered at the origin. For this reason, we break the cube up into two parts. The first part is the sphere, centered at the origin, with radius a/2. The second part of the region will be the cube minus the sphere. Thus, together, these regions add up to the original cube. The benefit of this approach will soon be apparent.

      Through the second region (cube minus sphere) we know that the divergence of the vector field is zero. Also, since we have removed the origin (it's part of the sphere) we can use the divergence theorem. This means that the flux through the surface of region two is zero.

      Through the first region, we will have to compute the flux integral. This is easy, because the vector field and the normal to the sphere are in the same direction. Furthermore, on the surface of the sphere, r = a/2, so we know . Thus,

      

10. To use Stokes' theorem, we first must compute the curl of this vector field. Taking partial derivatives, we find curl . The circle is the boundary of the surface S which is the disk in the plane z = f(x, y) = 2. Thus, we find that in polar coordinates. Substituting z = 2 into the vector field, we have