Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [160,200] | [150,159] | [130,149] | [120,129] | [100,119] | [90,99] | [0,89] |

Quoting from the formula sheet (even relevant formulas!) will not earn credit. Arithmetic errors and "small" errors (such as minor errors in differentiation) will be penalized minimally. The resulting work can be considered for full credit in the remainder of the problem unless the error makes the problem much easier.

**Problem 1** (15 points)

a) (7 points) Solve two of the equations: 5 points. Check the third: 2
points. If no valid method is shown, 0 points are earned.

b) (8 points) Normal vector: 4 points (2 for indicating the cross
product of the appropriate vectors and 2 for getting it correct), 2
points for a point on the plane (can be a point on any of the two
lines, or it can be the point found by the student in a), whether
correct or not). 2 points for the correct answer.

**Problem 2** (15 points)

a) (4 points) 1 point for each partial derivative, and 1 point for
the vector nature of the answer.

b) (6 points) 2 points for the gradient at p. 2 points for each answer.

c) (5 points on exams A and B) 2 points for a normal vector, 2 points
for a point on the line, and 1 point for the answer.

c) (5 points on exams C and D) 2 points for a normal vector, 2
points for a point on the plane, and 1 point for the answer.

**Problem 3** (15 points)

7 points for a correct setup in any order. 8 points for the
computation (2 points for the first antiderivative with limits, 3
points for the substitution in the integral, and 3 points for the
second antiderivative with limits and the answer). The student does
*not* earn the first 7 points if the setup is not correct.

If the student sets up an integral over a rectangle, *no* points
will be earned.

**Problem 4** (20 points)

a) (10 points) This is half a paraboloid opening down. Equations (a
total of 7 points): the paraboloid equation gets 3 points, and each
plane gets 2 points. 3 points for the sketch, which should be
reasonable.

b) (10 points) r integral done correctly: 3 points; integral
done correctly: 3 points; z integral done correctly: 2 points; the
answer: 2 points.

**Problem 5** (15 points)

a) (7 points) The partial derivatives each earn 2 points. Finding
the unique critical point: 3 points, including 1 for the solution.

b) (8 points) Computing the second partial derivatives earns 4
points. Evaluating them at the point found in a) earns 2 points. Using
the Second Derivative Test correctly: 2 points.

**Problem 6** (25 points)

(12 points) Computing div **T** correctly: 2 points. Computing the
triple integral of this function over the lower half of
the ball, 10 points (2 points for stating that the goal is computing
the triple integral of the divergence over the half ball, 2 points for
setting up the correct triple iterated integral, 1 point for the d integral, 2 points each for the
d rho and d phi integrals, and 1 point for the answer).

(10 points) Computing **T**·**n** on the bottom and
setting z=0: 2 points. Computing the double integral correctly, 8
points.

(3 points) Put everything together and get the correct answer.

**Problem 7** (20 points)

7. a) (10 points) 2 points for the answer alone, and 8 points for a
valid process. If only constants are shown in the antiderivatives with
no variables, 2 points are deducted. If the constant "functions" have
the same names, 1 point is deducted. If no constants are shown, then 4
points are deducted. Students who compute the curl of **F**, get 0,
and conclude that a potential function exists but do not find the
function earn 5 of the 10 points.

b) (10 points) 2 points for the answer alone, and 8 points for a valid
process: 3 points for stating or using
P(`END`)-P(`START`), 2 points each for start and end,
and 1 point for the correct answer. It is also possible to earn full
credit for a direct computation: parameterization, integration, and
evaluation.

**Problem 8** (15 points)

7 points for applying Green's Theorem correctly. 8 points for
evaluating the resulting double integral. Students who insist on using
a different method (such as parameterization) will earn an appropriate
amount of credit (there are 4 different pieces of the boundary curve
to parameterize).

**Problem 9** (20 points)

2 points for each level curve and 1 point for the label with the
function value: a total of 9 points. 1 point for computing the
gradient algebraically. 2 points for each gradient vector: a total of
10 points. The vector field should be drawn with correct directions and
magnitudes.

**Problem 10** (points)

a) 5 points for a correct sketch. 2 of these points are for some
specific labeling (with numbers) of the intersection points.

b) and c): 5 points for each correct integral (so one part will be
worth 10 points and one part will be worth 5). In each case, the
integrand which should be f(x,y) earns a point and each correct
boundary of the double integral earns a point.

**Problem 11** (12 points)

Finding and using a formula is 8 points. Identifying the t value for
each point is 1 point and computing the curvature is 1 point. Since
there are 2 points this is worth a total of 4 points.

**Problem 12** (8 points)

8 points for the computation. Each misuse of the chain rule or the
product rule loses 2 points. Simpler errors such as a floating minus
sign lose 1 point.