Translating number grades to letter grades

Discussion of the grading

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.