Mean = 131.0
Median = 139.5
Standard deviation = 18.5
Problem 1
Integrate (1/2)(2/theta)^2 between pi/6 and pi/3 to get 6/pi.
Some integrated r instead of (1/2)r^2, a serious mistake. I took off 10 points for this if everything else was correct.
Problem 2
(a) Replacing sec theta with 1/cos theta gives x=-3.
(b) r=2/(sin(theta)-3cos(theta)).
On part (a), several squared both sides (introducing an extraneous solution of x=+3), then simplified by clearing fractions (introducing another extraneous solution of x=0). I took off 5 points for both extraneous solutions and 3 points for only the first. I did not take off points for not writing (b) as r = f(theta).
Problem 3
(a) Using (7) on page 613, the curvature is -cos t.
(b) The function -cos t is minimum when t=0, and its value is -1.
(c) The velocity vector is (1, tan t), so the speed is sqrt(1+tan^2 t) = sec t. When t=0, sec t = 1.
I did not deduct points if part (a) wasn't simplified.
Problem 4
By using x=t as a convenient parameter in l_1, we get l_1 described parametrically as x=t, y=-1-3t, z=-t. Other parametrizations are possible. The direction vectors for l_1 and l_2 are then (1, -3, -1) and (-1/2, 0, 2). A normal for the parallel planes containing these two lines would then be the cross product of these two direction vectors, or (12, 3, 3). The plane for l_1 will then be 12x + 3y + 3z = -3 (by using the point (0,-1,0) on l_1). The plane for l_2 will be 12 x + 3y + 3z = 30 (using the point (3/2, 4, 0) on l_2). The distance between these planes is |30-(-3)|/|N| = 33/sqrt(162) =11/(3sqrt(2)) = 11sqrt(2)/6.
Many people made calculation mistakes, especially when taking the cross product. Some used the cross product to get a direction vector for l_1. Generally speaking, I awarded 5 points for the direction vector of l_1, 5 points for a point on l_1, 5 points for both direction vector and point on l_2, 10 points for the normal vector, and 5 points for the plane.
Problem 5
(a) Write z+2 = 2x^2+y^2, so that the curve is an elliptic paraboloid, translated down the z axis by 2.
(b) Substituting the parametric equations into the surface equation gives t=(2+sqrt(10))/3 and t=(2-sqrt(10))/3. Therefore the two points are ((sqrt(10)-1)/3, (sqrt(10)+2)/3, 2) and ((-sqrt(10)-1)/3, (-sqrt(10)+2)/3, 2).
(c) Setting z=0 gives the ellipse x^2 + y^2/2 = 1, so a = 1 and b = sqrt(2). Since the area of an ellipse is pi*a*b, the area is pi*sqrt(2). Alternatively, solve for y and integrate.
A few gave the area in part (c) in terms of z. I did not take points off for this if the answer was correct. However, technically, the xy-plane is the plane z=0.
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