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Midterm 1 Solutions Results for the midterm were varied, but generally good. Twenty- seven of the twenty-eight registered students in the class took the exam. The average was 19/25. Below is a table outlining the score distribu- tion.
Score Number of Students 25 4 24 3 23 1 22 1 21 2 20 2 19 1 18 2 17 1 16 2 15 4 14 1 13 2 9 1 I do not curve exams. I occasionally curve overall class scores. If we follow the preset scale outlined on the syllabus and only counted the midterm, then the grade scale would be as follows.
Score Grade 23-25 A+ , A, A- 19-22 B+, B, B- 17, 18 C+, C 15, 16 C- 12-14 D+, D, D- 0-11 E If these were the class scores, I would not curve.
1
The next table shows how many people attempted each problem. Problem Number who attempted 1 27 2 16 3 25 4 10 5 18 6 22 7 2 8 15
Thus, most people attempted problems 1 and 3, while problem 7 was especially unpopular. Below are solutions and commentary to each problem.
Problem #2 (5 points):
Let v = (4, 3 , 1) and u = (1, 6 , 3). Compute the vector projection of v onto u.
Solution Let p be the projection of v onto u. A formula for the vector pro- jection of v onto u is
p =
u · v |u|^2
u
12 + 6^2 + 3^2
p =
Commentary I think a lot of people got there u’s and v’s mixed up in the formula. The formula is not one I’ve bothered to commit to memory, since I can easily rederive it. However, rederiving it results in extra work, and for this reason, I thought this was the toughest problem on the exam.
Problem #3 (5 points):
Let P be the parallelogram generated by the vectors a = (1, 0 , 2) and b = (3, 4 , 5). Use the properties of the cross-product to compute the area of P.
Solution Let A be the area of P. Then A = |a × b|. Now a × b = (1, 0 , 2) × (3, 4 , 5) = ((0 · 2 − 4 · 5), (3 · 2 − 1 · 5), (1 · 4 − 3 · 0)) a × b = (− 20 , 1 , 4).
Thus,
(6) A = |a × b| =
202 + 1^2 + 4^2 =
Commentary
Most answers to this question were done well.
Problem #5 (5 points):
Let a, b ∈ R^3. Show that a × b is orthogonal to both a and b.
Solution Let us write a = (a 1 , a 2 , a 3 ) b = (b 1 , b 2 , b 3 ).
We must show that
(11) (a × b) · b = (a × b) · a = 0.
Now
(12) a × b = (a 2 b 3 − b 2 a 3 , a 3 b 1 − a 1 b 3 , a 1 b 2 − b 1 a 2 ).
Hence,
(a × b) · a = (a 2 b 3 − b 2 a 3 )a 1 + (a 3 b 1 − a 1 b 3 )a 2 + (a 1 b 2 − b 1 a 2 )a 3 = a 1 a 2 b 3 − a 1 b 2 a 3 + b 1 a 2 a 3 − a 1 a 2 b 3 + a 1 b 2 a 3 − b 1 a 2 a 3 (a × b) · a = 0.
Similarly,
(a × b) · b = (a 2 b 3 − b 2 a 3 )b 1 + (a 3 b 1 − a 1 b 3 )b 2 + (a 1 b 2 − b 1 a 2 )b 3 = b 1 a 2 b 3 − b 1 b 2 a 3 + b 1 b 2 a 3 − a 1 b 2 b 3 + a 1 b 2 b 3 − b 1 a 2 b 3 (a × b) · b = 0.
Commentary Answers here were usually of good quality.
π/
x
y
z
Figure 1. Illustration for Problem 6
Problem #6 (5 points):
Sketch the surface given in spherical coordinates by the equation θ = π 4. Carefully describe the surface in words also. For full credit, your sketch will be appropriately labeled.
Solution The surface is a vertical half-plane with the z-axis as a base line. The angle with the positive x-axis is π 4. In Cartesian coordinates, the
Problem #7 (5 points): Let I ⊂ R be an interval. Consider the function r : I → Rn. Care- fully define what it means for r to be differentiable at t ∈ I. (There will be a limit in your definition.)
Solution
The function, r, is differentiable at t ∈ I if lim h→ 0
r(t + h) − r(t) h
exists.
Commentary As is clear in the solution above, this question has a one line answer. Notice that continuity is not mentioned in the definition.
Problem #8 (5 points):
Let C be the space curve parameterized by r(t) = (t, e^2 t, sin(2t)). Find the tangent line to C when t = 0.
Solution
The tangent line, l, to the curve, C, at t = 0 is parallel to the vector r˙(0) and passes through r(0). Now
r˙(t) = (1, 2 e^2 t, 2 cos(2t)) r˙(0) = (1, 2 , 2) r(0) = (0, 1 , 0).
Thus, the equation of our tangent line is l(s) = (0, 1 , 0) + s(1, 2 , 2) where s ∈ R is a parameter.
Commentary
Several people misread the question and only found the tangent vec- tor.
