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Math 206 — Final Exam
December 11, 2012
Name: EXAM SOLUTIONS
Instructor: Section:
- Do not open this exam until you are told to do so.
- This exam has 10 pages including this cover AND IS DOUBLE SIDED. There are 11 prob- lems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.
- Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out when you hand in the exam.
- Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions.
- Show an appropriate amount of work (including appropriate explanation). Include units in your answer where that is appropriate. Time is of course a consideration, but do not provide no work except when specified.
- You may use any previously permitted calculator. However, you must state when you use it.
- If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph that you use.
- Turn off all cell phones and pagers, and remove all headphones and hats.
- Remember that this is a chance to show what you’ve learned, and that the questions are just prompts.
Problem Points Score
Total 100
Some formulae:
∇ × F =
dF 3 dy
dF 2 dz
dF 1 dz
dF 3 dx
dF 2 dx
dF 1 dy
(a 1 , a 2 , a 3 ) × (b 1 , b 2 , b 3 ) = (a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ) cos(2θ) = cos^2 (θ) − sin^2 (θ) sin(2θ) = 2 sin(θ) cos(θ) cos^2 (θ) + sin^2 (θ) = 1
- [13 points] In this problem, let
F (x, y, z) =
ex
3 ,
3 z y^2 + z^2
3 y y^2 + z^2
and let C be the curve parametrized by p(t) = (cos(t), sin(2t), cos(2t)) with 0 ≤ t ≤ 2 π. a. [5 points] Set up (but do not compute) the arc length integral for the curve C?
Solution: The formula for arclength is to integrate |p′(t)| from 0 to 2π. This is ∫ (^2) π
0
sin^2 (t) + 4dt.
b. [8 points] Compute (^) ∫
C
F · dL.
Solution: We need to compute
∫ (^2) π 0 F^ (p(t))^ ·^ p
′(t)dt. We have that
F (p(t)) =
ecos
(^3) (t) ,
3 cos(2t) sin^2 (2t) + cos^2 (2t)
3 sin(2t) sin^2 (2t) + cos^2 (2t)
We have that p′(t) = (− sin(t), 2 cos(2t), −2 sin(2t)). Combining all this we get ∫ (^2) π
0
− ecos
(^3) (t) sin(t) + 6dt = 12π,
after the u-sub u = cos(x) and du = − sin(x)dx.
- [10 points] Let f (x) and g(y) be functions of one variable with continuous second derivatives. Let u(x, y) = f (x + g(y)). Compute
uxuxy − uyuxx =
du dx
d^2 u dxdy
du dy
d^2 u dx^2
Solution: This is a chain rule problem. First let’s calculate ux and uy. We have that Jacu = Jacf ∗ Jach where h(x, y) = x + g(y). So:
Jac(h) = (1, g′(y)) Jac(f )|h(x,y) = f ′(x + g(y)) Jac(u) = (f ′(x + g(y)), f ′(x + g(y))g′(y)).
So ux = f ′(x + g(y)) and uy = f ′(x + g(y))g′(y). To compute uxy we take the derivative of uy with respect to x. This gives us (after a similar chain rule calculation) uxy = f ′′(x+g(y))g′(y). Finally, we get uxx by differentiating ux with respect to x and also get uxx = f ′′(x + g(y)). Putting all this together gives:
uxuxy − uy uxx = f ′(x + g(y))f ′′(x + g(y))g′^ (y) − f ′(x + g(y))g′^ (y)f ′′(x + g(y)) = 0.
- [12 points] Given a sheet of metal lying in the xy-plane with corners at (7, 0), (6, 0), (3, −4), (5, −8), and (7, −3) and mass density x kg/m. What is the total mass of this irregularly shaped metal?
- [15 points] Let S be the surface parametrized by p(s, t) = (4s cos(t), 4 s sin(t), 4 s^2 ), with 0 ≤ s ≤ 1 and 0 ≤ t ≤ 2 π, and let C be the boundary of S. Also let F (x, y, z) = (yz exyz^ , xz exyz^ −z, xy exyz^ −y + x). a. [5 points] What is ∇ × F? Is F a conservative vector field?
Solution: The curl of F is (0, − 1 , 0) so F is NOT conservative.
b. [5 points] What is the parametrization of C?
Solution: Since C is the boundary of S we can parametrize it by letting s = 1 in the parametrization of S. This gives C = f (t) = (4 cos(t), 4 sin(t), 4) which happens to be the circle of radius 4 parallel to the xy-plane at height z = 4.
c. [5 points] Use Stokes Theorem to compute
S ∇ ×^ F dA^ by finding another surface that also has C as its boundary. Solution: This is easier if we take the disc that is the fill in of that circle, D. This is parametrized by q(s, t) = (4s cos(t), 4 s sin(t), 4) and has qs × qt = (0, 0 , 4). From Stokes Theorem (since our new surface, D, and S share a boundary) we have that ∫ ∫
S
∇F dA =
D
∇F dA =
0
∫ (^2) π
0
(0, − 1 , 0) · (0, 0 , 4)dtds = 0.
- [10 points] True or False (no partial credit)
a. [2 points] It is true for any a, b, c ∈ R^3 we have that ((c × a) · c) + ((c × b) · c) = 0?
Solution: TRUE
b. [2 points] x^2 + y^2 − 2 y = z is the equation of a cone.
Solution: FALSE
c. [2 points] p(t) = (t − 1 , 2 t − 3 , 5 t) is the equation of a line through the point (1, 1 , 15).
Solution: FALSE
d. [2 points] tan(θ) = −1 in polar coordinates defines the same curve as x + y = 0 in rectilinear coordinates. Solution: TRUE
e. [2 points] if f (x, y) = xy and C : p(t) = (t, 1 + t^2 ) for 0 ≤ t ≤ 2 then
C ∇f dL^ = 10. Solution: TRUE
- [7 points] Find the critical points and classify them (local min, local max, saddle point) of the function f (x, y) = x + y − ln(xy).
Solution: We find the gradient of f (x, y) =
1 − (^1) x , 1 − (^1) y
. and set it equal to (0, 0). This only happens at x = y = 1. To figure out what type of critical point this is we compute the Hessian (^) ( 1 x^2 (^0) y^12
Since the determinants are both positive, this is a local minimum.
- [2 points] What was your favorite topic this semester?
Solution: It’s so hard to choose!
