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The instructions and problems for the final exam of the vector calculus course (appm 2350) held in spring 2006. The exam covers topics such as gradient, divergence, curl, line integrals, surface integrals, and the ideal gas law.
Typology: Exams
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INSTRUCTIONS: Computers, calculators, books, notes, flailing monkeys, etc. are not permitted. Some (possibly) useful formulae are attached. Write your name, your instruc- tor’s name and your recitation section (and/or TA’s name) on the front of your bluebook. Work all problems. Start each problem on a new page. Unless otherwise stated, show all your work clearly and box your final answer; a correct answer with incorrect or no supporting work will receive no credit, while an incorrect answer with relevant work may receive partial credit.
(a) ∇ · (∇f ) = 0. (b) ∇ × (∇f ) = 0. (c) ∇ · (∇ × F) = ∇ × (∇ · F). (d) If ∇ · F = 0 and ∇ × F = 0 then F = 0. (e) ∇ · (∇f × F) + ∇ · (F × ∇f ) = 0.
(a) Calculate the (counterclockwise) circulation of F = −y x^2 + y^2
ˆı + x x^2 + y^2
ˆ around a circle of radius a (centered at the origin). Clearly state any theorems you use and justify their use. (b) Calculate the (outward) flux of F = log(x^2 + y^2 )ˆı − log(4 − x^2 )ˆ through the path from (1, 1) down the line x = 1 to (1, −1) then back to (1, 1) around the right-hand portion of the circle x^2 + y^2 = 2. Clearly state any theorems you use and justify their use.
(a) A section of a bouncy-castle has a volume of 2 m^3 and is filled with gas at a temperature of 300 K and a pressure of 2 × 105 Pa. Is the pressure more sensitive to changes in temperature or to changes in volume? (Assume n and R are fixed.) (b) Calculus students celebrating the end of the semester jump on the bouncy-castle, compressing the volume to 1.8 m^3 ; the beautiful spring weather has also increased the temperature to 303 K. Estimate the change in the pressure. [Note: although you do not know n and R, you can determine the value of nR.]
(c) The calculus students are trying to guess how many moles of gas are in the bouncy-castle. To determine who has guessed correctly, they will measure the volume, pressure and temperature of the gas and calculate n using the ideal gas law. However, each of these measurements will have a small error associated with them. Assuming R is a known constant, show that the percentage error in the calculated value of n is the sum of the percentage errors in V , P and T.
(a) Being smart, you decide to make the ship’s computer do the donkey work. Set up the integral necessary to calculate the flux directly. In order for the computer to process your request, you need to give the integral set up completely in the most logical coordinate system (i.e. do everything except actually evaluate the integral). (b) Unfortunately, the computer is already somewhat overloaded with complex navi- gation calculations, so you will have to do the work yourself. Use your knowledge of Calc III to show how the flux through S can be related to the flux through P. [Hint: S and P together form the outer shell of the solid drives.] (c) Determine the flux through S, preferably before you get boarded by Klingons.
x = r cos(θ) y = r sin(θ) z = z
r = ρ sin(φ) z = ρ cos(φ) θ = θ
x = ρ sin(φ) cos(θ) y = ρ sin(φ) sin(θ) z = ρ cos(φ)
ρ =
x^2 + y^2 + z^2 r =
x^2 + y^2 θ = tan−^1 (y/x) φ = tan−^1 (r/z) dV = dx dy dz = r dr dθ dz = ρ^2 sin(φ) dρ dφ dθ
∫ ∫ ∫
V
f (x, y, z) dx dy dz =
V ′
f (x(u, v, w), y(u, v, w), z(u, v, w))|J(u, v, w)| du dv dw
J(u, v, w) =
∂x ∂u
∂x ∂v
∂x ∂y ∂w ∂u
∂y ∂v
∂y ∂z ∂w ∂u
∂z ∂v
∂z ∂w
Work:
C
F · T ds =
C
F · dr =
C
M dx + N dy
Flux:
C
F · n ds =
C
M dy − N dx
Conservative field: ∇ × F = 0 ⇔
∂P ∂y =^
∂N ∂z ∂P ∂x =^
∂M ∂N ∂z ∂x =^
∂M ∂y
For a surface given by g(x, y, z) = 0 :
S
f (x, y, z)dσ =
R
f (x, y, z) |∇g| |∇g · p| dA
Flux of F through surface given by g(x, y, z) = 0 :
S
F · n dσ =
R
±F · ∇g |∇g · p| dA
Green’s Theorem:
Circulation =
C
F · T ds =
R
(∇ × F) · ˆk dA =
R
∂x
∂y
dA
Flux =
C
F · n ds =
R
∇ · F dA =
R
∂x
∂y
dA
Stokes’s Theorem: ∮
C
F · dr =
C
M dx + N dy + P dz =
S
(∇ × F) · n dσ
Divergence Theorem of Gauss: ∫ ∫
S
F · n dσ =
D
∇ · F dV