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The instructions and problems for exam 1 in the vector calculus course appm 2350 held during the summer 2006 semester. The exam covers topics such as calculating equations for planes, finding distances between points and planes, proving statements related to vectors, and finding minimum and maximum values of velocities and accelerations for particles moving along curves.
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APPM 2350 Summer 2006 Exam 1 June 21, 2006 1
INSTRUCTIONS: Computers, calculators, books, notes, flying monkeys, etc. are not permitted.
Some (possibly) useful formulae are attached. Write your name, your instructor’s name, and the
color of your exam sheet on the front of your bluebook. Work all problems. Start each problem on
a new page. Show your work clearly and box your final answer. A correct answer with incorrect
or no supporting work may receive no credit, while an incorrect answer with relevant work may
receive partial credit.
(b) Calculate the distance between the point P (− 1 , − 1 , 0) and the plane from (a).
(c) Sketch and describe (in words) the curve of intersection between the plane x + y = 0
and the surface x
2
2 − z = 1.
(d) Find a parametrization of the curve in (c).
to be TRUE or FALSE):
(a) If A · B = C · B and B 6 = 0 , then A = C.
(b) (A × B) · A = 0.
(c) If an object moves with constant speed, its acceleration has no tangential component
and a constant normal component.
(a) Find the minimum and maximum magnitudes of its velocity and acceleration, i.e. the
minimum and maximum values of |v| and |a|.
(b) Find the curvature as a function of t.
(c) Find the minimum and maximum values of the curvature. What is the position of the
particle at these values?
(a) Calculate the tangential, normal and binormal directions of this curve, i.e. calculate
T, N and B.
(b) Calculate the curvature of the curve.
(c) Calculate the normal and tangential components of the acceleration.
— Over —
APPM 2350 Summer 2006 Exam 1 June 21, 2006 2
more equations than pictures, so five equations will be unused.) No work need be shown for
this problem.
a)
0
1
2
x
0
1
2
z
0
1
2
x
0
1
b)
0
1
2
x
0
1
2
y
0
1
z
0
1
c)
0
1
2
x
y
0
2
z
0
1
2
x
0
2
d)
-0.
0
1
x
-0. -0.
0
y
-0.
0
1
z
-0.
0
1
-1 x
-0.
0
(1) x
2
2 − z
2 = 1 (2) x
2 − 4 y
2 − z
2 = − 1 (3) 2 x(y − 1)z = 1
(4) 2 y = x
2 − 3 z
2 (5) 2 y(z − 1) = 1 (6) 2 x = y
2 − 3 z
2
(7) x
2 − 4 y
2 − z
2 = 1 (8) x
2
2
2 = 1 (9) z = x
2
2
— Useful and interesting formulae —
proj A
A d =
P S × v|
|v|
d =
n
|n|
dr
ds
v
|v|
d
T/ds
|d
T/ds|
d
T/dt
|d
T/dt|
κ =
d
ds
|v × a|
|v|
3
τ = −
dB
ds
a = a T
T + a N
N where a T
d
dt
|v|, a N
= κ|v|
|a|
2 − a
2
T